{"id":1764,"date":"2018-01-11T20:42:50","date_gmt":"2018-01-11T20:42:50","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/substitution\/"},"modified":"2018-01-31T20:57:32","modified_gmt":"2018-01-31T20:57:32","slug":"substitution","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/chapter\/substitution\/","title":{"raw":"5.5 Substitution","rendered":"5.5 Substitution"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Use substitution to evaluate indefinite integrals.<\/li>\r\n \t<li>Use substitution to evaluate definite integrals.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1170573367590\">The Fundamental Theorem of Calculus gave us a method to evaluate integrals without using Riemann sums. The drawback of this method, though, is that we must be able to find an antiderivative, and this is not always easy. In this section we examine a technique, called<strong> integration by substitution<\/strong>, to help us find antiderivatives. Specifically, this method helps us find antiderivatives when the integrand is the result of a chain-rule derivative.<\/p>\r\n<p id=\"fs-id1170573502792\">At first, the approach to the substitution procedure may not appear very obvious. However, it is primarily a visual task\u2014that is, the integrand shows you what to do; it is a matter of recognizing the form of the function. So, what are we supposed to see? We are looking for an integrand of the form [latex]f\\left[g(x)\\right]{g}^{\\prime }(x)dx.[\/latex] For example, in the integral [latex]\\int {({x}^{2}-3)}^{3}2xdx,[\/latex] we have [latex]f(x)={x}^{3},g(x)={x}^{2}-3,[\/latex] and [latex]g\\text{'}(x)=2x.[\/latex] Then,<\/p>\r\n\r\n<div id=\"fs-id1170573442602\" class=\"equation unnumbered\">[latex]f\\left[g(x)\\right]{g}^{\\prime }(x)={({x}^{2}-3)}^{3}(2x),[\/latex]<\/div>\r\n<p id=\"fs-id1170573209422\">and we see that our integrand is in the correct form.<\/p>\r\n<p id=\"fs-id1170573431694\">The method is called <em>substitution<\/em> because we substitute part of the integrand with the variable [latex]u[\/latex] and part of the integrand with <em>du<\/em>. It is also referred to as <strong>change of variables<\/strong> because we are changing variables to obtain an expression that is easier to work with for applying the integration rules.<\/p>\r\n\r\n<div id=\"fs-id1170573406868\" class=\"textbox key-takeaways theorem\">\r\n<h3>Substitution with Indefinite Integrals<\/h3>\r\n<p id=\"fs-id1170571193831\">Let [latex]u=g(x),,[\/latex] where [latex]{g}^{\\prime }(x)[\/latex] is continuous over an interval, let [latex]f(x)[\/latex] be continuous over the corresponding range of [latex]g[\/latex], and let [latex]F(x)[\/latex] be an antiderivative of [latex]f(x).[\/latex] Then,<\/p>\r\n\r\n<div id=\"fs-id1170573306241\" class=\"equation\">[latex]\\begin{array}{cc}\\int f\\left[g(x)\\right]{g}^{\\prime }(x)dx\\hfill &amp; =\\int f(u)du\\hfill \\\\ &amp; =F(u)+C\\hfill \\\\ &amp; =F(g(x))+C.\\hfill \\end{array}[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573413133\" class=\"bc-section section\">\r\n<h1>Proof<\/h1>\r\n<p id=\"fs-id1170573351199\">Let [latex]f[\/latex], [latex]g[\/latex], [latex]u[\/latex], and <em>F<\/em> be as specified in the theorem. Then<\/p>\r\n\r\n<div id=\"fs-id1170573534008\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}\\frac{d}{dx}F(g(x))\\hfill &amp; ={F}^{\\prime }(g(x)){g}^{\\prime }(x)\\hfill \\\\ &amp; =f\\left[g(x)\\right]{g}^{\\prime }(x).\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170573366538\">Integrating both sides with respect to [latex]x[\/latex], we see that<\/p>\r\n\r\n<div id=\"fs-id1170573355959\" class=\"equation unnumbered\">[latex]\\int f\\left[g(x)\\right]{g}^{\\prime }(x)dx=F(g(x))+C.[\/latex]<\/div>\r\n<p id=\"fs-id1170573362143\">If we now substitute [latex]u=g(x),[\/latex] and [latex]du=g\\text{'}(x)dx,[\/latex] we get<\/p>\r\n\r\n<div id=\"fs-id1170573437846\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}\\int f\\left[g(x)\\right]{g}^{\\prime }(x)dx\\hfill &amp; =\\int f(u)du\\hfill \\\\ &amp; =F(u)+C\\hfill \\\\ &amp; =F(g(x))+C.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170573331564\">\u25a1<\/p>\r\n<p id=\"fs-id1170573368234\">Returning to the problem we looked at originally, we let [latex]u={x}^{2}-3[\/latex] and then [latex]du=2xdx.[\/latex] Rewrite the integral in terms of [latex]u[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1170573366731\" class=\"equation unnumbered\">[latex]{\\int \\underset{u}{\\underbrace{({x}^{2}-3)}}}^{3}\\underset{du}{\\underbrace{(2xdx)}}=\\int {u}^{3}du.[\/latex]<\/div>\r\nUsing the power rule for integrals, we have\r\n<div id=\"fs-id1170571257696\" class=\"equation unnumbered\">[latex]\\int {u}^{3}du=\\frac{{u}^{4}}{4}+C.[\/latex]<\/div>\r\n<p id=\"fs-id1170571099748\">Substitute the original expression for [latex]x[\/latex] back into the solution:<\/p>\r\n\r\n<div id=\"fs-id1170573410936\" class=\"equation unnumbered\">[latex]\\frac{{u}^{4}}{4}+C=\\frac{{({x}^{2}-3)}^{4}}{4}+C.[\/latex]<\/div>\r\n<p id=\"fs-id1170573363034\">We can generalize the procedure in the following Problem-Solving Strategy.<\/p>\r\n\r\n<div id=\"fs-id1170573435531\" class=\"textbox key-takeaways problem-solving\">\r\n<h3>Problem-Solving Strategy: Integration by Substitution<\/h3>\r\n<ol id=\"fs-id1170570997711\">\r\n \t<li>Look carefully at the integrand and select an expression [latex]g(x)[\/latex] within the integrand to set equal to [latex]u[\/latex]. Let\u2019s select [latex]g(x).[\/latex] such that [latex]{g}^{\\prime }(x)[\/latex] is also part of the integrand.<\/li>\r\n \t<li>Substitute [latex]u=g(x)[\/latex] and [latex]du={g}^{\\prime }(x)dx.[\/latex] into the integral.<\/li>\r\n \t<li>We should now be able to evaluate the integral with respect to [latex]u[\/latex]. If the integral can\u2019t be evaluated we need to go back and select a different expression to use as [latex]u[\/latex].<\/li>\r\n \t<li>Evaluate the integral in terms of [latex]u[\/latex].<\/li>\r\n \t<li>Write the result in terms of [latex]x[\/latex] and the expression [latex]g(x).[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1170573385785\" class=\"textbox examples\">\r\n<div class=\"exercise\">\r\n<h3>Using Substitution to Find an Antiderivative<\/h3>\r\n<div class=\"textbox\">\r\n<p id=\"fs-id1170573386292\">Use substitution to find the antiderivative of [latex]\\int 6x{(3{x}^{2}+4)}^{4}dx.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573569711\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573569711\"]\r\n<p id=\"fs-id1170573569711\">The first step is to choose an expression for [latex]u[\/latex]. We choose [latex]u=3{x}^{2}+4.[\/latex] because then [latex]du=6xdx.,[\/latex] and we already have <em>du<\/em> in the integrand. Write the integral in terms of [latex]u[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1170570995689\" class=\"equation unnumbered\">[latex]\\int 6x{(3{x}^{2}+4)}^{4}dx=\\int {u}^{4}du.[\/latex]<\/div>\r\n<p id=\"fs-id1170573398341\">Remember that <em>du<\/em> is the derivative of the expression chosen for [latex]u[\/latex], regardless of what is inside the integrand. Now we can evaluate the integral with respect to [latex]u[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1170571000121\" class=\"equation unnumbered\">[latex]\\begin{array}{ll}\\int {u}^{4}du\\hfill &amp; =\\frac{{u}^{5}}{5}+C\\hfill \\\\ \\\\ \\\\ &amp; =\\frac{{(3{x}^{2}+4)}^{5}}{5}+C.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170573208347\"><strong>Analysis<\/strong><\/p>\r\n<p id=\"fs-id1170573569153\">We can check our answer by taking the derivative of the result of integration. We should obtain the integrand. Picking a value for <em>C<\/em> of 1, we let [latex]y=\\frac{1}{5}{(3{x}^{2}+4)}^{5}+1.[\/latex] We have<\/p>\r\n\r\n<div id=\"fs-id1170573336352\" class=\"equation unnumbered\">[latex]y=\\frac{1}{5}{(3{x}^{2}+4)}^{5}+1,[\/latex]<\/div>\r\n<p id=\"fs-id1170573411738\">so<\/p>\r\n\r\n<div id=\"fs-id1170573408431\" class=\"equation unnumbered\">[latex]\\begin{array}{}\\\\ \\hfill {y}^{\\prime }&amp; =(\\frac{1}{5})5{(3{x}^{2}+4)}^{4}6x\\hfill \\\\ &amp; =6x{(3{x}^{2}+4)}^{4}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170573371003\">This is exactly the expression we started with inside the integrand.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1170573359652\" class=\"exercise\">\r\n<div id=\"fs-id1170571227254\" class=\"textbox\">\r\n<p id=\"fs-id1170571334089\">Use substitution to find the antiderivative of [latex]\\int 3{x}^{2}{({x}^{3}-3)}^{2}dx.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"solution\">\r\n<p id=\"fs-id1170573405086\">[latex]\\int 3{x}^{2}{({x}^{3}-3)}^{2}dx=\\frac{1}{3}{({x}^{3}-3)}^{3}+C[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170571728784\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170573396896\">Let [latex]u={x}^{3}-3.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170571121618\">Sometimes we need to adjust the constants in our integral if they don\u2019t match up exactly with the expressions we are substituting.<\/p>\r\n\r\n<div id=\"fs-id1170571334083\" class=\"textbox examples\">\r\n<h3>Using Substitution with Alteration<\/h3>\r\n<div id=\"fs-id1170573255224\" class=\"exercise\">\r\n<div id=\"fs-id1170571027516\" class=\"textbox\">\r\n<p id=\"fs-id1170573384093\">Use substitution to find the antiderivative of [latex]\\int z\\sqrt{{z}^{2}-5}dz.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573391210\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573391210\"]\r\n<p id=\"fs-id1170573391210\">Rewrite the integral as [latex]\\int z{({z}^{2}-5)}^{1\\text{\/}2}dz.[\/latex] Let [latex]u={z}^{2}-5[\/latex] and [latex]du=2zdz.[\/latex] Now we have a problem because [latex]du=2zdz[\/latex] and the original expression has only [latex]zdz.[\/latex] We have to alter our expression for <em>du<\/em> or the integral in [latex]u[\/latex] will be twice as large as it should be. If we multiply both sides of the <em>du<\/em> equation by [latex]\\frac{1}{2}.[\/latex] we can solve this problem. Thus,<\/p>\r\n\r\n<div id=\"fs-id1170573430996\" class=\"equation unnumbered\">[latex]\\begin{array}{}\\\\ \\hfill u&amp; ={z}^{2}-5\\hfill \\\\ \\hfill du&amp; =2zdz\\hfill \\\\ \\hfill \\frac{1}{2}du&amp; =\\frac{1}{2}(2z)dz=zdz.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170570976225\">Write the integral in terms of [latex]u[\/latex], but pull the [latex]\\frac{1}{2}[\/latex] outside the integration symbol:<\/p>\r\n\r\n<div id=\"fs-id1170573533836\" class=\"equation unnumbered\">[latex]\\int z{({z}^{2}-5)}^{1\\text{\/}2}dz=\\frac{1}{2}\\int {u}^{1\\text{\/}2}du.[\/latex]<\/div>\r\n<p id=\"fs-id1170573586341\">Integrate the expression in [latex]u[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1170570994415\" class=\"equation unnumbered\">[latex]\\begin{array}{}\\\\ \\frac{1}{2}\\int {u}^{1\\text{\/}2}du\\hfill &amp; =(\\frac{1}{2})\\frac{{u}^{3\\text{\/}2}}{\\frac{3}{2}}+C\\hfill \\\\ \\\\ &amp; =(\\frac{1}{2})(\\frac{2}{3}){u}^{3\\text{\/}2}+C\\hfill \\\\ &amp; =\\frac{1}{3}{u}^{3\\text{\/}2}+C\\hfill \\\\ &amp; =\\frac{1}{3}{({z}^{2}-5)}^{3\\text{\/}2}+C.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573362318\" class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1170571254579\" class=\"exercise\">\r\n<div id=\"fs-id1170571381835\" class=\"textbox\">\r\n<p id=\"fs-id1170573759886\">Use substitution to find the antiderivative of [latex]\\int {x}^{2}{({x}^{3}+5)}^{9}dx.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573361491\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573361491\"]\r\n<p id=\"fs-id1170573361491\">[latex]\\frac{{({x}^{3}+5)}^{10}}{30}+C[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572231096\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170571138635\">Multiply the <em>du<\/em> equation by [latex]\\frac{1}{3}.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573573975\" class=\"textbox examples\">\r\n<h3>Using Substitution with Integrals of Trigonometric Functions<\/h3>\r\n<div class=\"exercise\">\r\n<div id=\"fs-id1170573333636\" class=\"textbox\">\r\n<p id=\"fs-id1170573385625\">Use substitution to evaluate the integral [latex]\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573437523\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573437523\"]\r\n<p id=\"fs-id1170573437523\">We know the derivative of [latex] \\cos t[\/latex] is [latex]\\text{\u2212} \\sin t,[\/latex] so we set [latex]u= \\cos t.[\/latex] Then [latex]du=\\text{\u2212} \\sin tdt.[\/latex] Substituting into the integral, we have<\/p>\r\n\r\n<div id=\"fs-id1170573337947\" class=\"equation unnumbered\">[latex]\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt=\\text{\u2212}\\int \\frac{du}{{u}^{3}}.[\/latex]<\/div>\r\n<p id=\"fs-id1170573364147\">Evaluating the integral, we get<\/p>\r\n\r\n<div id=\"fs-id1170571136365\" class=\"equation unnumbered\">[latex]\\begin{array}{}\\\\ \\\\ \\text{\u2212}\\int \\frac{du}{{u}^{3}}\\hfill &amp; =\\text{\u2212}\\int {u}^{-3}du\\hfill \\\\ &amp; =\\text{\u2212}(-\\frac{1}{2}){u}^{-2}+C.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170573409376\">Putting the answer back in terms of [latex]t[\/latex], we get<\/p>\r\n\r\n<div id=\"fs-id1170573587515\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt\\hfill &amp; =\\frac{1}{2{u}^{2}}+C\\hfill \\\\ \\\\ &amp; =\\frac{1}{2{ \\cos }^{2}t}+C.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573733746\" class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1170573387947\" class=\"exercise\">\r\n<div id=\"fs-id1170573717717\" class=\"textbox\">\r\n<p id=\"fs-id1170571198076\">Use substitution to evaluate the integral [latex]\\int \\frac{ \\cos t}{{ \\sin }^{2}t}dt.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571171177\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571171177\"]\r\n<p id=\"fs-id1170571171177\">[latex]-\\frac{1}{ \\sin t}+C[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572333217\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170571258771\">Use the process from <a class=\"autogenerated-content\" href=\"#fs-id1170573573975\">(Figure)<\/a> to solve the problem.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170573413569\">Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a constant. We need to eliminate all the expressions within the integrand that are in terms of the original variable. When we are done, [latex]u[\/latex] should be the only variable in the integrand. In some cases, this means solving for the original variable in terms of [latex]u[\/latex]. This technique should become clear in the next example.<\/p>\r\n\r\n<div id=\"fs-id1170573426799\" class=\"textbox examples\">\r\n<h3>Finding an Antiderivative Using [latex]u[\/latex]-Substitution<\/h3>\r\n<div id=\"fs-id1170573417818\" class=\"exercise\">\r\n<div id=\"fs-id1170573533908\" class=\"textbox\">\r\n<p id=\"fs-id1170573328644\">Use substitution to find the antiderivative of [latex]\\int \\frac{x}{\\sqrt{x-1}}dx.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573435928\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573435928\"]\r\n<p id=\"fs-id1170573435928\">If we let [latex]u=x-1,[\/latex] then [latex]du=dx.[\/latex] But this does not account for the [latex]x[\/latex] in the numerator of the integrand. We need to express [latex]x[\/latex] in terms of [latex]u[\/latex]. If [latex]u=x-1,[\/latex] then [latex]x=u+1.[\/latex] Now we can rewrite the integral in terms of [latex]u[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1170573333822\" class=\"equation unnumbered\">[latex]\\begin{array}{ll}\\int \\frac{x}{\\sqrt{x-1}}dx\\hfill &amp; =\\int \\frac{u+1}{\\sqrt{u}}du\\hfill \\\\ \\\\ &amp; =\\int \\sqrt{u}+\\frac{1}{\\sqrt{u}}du\\hfill \\\\ &amp; =\\int ({u}^{1\\text{\/}2}+{u}^{-1\\text{\/}2})du.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170573544033\">Then we integrate in the usual way, replace [latex]u[\/latex] with the original expression, and factor and simplify the result. Thus,<\/p>\r\n\r\n<div id=\"fs-id1170570997027\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}\\int ({u}^{1\\text{\/}2}+{u}^{-1\\text{\/}2})du\\hfill &amp; =\\frac{2}{3}{u}^{3\\text{\/}2}+2{u}^{1\\text{\/}2}+C\\hfill \\\\ \\\\ &amp; =\\frac{2}{3}{(x-1)}^{3\\text{\/}2}+2{(x-1)}^{1\\text{\/}2}+C\\hfill \\\\ &amp; ={(x-1)}^{1\\text{\/}2}\\left[\\frac{2}{3}(x-1)+2\\right]+C\\hfill \\\\ &amp; ={(x-1)}^{1\\text{\/}2}(\\frac{2}{3}x-\\frac{2}{3}+\\frac{6}{3})\\hfill \\\\ &amp; ={(x-1)}^{1\\text{\/}2}(\\frac{2}{3}x+\\frac{4}{3})\\hfill \\\\ &amp; =\\frac{2}{3}{(x-1)}^{1\\text{\/}2}(x+2)+C.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571115891\" class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1170571283562\" class=\"exercise\">\r\n<div id=\"fs-id1170571189256\" class=\"textbox\">\r\n<p id=\"fs-id1170571087102\">Use substitution to evaluate the indefinite integral [latex]\\int { \\cos }^{3}t \\sin tdt.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571285848\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571285848\"]\r\n<p id=\"fs-id1170571285848\">[latex]-\\frac{{ \\cos }^{4}t}{4}+C[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572295331\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170573501914\">Use the process from <a class=\"autogenerated-content\" href=\"#fs-id1170573426799\">(Figure)<\/a> to solve the problem.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573399047\" class=\"bc-section section\">\r\n<h1>Substitution for Definite Integrals<\/h1>\r\n<p id=\"fs-id1170573387843\">Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.<\/p>\r\n\r\n<div id=\"fs-id1170573497309\" class=\"textbox key-takeaways theorem\">\r\n<h3>Substitution with Definite Integrals<\/h3>\r\n<p id=\"fs-id1170570999523\">Let [latex]u=g(x)[\/latex] and let [latex]{g}^{\\text{\u2032}}[\/latex] be continuous over an interval [latex]\\left[a,b\\right],[\/latex] and let [latex]f[\/latex] be continuous over the range of [latex]u=g(x).[\/latex] Then,<\/p>\r\n\r\n<div id=\"fs-id1170571189407\" class=\"equation unnumbered\">[latex]{\\int }_{a}^{b}f(g(x)){g}^{\\prime }(x)dx={\\int }_{g(a)}^{g(b)}f(u)du.[\/latex]<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170573502228\">Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if [latex]F(x)[\/latex] is an antiderivative of [latex]f(x),[\/latex] we have<\/p>\r\n\r\n<div id=\"fs-id1170571168095\" class=\"equation unnumbered\">[latex]\\int f(g(x)){g}^{\\prime }(x)dx=F(g(x))+C.[\/latex]<\/div>\r\n<p id=\"fs-id1170573420715\">Then<\/p>\r\n\r\n<div id=\"fs-id1170573388911\" class=\"equation\">[latex]\\begin{array}{cc}{\\int }_{a}^{b}f\\left[g(x)\\right]{g}^{\\prime }(x)dx\\hfill &amp; ={F(g(x))|}_{x=a}^{x=b}\\hfill \\\\ &amp; =F(g(b))-F(g(a))\\hfill \\\\ &amp; ={F(u)|}_{u=g(a)}^{u=g(b)}\\hfill \\\\ \\\\ \\\\ &amp; ={\\int }_{g(a)}^{g(b)}f(u)du,\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170571100165\">and we have the desired result.<\/p>\r\n\r\n<div id=\"fs-id1170571215481\" class=\"textbox examples\">\r\n<h3>Using Substitution to Evaluate a Definite Integral<\/h3>\r\n<div id=\"fs-id1170571215483\" class=\"exercise\">\r\n<div id=\"fs-id1170573750246\" class=\"textbox\">\r\n<p id=\"fs-id1170571382554\">Use substitution to evaluate [latex]{\\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573440252\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573440252\"]\r\n<p id=\"fs-id1170573440252\">Let [latex]u=1+2{x}^{3},[\/latex] so [latex]du=6{x}^{2}dx.[\/latex] Since the original function includes one factor of [latex]x[\/latex]<sup>2<\/sup> and [latex]du=6{x}^{2}dx,[\/latex] multiply both sides of the <em>du<\/em> equation by [latex]1\\text{\/}6.[\/latex] Then,<\/p>\r\n\r\n<div id=\"fs-id1170573362498\" class=\"equation unnumbered\">[latex]\\begin{array}{ccc}du\\hfill &amp; =\\hfill &amp; 6{x}^{2}dx\\hfill \\\\ \\frac{1}{6}du\\hfill &amp; =\\hfill &amp; {x}^{2}dx.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170571157085\">To adjust the limits of integration, note that when [latex]x=0,u=1+2(0)=1,[\/latex] and when [latex]x=1,u=1+2(1)=3.[\/latex] Then<\/p>\r\n\r\n<div id=\"fs-id1170571303367\" class=\"equation unnumbered\">[latex]{\\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx=\\frac{1}{6}{\\int }_{1}^{3}{u}^{5}du.[\/latex]<\/div>\r\n<p id=\"fs-id1170571285285\">Evaluating this expression, we get<\/p>\r\n\r\n<div id=\"fs-id1170571285289\" class=\"equation unnumbered\">[latex]\\begin{array}{}\\\\ \\\\ \\frac{1}{6}{\\int }_{1}^{3}{u}^{5}du\\hfill &amp; =(\\frac{1}{6})(\\frac{{u}^{6}}{6}){|}_{1}^{3}\\hfill \\\\ &amp; =\\frac{1}{36}\\left[{(3)}^{6}-{(1)}^{6}\\right]\\hfill \\\\ &amp; =\\frac{182}{9}.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571101513\" class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1170571061496\" class=\"exercise\">\r\n<div id=\"fs-id1170571061498\" class=\"textbox\">\r\n<p id=\"fs-id1170571061500\">Use substitution to evaluate the definite integral [latex]{\\int }_{-1}^{0}y{(2{y}^{2}-3)}^{5}dy.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573518898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573518898\"]\r\n<p id=\"fs-id1170573518898\">[latex]\\frac{91}{3}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170571619018\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170573593922\">Use the steps from <a class=\"autogenerated-content\" href=\"#fs-id1170571215481\">(Figure)<\/a> to solve the problem.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573593895\" class=\"textbox examples\">\r\n<h3>Using Substitution with an Exponential Function<\/h3>\r\n<div id=\"fs-id1170573593897\" class=\"exercise\">\r\n<div id=\"fs-id1170571158819\" class=\"textbox\">\r\n<p id=\"fs-id1170571158824\">Use substitution to evaluate [latex]{\\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571347206\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571347206\"]\r\n<p id=\"fs-id1170571347206\">Let [latex]u=4{x}^{3}+3.[\/latex] Then, [latex]du=8xdx.[\/latex] To adjust the limits of integration, we note that when [latex]x=0,u=3,[\/latex] and when [latex]x=1,u=7.[\/latex] So our substitution gives<\/p>\r\n\r\n<div id=\"fs-id1170571052992\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}{\\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx\\hfill &amp; =\\frac{1}{8}{\\int }_{3}^{7}{e}^{u}du\\hfill \\\\ \\\\ &amp; =\\frac{1}{8}{e}^{u}{|}_{3}^{7}\\hfill \\\\ &amp; =\\frac{{e}^{7}-{e}^{3}}{8}\\hfill \\\\ &amp; \\approx 134.568.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170570976160\" class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1170571099763\" class=\"exercise\">\r\n<div id=\"fs-id1170571099765\" class=\"textbox\">\r\n<p id=\"fs-id1170571099767\">Use substitution to evaluate [latex]{\\int }_{0}^{1}{x}^{2} \\cos (\\frac{\\pi }{2}{x}^{3})dx.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573438316\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573438316\"]\r\n<p id=\"fs-id1170573438316\">[latex]\\frac{2}{3\\pi }\\approx 0.2122[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170573592042\">Use the process from <a class=\"autogenerated-content\" href=\"#fs-id1170573593895\">(Figure)<\/a> to solve the problem.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170571289709\">Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we can apply substitution. Also, we have the option of replacing the original expression for [latex]u[\/latex] after we find the antiderivative, which means that we do not have to change the limits of integration. These two approaches are shown in <a class=\"autogenerated-content\" href=\"#fs-id1170570994085\">(Figure)<\/a>.<\/p>\r\n\r\n<div id=\"fs-id1170570994085\" class=\"textbox examples\">\r\n<h3>Using Substitution to Evaluate a Trigonometric Integral<\/h3>\r\n<div id=\"fs-id1170571213222\" class=\"exercise\">\r\n<div id=\"fs-id1170571213224\" class=\"textbox\">\r\n<p id=\"fs-id1170571219853\">Use substitution to evaluate [latex]{\\int }_{0}^{\\pi \\text{\/}2}{ \\cos }^{2}\\theta d\\theta .[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571030577\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571030577\"]\r\n<p id=\"fs-id1170571030577\">Let us first use a trigonometric identity to rewrite the integral. The trig identity [latex]{ \\cos }^{2}\\theta =\\frac{1+ \\cos 2\\theta }{2}[\/latex] allows us to rewrite the integral as<\/p>\r\n\r\n<div id=\"fs-id1170571131164\" class=\"equation unnumbered\">[latex]{\\int }_{0}^{\\pi \\text{\/}2}{ \\cos }^{2}\\theta d\\theta ={\\int }_{0}^{\\pi \\text{\/}2}\\frac{1+ \\cos 2\\theta }{2}d\\theta .[\/latex]<\/div>\r\n<p id=\"fs-id1170573351678\">Then,<\/p>\r\n\r\n<div id=\"fs-id1170573581363\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}{\\int }_{0}^{\\pi \\text{\/}2}(\\frac{1+ \\cos 2\\theta }{2})d\\theta \\hfill &amp; ={\\int }_{0}^{\\pi \\text{\/}2}(\\frac{1}{2}+\\frac{1}{2} \\cos 2\\theta )d\\theta \\hfill \\\\ \\\\ \\\\ &amp; =\\frac{1}{2}{\\int }_{0}^{\\pi \\text{\/}2}d\\theta +{\\int }_{0}^{\\pi \\text{\/}2} \\cos 2\\theta d\\theta .\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170571374921\">We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let [latex]u=2\\theta .[\/latex] Then, [latex]du=2d\\theta ,[\/latex] or [latex]\\frac{1}{2}du=d\\theta .[\/latex] Also, when [latex]\\theta =0,u=0,[\/latex] and when [latex]\\theta =\\pi \\text{\/}2,u=\\pi .[\/latex] Expressing the second integral in terms of [latex]u[\/latex], we have<\/p>\r\n\r\n<div id=\"fs-id1170571244772\" class=\"equation unnumbered\">[latex]\\begin{array}{}\\\\ \\\\ \\frac{1}{2}{\\int }_{0}^{\\pi \\text{\/}2}d\\theta +\\frac{1}{2}{\\int }_{0}^{\\pi \\text{\/}2} \\cos 2\\theta d\\theta \\hfill &amp; =\\frac{1}{2}{\\int }_{0}^{\\pi \\text{\/}2}d\\theta +\\frac{1}{2}(\\frac{1}{2}){\\int }_{0}^{\\pi } \\cos udu\\hfill \\\\ &amp; =\\frac{\\theta }{2}{|}_{\\theta =0}^{\\theta =\\pi \\text{\/}2}+\\frac{1}{4} \\sin u{|}_{u=0}^{u=\\theta }\\hfill \\\\ &amp; =(\\frac{\\pi }{4}-0)+(0-0)=\\frac{\\pi }{4}.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571220845\" class=\"textbox key-takeaways\">\r\n<h3>Key Concepts<\/h3>\r\n<ul id=\"fs-id1170571098317\">\r\n \t<li>Substitution is a technique that simplifies the integration of functions that are the result of a chain-rule derivative. The term \u2018substitution\u2019 refers to changing variables or substituting the variable [latex]u[\/latex] and <em>du<\/em> for appropriate expressions in the integrand.<\/li>\r\n \t<li>When using substitution for a definite integral, we also have to change the limits of integration.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1170573581296\" class=\"key-equations\">\r\n<h1>Key Equations<\/h1>\r\n<ul id=\"fs-id1170573390237\">\r\n \t<li><strong>Substitution with Indefinite Integrals<\/strong>\r\n[latex]\\int f\\left[g(x)\\right]{g}^{\\prime }(x)dx=\\int f(u)du=F(u)+C=F(g(x))+C[\/latex]<\/li>\r\n \t<li><strong>Substitution with Definite Integrals<\/strong>\r\n[latex]{\\int }_{a}^{b}f(g(x))g\\text{'}(x)dx={\\int }_{g(a)}^{g(b)}f(u)du[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1170571209484\" class=\"textbox exercises\">\r\n<div id=\"fs-id1170573549514\" class=\"exercise\">\r\n<div id=\"fs-id1170573549516\" class=\"textbox\">\r\n<p id=\"fs-id1170573549518\">Why is [latex]u[\/latex]-substitution referred to as <em>change of variable<\/em>?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573575165\" class=\"exercise\">\r\n<div id=\"fs-id1170573575167\" class=\"textbox\">\r\n<p id=\"fs-id1170573575169\">2. If [latex]f=g\\circ h,[\/latex] when reversing the chain rule, [latex]\\frac{d}{dx}(g\\circ h)(x)={g}^{\\prime }(h(x)){h}^{\\prime }(x),[\/latex] should you take [latex]u=g(x)[\/latex] or [latex]u=h(x)?[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573501482\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573501482\"]\r\n<p id=\"fs-id1170573501482\">[latex]u=h(x)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170571118049\">In the following exercises, verify each identity using differentiation. Then, using the indicated [latex]u[\/latex]-substitution, identify [latex]f[\/latex] such that the integral takes the form [latex]\\int f(u)du.[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1170571033648\" class=\"exercise\">\r\n<div id=\"fs-id1170571033650\" class=\"textbox\">\r\n<p id=\"fs-id1170571033652\">[latex]\\int x\\sqrt{x+1}dx=\\frac{2}{15}{(x+1)}^{3\\text{\/}2}(3x-2)+C;u=x+1[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573718856\" class=\"exercise\">\r\n<div id=\"fs-id1170573718859\" class=\"textbox\">\r\n<p id=\"fs-id1170573750112\">[latex]\\int \\frac{{x}^{2}}{\\sqrt{x-1}}dx(x&gt;1)=\\frac{2}{15}\\sqrt{x-1}(3{x}^{2}+4x+8)+C;u=x-1[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573398685\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573398685\"]\r\n<p id=\"fs-id1170573398685\">[latex]f(u)=\\frac{{(u+1)}^{2}}{\\sqrt{u}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571303033\" class=\"exercise\">\r\n<div id=\"fs-id1170571303035\" class=\"textbox\">\r\n<p id=\"fs-id1170571248540\">[latex]\\int x\\sqrt{4{x}^{2}+9}dx=\\frac{1}{12}{(4{x}^{2}+9)}^{3\\text{\/}2}+C;u=4{x}^{2}+9[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571206963\" class=\"exercise\">\r\n<div id=\"fs-id1170571305052\" class=\"textbox\">\r\n<p id=\"fs-id1170571305054\">[latex]\\int \\frac{x}{\\sqrt{4{x}^{2}+9}}dx=\\frac{1}{4}\\sqrt{4{x}^{2}+9}+C;u=4{x}^{2}+9[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573436276\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573436276\"]\r\n<p id=\"fs-id1170573436276\">[latex]du=8xdx;f(u)=\\frac{1}{8\\sqrt{u}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573759850\" class=\"exercise\">\r\n<div id=\"fs-id1170573759852\" class=\"textbox\">\r\n<p id=\"fs-id1170573759854\">[latex]\\int \\frac{x}{{(4{x}^{2}+9)}^{2}}dx=-\\frac{1}{8(4{x}^{2}+9)};u=4{x}^{2}+9[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170570992528\">In the following exercises, find the antiderivative using the indicated substitution.<\/p>\r\n\r\n<div id=\"fs-id1170571119993\" class=\"exercise\">\r\n<div id=\"fs-id1170571119995\" class=\"textbox\">\r\n<p id=\"fs-id1170571119997\">[latex]\\int {(x+1)}^{4}dx;u=x+1[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573385475\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573385475\"]\r\n<p id=\"fs-id1170573385475\">[latex]\\frac{1}{5}{(x+1)}^{5}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170570996054\" class=\"exercise\">\r\n<div id=\"fs-id1170570996056\" class=\"textbox\">\r\n<p id=\"fs-id1170570996058\">[latex]\\int {(x-1)}^{5}dx;u=x-1[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571056458\" class=\"exercise\">\r\n<div id=\"fs-id1170571240500\" class=\"textbox\">\r\n<p id=\"fs-id1170571240502\">[latex]\\int {(2x-3)}^{-7}dx;u=2x-3[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573518184\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573518184\"]\r\n<p id=\"fs-id1170573518184\">[latex]-\\frac{1}{12{(3-2x)}^{6}}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573760680\" class=\"exercise\">\r\n<div id=\"fs-id1170573760682\" class=\"textbox\">\r\n<p id=\"fs-id1170573760685\">[latex]\\int {(3x-2)}^{-11}dx;u=3x-2[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573417176\" class=\"exercise\">\r\n<div id=\"fs-id1170573417178\" class=\"textbox\">\r\n<p id=\"fs-id1170573417180\">[latex]\\int \\frac{x}{\\sqrt{{x}^{2}+1}}dx;u={x}^{2}+1[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573535993\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573535993\"]\r\n<p id=\"fs-id1170573535993\">[latex]\\sqrt{{x}^{2}+1}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573542517\" class=\"exercise\">\r\n<div id=\"fs-id1170573542519\" class=\"textbox\">\r\n<p id=\"fs-id1170573542521\">[latex]\\int \\frac{x}{\\sqrt{1-{x}^{2}}}dx;u=1-{x}^{2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571123643\" class=\"exercise\">\r\n<div id=\"fs-id1170573546495\" class=\"textbox\">\r\n<p id=\"fs-id1170573546497\">[latex]\\int (x-1){({x}^{2}-2x)}^{3}dx;u={x}^{2}-2x[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573726592\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573726592\"]\r\n<p id=\"fs-id1170573726592\">[latex]\\frac{1}{8}{({x}^{2}-2x)}^{4}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571145634\" class=\"exercise\">\r\n<div id=\"fs-id1170571145636\" class=\"textbox\">\r\n<p id=\"fs-id1170571145638\">[latex]\\int ({x}^{2}-2x){({x}^{3}-3{x}^{2})}^{2}dx;u={x}^{3}=3{x}^{2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573623732\" class=\"exercise\">\r\n<div id=\"fs-id1170573623734\" class=\"textbox\">\r\n<p id=\"fs-id1170573623736\">[latex]\\int { \\cos }^{3}\\theta d\\theta ;u= \\sin \\theta [\/latex]([latex]Hint\\text{:}{ \\cos }^{2}\\theta =1-{ \\sin }^{2}\\theta [\/latex])<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573618433\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573618433\"]\r\n<p id=\"fs-id1170573618433\">[latex] \\sin \\theta -\\frac{{ \\sin }^{3}\\theta }{3}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571227343\" class=\"exercise\">\r\n<div id=\"fs-id1170571227346\" class=\"textbox\">\r\n<p id=\"fs-id1170571227348\">[latex]\\int { \\sin }^{3}\\theta d\\theta ;u= \\cos \\theta [\/latex][latex](Hint\\text{:}{ \\sin }^{2}\\theta =1-{ \\cos }^{2}\\theta)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170571136653\">In the following exercises, use a suitable change of variables to determine the indefinite integral.<\/p>\r\n\r\n<div id=\"fs-id1170571136656\" class=\"exercise\">\r\n<div id=\"fs-id1170571136658\" class=\"textbox\">\r\n<p id=\"fs-id1170571136660\">[latex]\\int x{(1-x)}^{99}dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571098395\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571098395\"]\r\n<p id=\"fs-id1170571098395\">[latex]\\frac{{(1-x)}^{101}}{101}-\\frac{{(1-x)}^{100}}{100}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571304981\" class=\"exercise\">\r\n<div id=\"fs-id1170571304983\" class=\"textbox\">\r\n<p id=\"fs-id1170571304985\">[latex]\\int t{(1-{t}^{2})}^{10}dt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571275277\" class=\"exercise\">\r\n<div id=\"fs-id1170571275279\" class=\"textbox\">\r\n<p id=\"fs-id1170571275281\">[latex]\\int {(11x-7)}^{-3}dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573732555\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573732555\"]\r\n<p id=\"fs-id1170573732555\">[latex]-\\frac{1}{22(7-11{x}^{2})}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573545918\" class=\"exercise\">\r\n<div id=\"fs-id1170573422123\" class=\"textbox\">\r\n<p id=\"fs-id1170573422126\">[latex]\\int {(7x-11)}^{4}dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573727306\" class=\"exercise\">\r\n<div id=\"fs-id1170573727308\" class=\"textbox\">\r\n<p id=\"fs-id1170573727310\">[latex]\\int { \\cos }^{3}\\theta \\sin \\theta d\\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573404920\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573404920\"]\r\n<p id=\"fs-id1170573404920\">[latex]-\\frac{{ \\cos }^{4}\\theta }{4}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571094884\" class=\"exercise\">\r\n<div id=\"fs-id1170571094886\" class=\"textbox\">\r\n<p id=\"fs-id1170571094888\">[latex]\\int { \\sin }^{7}\\theta \\cos \\theta d\\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571158870\" class=\"exercise\">\r\n<div id=\"fs-id1170573726094\" class=\"textbox\">\r\n<p id=\"fs-id1170573726096\">[latex]\\int { \\cos }^{2}(\\pi t) \\sin (\\pi t)dt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571227213\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571227213\"]\r\n<p id=\"fs-id1170571227213\">[latex]-\\frac{{ \\cos }^{3}(\\pi t)}{3\\pi }+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573759909\" class=\"exercise\">\r\n<div id=\"fs-id1170573759911\" class=\"textbox\">\r\n<p id=\"fs-id1170573759913\">[latex]\\int { \\sin }^{2}x{ \\cos }^{3}xdx[\/latex][latex](Hint\\text{:}{ \\sin }^{2}x+{ \\cos }^{2}x=1)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571340505\" class=\"exercise\">\r\n<div id=\"fs-id1170571340508\" class=\"textbox\">\r\n<p id=\"fs-id1170571340510\">[latex]\\int t \\sin ({t}^{2}) \\cos ({t}^{2})dt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573727339\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573727339\"]\r\n<p id=\"fs-id1170573727339\">[latex]-\\frac{1}{4}\\phantom{\\rule{0.05em}{0ex}}{ \\cos }^{2}({t}^{2})+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571340409\" class=\"exercise\">\r\n<div id=\"fs-id1170571340411\" class=\"textbox\">\r\n<p id=\"fs-id1170571340413\">[latex]\\int {t}^{2}{ \\cos }^{2}({t}^{3}) \\sin ({t}^{3})dt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571232630\" class=\"exercise\">\r\n<div id=\"fs-id1170571232632\" class=\"textbox\">\r\n<p id=\"fs-id1170571232634\">[latex]\\int \\frac{{x}^{2}}{{({x}^{3}-3)}^{2}}dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571158899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571158899\"]\r\n<p id=\"fs-id1170571158899\">[latex]-\\frac{1}{3({x}^{3}-3)}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170570999552\" class=\"exercise\">\r\n<div id=\"fs-id1170570999554\" class=\"textbox\">\r\n<p id=\"fs-id1170570999556\">[latex]\\int \\frac{{x}^{3}}{\\sqrt{1-{x}^{2}}}dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571057347\" class=\"exercise\">\r\n<div id=\"fs-id1170571057349\" class=\"textbox\">\r\n<p id=\"fs-id1170571057351\">[latex]\\int \\frac{{y}^{5}}{{(1-{y}^{3})}^{3\\text{\/}2}}dy[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573525700\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573525700\"]\r\n<p id=\"fs-id1170573525700\">[latex]-\\frac{2({y}^{3}-2)}{3\\sqrt{1-{y}^{3}}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571269414\" class=\"exercise\">\r\n<div id=\"fs-id1170571269416\" class=\"textbox\">\r\n<p id=\"fs-id1170571269419\">[latex]{\\int \\cos \\theta (1- \\cos \\theta )}^{99} \\sin \\theta d\\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571160788\" class=\"exercise\">\r\n<div id=\"fs-id1170571160790\" class=\"textbox\">\r\n<p id=\"fs-id1170571160792\">[latex]{\\int (1-{ \\cos }^{3}\\theta )}^{10}{ \\cos }^{2}\\theta \\sin \\theta d\\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571198030\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571198030\"]\r\n<p id=\"fs-id1170571198030\">[latex]\\frac{1}{33}{(1-{ \\cos }^{3}\\theta )}^{11}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571379754\" class=\"exercise\">\r\n<div id=\"fs-id1170571379756\" class=\"textbox\">\r\n<p id=\"fs-id1170571379758\">[latex]\\int ( \\cos \\theta -1){({ \\cos }^{2}\\theta -2 \\cos \\theta )}^{3} \\sin \\theta d\\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571030683\" class=\"exercise\">\r\n<div id=\"fs-id1170571030685\" class=\"textbox\">\r\n<p id=\"fs-id1170571030687\">[latex]\\int ({ \\sin }^{2}\\theta -2 \\sin \\theta ){({ \\sin }^{3}\\theta -3{ \\sin }^{2}\\theta )}^{3} \\cos \\theta d\\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573633894\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573633894\"]\r\n<p id=\"fs-id1170573633894\">[latex]\\frac{1}{12}{({ \\sin }^{3}\\theta -3{ \\sin }^{2}\\theta )}^{4}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170571114997\">In the following exercises, use a calculator to estimate the area under the curve using left Riemann sums with 50 terms, then use substitution to solve for the exact answer.<\/p>\r\n\r\n<div id=\"fs-id1170571115002\" class=\"exercise\">\r\n<div id=\"fs-id1170571115004\" class=\"textbox\">\r\n<p id=\"fs-id1170571115006\"><strong>[T]<\/strong>[latex]y=3{(1-x)}^{2}[\/latex] over [latex]\\left[0,2\\right][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571158789\" class=\"exercise\">\r\n<div id=\"fs-id1170571158791\" class=\"textbox\">\r\n<p id=\"fs-id1170571158793\"><strong>[T]<\/strong>[latex]y=x{(1-{x}^{2})}^{3}[\/latex] over [latex]\\left[-1,2\\right][\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571269340\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571269340\"]\r\n<p id=\"fs-id1170571269340\">[latex]{L}_{50}=-8.5779.[\/latex] The exact area is [latex]\\frac{-81}{8}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571337335\" class=\"exercise\">\r\n<div id=\"fs-id1170571337337\" class=\"textbox\">\r\n<p id=\"fs-id1170571337339\"><strong>[T]<\/strong>[latex]y= \\sin x{(1- \\cos x)}^{2}[\/latex] over [latex]\\left[0,\\pi \\right][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571159912\" class=\"exercise\">\r\n<div id=\"fs-id1170571159914\" class=\"textbox\">\r\n<p id=\"fs-id1170571159917\"><strong>[T]<\/strong>[latex]y=\\frac{x}{{({x}^{2}+1)}^{2}}[\/latex] over [latex]\\left[-1,1\\right][\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571057417\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571057417\"]\r\n<p id=\"fs-id1170571057417\">[latex]{L}_{50}=-0.006399[\/latex] \u2026 The exact area is 0.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170571057435\">In the following exercises, use a change of variables to evaluate the definite integral.<\/p>\r\n\r\n<div id=\"fs-id1170571057439\" class=\"exercise\">\r\n<div id=\"fs-id1170571057441\" class=\"textbox\">\r\n<p id=\"fs-id1170571057443\">[latex]{\\int }_{0}^{1}x\\sqrt{1-{x}^{2}}dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571026374\" class=\"exercise\">\r\n<div id=\"fs-id1170571026376\" class=\"textbox\">\r\n<p id=\"fs-id1170571026378\">[latex]{\\int }_{0}^{1}\\frac{x}{\\sqrt{1+{x}^{2}}}dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573618361\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573618361\"]\r\n<p id=\"fs-id1170573618361\">[latex]u=1+{x}^{2},du=2xdx,\\frac{1}{2}{\\int }_{1}^{2}{u}^{-1\\text{\/}2}du=\\sqrt{2}-1[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571037384\" class=\"exercise\">\r\n<div id=\"fs-id1170571037387\" class=\"textbox\">\r\n<p id=\"fs-id1170571037389\">[latex]{\\int }_{0}^{2}\\frac{t}{\\sqrt{5+{t}^{2}}}dt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571329548\" class=\"exercise\">\r\n<div id=\"fs-id1170571329550\" class=\"textbox\">\r\n<p id=\"fs-id1170571150593\">[latex]{\\int }_{0}^{1}\\frac{t}{\\sqrt{1+{t}^{3}}}dt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571150634\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571150634\"]\r\n<p id=\"fs-id1170571150634\">[latex]u=1+{t}^{3},du=3{t}^{2},\\frac{1}{3}{\\int }_{1}^{2}{u}^{-1\\text{\/}2}du=\\frac{2}{3}(\\sqrt{2}-1)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571160643\" class=\"exercise\">\r\n<div id=\"fs-id1170571160646\" class=\"textbox\">\r\n<p id=\"fs-id1170571160648\">[latex]{\\int }_{0}^{\\pi \\text{\/}4}{ \\sec }^{2}\\theta \\tan \\theta d\\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571078760\" class=\"exercise\">\r\n<div id=\"fs-id1170571078762\" class=\"textbox\">\r\n<p id=\"fs-id1170571078764\">[latex]{\\int }_{0}^{\\pi \\text{\/}4}\\frac{ \\sin \\theta }{{ \\cos }^{4}\\theta }d\\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573704614\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573704614\"]\r\n<p id=\"fs-id1170573704614\">[latex]u= \\cos \\theta ,du=\\text{\u2212} \\sin \\theta d\\theta ,{\\int }_{1\\text{\/}\\sqrt{2}}^{1}{u}^{-4}du=\\frac{1}{3}(2\\sqrt{2}-1)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170571377045\">In the following exercises, evaluate the indefinite integral [latex]\\int f(x)dx[\/latex] with constant [latex]C=0[\/latex] using [latex]u[\/latex]-substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of <em>C<\/em> that would need to be added to the antiderivative to make it equal to the definite integral [latex]F(x)={\\int }_{a}^{x}f(t)dt,[\/latex] with [latex]a[\/latex] the left endpoint of the given interval.<\/p>\r\n\r\n<div id=\"fs-id1170571303693\" class=\"exercise\">\r\n<div id=\"fs-id1170571303695\" class=\"textbox\">\r\n<p id=\"fs-id1170571303698\"><strong>[T]<\/strong>[latex]\\int (2x+1){e}^{{x}^{2}+x-6}dx[\/latex] over [latex]\\left[-3,2\\right][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573762457\" class=\"exercise\">\r\n<div id=\"fs-id1170573762459\" class=\"textbox\">\r\n<p id=\"fs-id1170573762461\"><strong>[T]<\/strong>[latex]\\int \\frac{ \\cos (\\text{ln}(2x))}{x}dx[\/latex] on [latex]\\left[0,2\\right][\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573762526\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573762526\"]<span id=\"fs-id1170573762529\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204236\/CNX_Calc_Figure_05_05_204.jpg\" alt=\"Two graphs. The first shows the function f(x) = cos(ln(2x)) \/ x, which increases sharply over the approximate interval (0,.25) and then decreases gradually to the x axis. The second shows the function f(x) = sin(ln(2x)), which decreases sharply on the approximate interval (0, .25) and then increases in a gently curve into the first quadrant.\" \/><\/span>\r\nThe antiderivative is [latex]y= \\sin (\\text{ln}(2x)).[\/latex] Since the antiderivative is not continuous at [latex]x=0,[\/latex] one cannot find a value of <em>C<\/em> that would make [latex]y= \\sin (\\text{ln}(2x))-C[\/latex] work as a definite integral.[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571021797\" class=\"exercise\">\r\n<div id=\"fs-id1170571021799\" class=\"textbox\">\r\n<p id=\"fs-id1170571021802\"><strong>[T]<\/strong>[latex]\\int \\frac{3{x}^{2}+2x+1}{\\sqrt{{x}^{3}+{x}^{2}+x+4}}dx[\/latex] over [latex]\\left[-1,2\\right][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573542457\" class=\"exercise\">\r\n<div id=\"fs-id1170573542459\" class=\"textbox\">\r\n<p id=\"fs-id1170573542461\"><strong>[T]<\/strong>[latex]\\int \\frac{ \\sin x}{{ \\cos }^{3}x}dx[\/latex] over [latex]\\left[-\\frac{\\pi }{3},\\frac{\\pi }{3}\\right][\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571340048\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571340048\"]<span id=\"fs-id1170571340054\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204239\/CNX_Calc_Figure_05_05_206.jpg\" alt=\"Two graphs. The first is the function f(x) = sin(x) \/ cos(x)^3 over [-5pi\/16, 5pi\/16]. It is an increasing concave down function for values less than zero and an increasing concave up function for values greater than zero. The second is the fuction f(x) = \u00bd sec(x)^2 over the same interval. It is a wide, concave up curve which decreases for values less than zero and increases for values greater than zero.\" \/><\/span>\r\nThe antiderivative is [latex]y=\\frac{1}{2}\\phantom{\\rule{0.05em}{0ex}}{ \\sec }^{2}x.[\/latex] You should take [latex]C=-2[\/latex] so that [latex]F(-\\frac{\\pi }{3})=0.[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571340126\" class=\"exercise\">\r\n<div id=\"fs-id1170571340128\" class=\"textbox\">\r\n<p id=\"fs-id1170571340131\"><strong>[T]<\/strong>[latex]\\int (x+2){e}^{\\text{\u2212}{x}^{2}-4x+3}dx[\/latex] over [latex]\\left[-5,1\\right][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571098887\" class=\"exercise\">\r\n<div id=\"fs-id1170571098889\" class=\"textbox\">\r\n<p id=\"fs-id1170571098892\"><strong>[T]<\/strong>[latex]\\int 3{x}^{2}\\sqrt{2{x}^{3}+1}dx[\/latex] over [latex]\\left[0,1\\right][\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571098951\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571098951\"]<span id=\"fs-id1170571098958\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204243\/CNX_Calc_Figure_05_05_208.jpg\" alt=\"Two graphs. The first shows the function f(x) = 3x^2 * sqrt(2x^3 + 1). It is an increasing concave up curve starting at the origin. The second shows the function f(x) = 1\/3 * (2x^3 + 1)^(1\/3). It is an increasing concave up curve starting at about 0.3.\" \/><\/span>\r\nThe antiderivative is [latex]y=\\frac{1}{3}{(2{x}^{3}+1)}^{3\\text{\/}2}.[\/latex] One should take [latex]C=-\\frac{1}{3}.[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571160883\" class=\"exercise\">\r\n<div id=\"fs-id1170571160885\" class=\"textbox\">\r\n<p id=\"fs-id1170571160887\">If [latex]h(a)=h(b)[\/latex] in [latex]{\\int }_{a}^{b}g\\text{'}(h(x))h(x)dx,[\/latex] what can you say about the value of the integral?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571160975\" class=\"exercise\">\r\n<div id=\"fs-id1170571160977\" class=\"textbox\">\r\n<p id=\"fs-id1170571160979\">Is the substitution [latex]u=1-{x}^{2}[\/latex] in the definite integral [latex]{\\int }_{0}^{2}\\frac{x}{1-{x}^{2}}dx[\/latex] okay? If not, why not?<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571291124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571291124\"]\r\n<p id=\"fs-id1170571291124\">No, because the integrand is discontinuous at [latex]x=1.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170571291140\">In the following exercises, use a change of variables to show that each definite integral is equal to zero.<\/p>\r\n\r\n<div id=\"fs-id1170571291144\" class=\"exercise\">\r\n<div id=\"fs-id1170571291146\" class=\"textbox\">\r\n<p id=\"fs-id1170571291148\">[latex]{\\int }_{0}^{\\pi }{ \\cos }^{2}(2\\theta ) \\sin (2\\theta )d\\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571158577\" class=\"exercise\">\r\n<div id=\"fs-id1170571158579\" class=\"textbox\">\r\n<p id=\"fs-id1170571158582\">[latex]{\\int }_{0}^{\\sqrt{\\pi }}t \\cos ({t}^{2}) \\sin ({t}^{2})dt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571158639\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571158639\"]\r\n<p id=\"fs-id1170571158639\">[latex]u= \\sin ({t}^{2});[\/latex] the integral becomes [latex]\\frac{1}{2}{\\int }_{0}^{0}udu.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571046470\" class=\"exercise\">\r\n<div id=\"fs-id1170571046472\" class=\"textbox\">\r\n<p id=\"fs-id1170571046474\">[latex]{\\int }_{0}^{1}(1-2t)dt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571046569\" class=\"exercise\">\r\n<div id=\"fs-id1170571046572\" class=\"textbox\">\r\n<p id=\"fs-id1170571046574\">[latex]{\\int }_{0}^{1}\\frac{1-2t}{(1+{(t-\\frac{1}{2})}^{2})}dt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571327399\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571327399\"]\r\n<p id=\"fs-id1170571327399\">[latex]u=(1+{(t-\\frac{1}{2})}^{2});[\/latex] the integral becomes [latex]\\text{\u2212}{\\int }_{5\\text{\/}4}^{5\\text{\/}4}\\frac{1}{u}du.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571327483\" class=\"exercise\">\r\n<div id=\"fs-id1170571327486\" class=\"textbox\">\r\n<p id=\"fs-id1170571327488\">[latex]{\\int }_{0}^{\\pi } \\sin ({(t-\\frac{\\pi }{2})}^{3}) \\cos (t-\\frac{\\pi }{2})dt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571371143\" class=\"exercise\">\r\n<div id=\"fs-id1170571371145\" class=\"textbox\">\r\n<p id=\"fs-id1170571371147\">[latex]{\\int }_{0}^{2}(1-t) \\cos (\\pi t)dt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571050111\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571050111\"]\r\n<p id=\"fs-id1170571050111\">[latex]u=1-t;[\/latex] the integral becomes<\/p>\r\n[latex]\\begin{array}{l}{\\int }_{1}^{-1}u \\cos (\\pi (1-u))du\\hfill \\\\ ={\\int }_{1}^{-1}u\\left[ \\cos \\pi \\cos u- \\sin \\pi \\sin u\\right]du\\hfill \\\\ =\\text{\u2212}{\\int }_{1}^{-1}u \\cos udu\\hfill \\\\ ={\\int }_{-1}^{1}u \\cos udu=0\\hfill \\end{array}[\/latex]\r\nsince the integrand is odd.[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571134790\" class=\"exercise\">\r\n<div id=\"fs-id1170571134792\" class=\"textbox\">\r\n<p id=\"fs-id1170571134794\">[latex]{\\int }_{\\pi \\text{\/}4}^{3\\pi \\text{\/}4}{ \\sin }^{2}t \\cos tdt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571276371\" class=\"exercise\">\r\n<div id=\"fs-id1170571276373\" class=\"textbox\">\r\n<p id=\"fs-id1170571276375\">Show that the average value of [latex]f(x)[\/latex] over an interval [latex]\\left[a,b\\right][\/latex] is the same as the average value of [latex]f(cx)[\/latex] over the interval [latex]\\left[\\frac{a}{c},\\frac{b}{c}\\right][\/latex] for [latex]c&gt;0.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571276459\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571276459\"]\r\n<p id=\"fs-id1170571276459\">Setting [latex]u=cx[\/latex] and [latex]du=cdx[\/latex] gets you [latex]\\frac{1}{\\frac{b}{c}-\\frac{a}{c}}{\\int }_{a\\text{\/}c}^{b\\text{\/}c}f(cx)dx=\\frac{c}{b-a}{\\int }_{u=a}^{u=b}f(u)\\frac{du}{c}=\\frac{1}{b-a}{\\int }_{a}^{b}f(u)du.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571273653\" class=\"exercise\">\r\n<div id=\"fs-id1170571273655\" class=\"textbox\">\r\n<p id=\"fs-id1170571273657\">Find the area under the graph of [latex]f(t)=\\frac{t}{{(1+{t}^{2})}^{a}}[\/latex] between [latex]t=0[\/latex] and [latex]t=x[\/latex] where [latex]a&gt;0[\/latex] and [latex]a\\ne 1[\/latex] is fixed, and evaluate the limit as [latex]x\\to \\infty .[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571132654\" class=\"exercise\">\r\n<div id=\"fs-id1170571132656\" class=\"textbox\">\r\n<p id=\"fs-id1170571132658\">Find the area under the graph of [latex]g(t)=\\frac{t}{{(1-{t}^{2})}^{a}}[\/latex] between [latex]t=0[\/latex] and [latex]t=x,[\/latex] where [latex]0&lt;x&lt;1[\/latex] and [latex]a&gt;0[\/latex] is fixed. Evaluate the limit as [latex]x\\to 1.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571245712\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571245712\"]\r\n<p id=\"fs-id1170571245712\">[latex]{\\int }_{0}^{x}g(t)dt=\\frac{1}{2}{\\int }_{u=1-{x}^{2}}^{1}\\frac{du}{{u}^{a}}=\\frac{1}{2(1-a)}{u}^{1-a}{|}_{u=1-{x}^{2}}^{1}=\\frac{1}{2(1-a)}(1-{(1-{x}^{2})}^{1-a}).[\/latex] As [latex]x\\to 1[\/latex] the limit is [latex]\\frac{1}{2(1-a)}[\/latex] if [latex]a&lt;1,[\/latex] and the limit diverges to +\u221e if [latex]a&gt;1.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573713777\" class=\"exercise\">\r\n<div id=\"fs-id1170573713779\" class=\"textbox\">\r\n<p id=\"fs-id1170573713781\">The area of a semicircle of radius 1 can be expressed as [latex]{\\int }_{-1}^{1}\\sqrt{1-{x}^{2}}dx.[\/latex] Use the substitution [latex]x= \\cos t[\/latex] to express the area of a semicircle as the integral of a trigonometric function. You do not need to compute the integral.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571169773\" class=\"exercise\">\r\n<div id=\"fs-id1170571169775\" class=\"textbox\">\r\n<p id=\"fs-id1170571169777\">The area of the top half of an ellipse with a major axis that is the [latex]x[\/latex]-axis from [latex]x=-1[\/latex] to [latex]a[\/latex] and with a minor axis that is the [latex]y[\/latex]-axis from [latex]y=\\text{\u2212}b[\/latex] to [latex]b[\/latex] can be written as [latex]{\\int }_{\\text{\u2212}a}^{a}b\\sqrt{1-\\frac{{x}^{2}}{{a}^{2}}}dx.[\/latex] Use the substitution [latex]x=a \\cos t[\/latex] to express this area in terms of an integral of a trigonometric function. You do not need to compute the integral.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571169897\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571169897\"]\r\n<p id=\"fs-id1170571169897\">[latex]{\\int }_{t=\\pi }^{0}b\\sqrt{1-{ \\cos }^{2}t}\u00d7(\\text{\u2212}a \\sin t)dt={\\int }_{t=0}^{\\pi }ab{ \\sin }^{2}tdt[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571068045\" class=\"exercise\">\r\n<div id=\"fs-id1170571068047\" class=\"textbox\">\r\n<p id=\"fs-id1170571068049\"><strong>[T]<\/strong> The following graph is of a function of the form [latex]f(t)=a \\sin (nt)+b \\sin (mt).[\/latex] Estimate the coefficients [latex]a[\/latex] and [latex]b[\/latex], and the frequency parameters [latex]n[\/latex] and [latex]m[\/latex]. Use these estimates to approximate [latex]{\\int }_{0}^{\\pi }f(t)dt.[\/latex]<\/p>\r\n<span id=\"fs-id1170571060823\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204247\/CNX_Calc_Figure_05_05_201.jpg\" alt=\"A graph of a function of the given form over [0, 2pi], which has six turning points. They are located at just before pi\/4, just after pi\/2, between 3pi\/4 and pi, between pi and 5pi\/4, just before 3pi\/2, and just after 7pi\/4 at about 3, -2, 1, -1, 2, and -3. It begins at the origin and ends at (2pi, 0). It crosses the x axis between pi\/4 and pi\/2, just before 3pi\/4, pi, just after 5pi\/4, and between 3pi\/2 and 4pi\/4.\" \/><\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571249463\" class=\"exercise\">\r\n<div id=\"fs-id1170571249465\" class=\"textbox\">\r\n<p id=\"fs-id1170571249467\"><strong>[T]<\/strong> The following graph is of a function of the form [latex]f(x)=a \\cos (nt)+b \\cos (mt).[\/latex] Estimate the coefficients [latex]a[\/latex] and [latex]b[\/latex] and the frequency parameters [latex]n[\/latex] and [latex]m[\/latex]. Use these estimates to approximate [latex]{\\int }_{0}^{\\pi }f(t)dt.[\/latex]<\/p>\r\n<span id=\"fs-id1170571249581\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204249\/CNX_Calc_Figure_05_05_202.jpg\" alt=\"The graph of a function of the given form over [0, 2pi]. It begins at (0,1) and ends at (2pi, 1). It has five turning points, located just after pi\/4, between pi\/2 and 3pi\/4, pi, between 5pi\/4 and 3pi\/2, and just before 7pi\/4 at about -1.5, 2.5, -3, 2.5, and -1. It crosses the x axis between 0 and pi\/4, just before pi\/2, just after 3pi\/4, just before 5pi\/4, just after 3pi\/2, and between 7pi\/4 and 2pi.\" \/><\/span>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573497145\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573497145\"]\r\n<p id=\"fs-id1170573497145\">[latex]f(t)=2 \\cos (3t)- \\cos (2t);{\\int }_{0}^{\\pi \\text{\/}2}(2 \\cos (3t)- \\cos (2t))=-\\frac{2}{3}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1170573497269\" class=\"definition\">\r\n \t<dt>change of variables<\/dt>\r\n \t<dd id=\"fs-id1170573497274\">the substitution of a variable, such as [latex]u[\/latex], for an expression in the integrand<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1170573497284\" class=\"definition\">\r\n \t<dt>integration by substitution<\/dt>\r\n \t<dd id=\"fs-id1170573497289\">a technique for integration that allows integration of functions that are the result of a chain-rule derivative<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Use substitution to evaluate indefinite integrals.<\/li>\n<li>Use substitution to evaluate definite integrals.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1170573367590\">The Fundamental Theorem of Calculus gave us a method to evaluate integrals without using Riemann sums. The drawback of this method, though, is that we must be able to find an antiderivative, and this is not always easy. In this section we examine a technique, called<strong> integration by substitution<\/strong>, to help us find antiderivatives. Specifically, this method helps us find antiderivatives when the integrand is the result of a chain-rule derivative.<\/p>\n<p id=\"fs-id1170573502792\">At first, the approach to the substitution procedure may not appear very obvious. However, it is primarily a visual task\u2014that is, the integrand shows you what to do; it is a matter of recognizing the form of the function. So, what are we supposed to see? We are looking for an integrand of the form [latex]f\\left[g(x)\\right]{g}^{\\prime }(x)dx.[\/latex] For example, in the integral [latex]\\int {({x}^{2}-3)}^{3}2xdx,[\/latex] we have [latex]f(x)={x}^{3},g(x)={x}^{2}-3,[\/latex] and [latex]g\\text{'}(x)=2x.[\/latex] Then,<\/p>\n<div id=\"fs-id1170573442602\" class=\"equation unnumbered\">[latex]f\\left[g(x)\\right]{g}^{\\prime }(x)={({x}^{2}-3)}^{3}(2x),[\/latex]<\/div>\n<p id=\"fs-id1170573209422\">and we see that our integrand is in the correct form.<\/p>\n<p id=\"fs-id1170573431694\">The method is called <em>substitution<\/em> because we substitute part of the integrand with the variable [latex]u[\/latex] and part of the integrand with <em>du<\/em>. It is also referred to as <strong>change of variables<\/strong> because we are changing variables to obtain an expression that is easier to work with for applying the integration rules.<\/p>\n<div id=\"fs-id1170573406868\" class=\"textbox key-takeaways theorem\">\n<h3>Substitution with Indefinite Integrals<\/h3>\n<p id=\"fs-id1170571193831\">Let [latex]u=g(x),,[\/latex] where [latex]{g}^{\\prime }(x)[\/latex] is continuous over an interval, let [latex]f(x)[\/latex] be continuous over the corresponding range of [latex]g[\/latex], and let [latex]F(x)[\/latex] be an antiderivative of [latex]f(x).[\/latex] Then,<\/p>\n<div id=\"fs-id1170573306241\" class=\"equation\">[latex]\\begin{array}{cc}\\int f\\left[g(x)\\right]{g}^{\\prime }(x)dx\\hfill & =\\int f(u)du\\hfill \\\\ & =F(u)+C\\hfill \\\\ & =F(g(x))+C.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1170573413133\" class=\"bc-section section\">\n<h1>Proof<\/h1>\n<p id=\"fs-id1170573351199\">Let [latex]f[\/latex], [latex]g[\/latex], [latex]u[\/latex], and <em>F<\/em> be as specified in the theorem. Then<\/p>\n<div id=\"fs-id1170573534008\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}\\frac{d}{dx}F(g(x))\\hfill & ={F}^{\\prime }(g(x)){g}^{\\prime }(x)\\hfill \\\\ & =f\\left[g(x)\\right]{g}^{\\prime }(x).\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170573366538\">Integrating both sides with respect to [latex]x[\/latex], we see that<\/p>\n<div id=\"fs-id1170573355959\" class=\"equation unnumbered\">[latex]\\int f\\left[g(x)\\right]{g}^{\\prime }(x)dx=F(g(x))+C.[\/latex]<\/div>\n<p id=\"fs-id1170573362143\">If we now substitute [latex]u=g(x),[\/latex] and [latex]du=g\\text{'}(x)dx,[\/latex] we get<\/p>\n<div id=\"fs-id1170573437846\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}\\int f\\left[g(x)\\right]{g}^{\\prime }(x)dx\\hfill & =\\int f(u)du\\hfill \\\\ & =F(u)+C\\hfill \\\\ & =F(g(x))+C.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170573331564\">\u25a1<\/p>\n<p id=\"fs-id1170573368234\">Returning to the problem we looked at originally, we let [latex]u={x}^{2}-3[\/latex] and then [latex]du=2xdx.[\/latex] Rewrite the integral in terms of [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170573366731\" class=\"equation unnumbered\">[latex]{\\int \\underset{u}{\\underbrace{({x}^{2}-3)}}}^{3}\\underset{du}{\\underbrace{(2xdx)}}=\\int {u}^{3}du.[\/latex]<\/div>\n<p>Using the power rule for integrals, we have<\/p>\n<div id=\"fs-id1170571257696\" class=\"equation unnumbered\">[latex]\\int {u}^{3}du=\\frac{{u}^{4}}{4}+C.[\/latex]<\/div>\n<p id=\"fs-id1170571099748\">Substitute the original expression for [latex]x[\/latex] back into the solution:<\/p>\n<div id=\"fs-id1170573410936\" class=\"equation unnumbered\">[latex]\\frac{{u}^{4}}{4}+C=\\frac{{({x}^{2}-3)}^{4}}{4}+C.[\/latex]<\/div>\n<p id=\"fs-id1170573363034\">We can generalize the procedure in the following Problem-Solving Strategy.<\/p>\n<div id=\"fs-id1170573435531\" class=\"textbox key-takeaways problem-solving\">\n<h3>Problem-Solving Strategy: Integration by Substitution<\/h3>\n<ol id=\"fs-id1170570997711\">\n<li>Look carefully at the integrand and select an expression [latex]g(x)[\/latex] within the integrand to set equal to [latex]u[\/latex]. Let\u2019s select [latex]g(x).[\/latex] such that [latex]{g}^{\\prime }(x)[\/latex] is also part of the integrand.<\/li>\n<li>Substitute [latex]u=g(x)[\/latex] and [latex]du={g}^{\\prime }(x)dx.[\/latex] into the integral.<\/li>\n<li>We should now be able to evaluate the integral with respect to [latex]u[\/latex]. If the integral can\u2019t be evaluated we need to go back and select a different expression to use as [latex]u[\/latex].<\/li>\n<li>Evaluate the integral in terms of [latex]u[\/latex].<\/li>\n<li>Write the result in terms of [latex]x[\/latex] and the expression [latex]g(x).[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1170573385785\" class=\"textbox examples\">\n<div class=\"exercise\">\n<h3>Using Substitution to Find an Antiderivative<\/h3>\n<div class=\"textbox\">\n<p id=\"fs-id1170573386292\">Use substitution to find the antiderivative of [latex]\\int 6x{(3{x}^{2}+4)}^{4}dx.[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573569711\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573569711\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573569711\">The first step is to choose an expression for [latex]u[\/latex]. We choose [latex]u=3{x}^{2}+4.[\/latex] because then [latex]du=6xdx.,[\/latex] and we already have <em>du<\/em> in the integrand. Write the integral in terms of [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170570995689\" class=\"equation unnumbered\">[latex]\\int 6x{(3{x}^{2}+4)}^{4}dx=\\int {u}^{4}du.[\/latex]<\/div>\n<p id=\"fs-id1170573398341\">Remember that <em>du<\/em> is the derivative of the expression chosen for [latex]u[\/latex], regardless of what is inside the integrand. Now we can evaluate the integral with respect to [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170571000121\" class=\"equation unnumbered\">[latex]\\begin{array}{ll}\\int {u}^{4}du\\hfill & =\\frac{{u}^{5}}{5}+C\\hfill \\\\ \\\\ \\\\ & =\\frac{{(3{x}^{2}+4)}^{5}}{5}+C.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170573208347\"><strong>Analysis<\/strong><\/p>\n<p id=\"fs-id1170573569153\">We can check our answer by taking the derivative of the result of integration. We should obtain the integrand. Picking a value for <em>C<\/em> of 1, we let [latex]y=\\frac{1}{5}{(3{x}^{2}+4)}^{5}+1.[\/latex] We have<\/p>\n<div id=\"fs-id1170573336352\" class=\"equation unnumbered\">[latex]y=\\frac{1}{5}{(3{x}^{2}+4)}^{5}+1,[\/latex]<\/div>\n<p id=\"fs-id1170573411738\">so<\/p>\n<div id=\"fs-id1170573408431\" class=\"equation unnumbered\">[latex]\\begin{array}{}\\\\ \\hfill {y}^{\\prime }& =(\\frac{1}{5})5{(3{x}^{2}+4)}^{4}6x\\hfill \\\\ & =6x{(3{x}^{2}+4)}^{4}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170573371003\">This is exactly the expression we started with inside the integrand.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1170573359652\" class=\"exercise\">\n<div id=\"fs-id1170571227254\" class=\"textbox\">\n<p id=\"fs-id1170571334089\">Use substitution to find the antiderivative of [latex]\\int 3{x}^{2}{({x}^{3}-3)}^{2}dx.[\/latex]<\/p>\n<\/div>\n<div class=\"solution\">\n<p id=\"fs-id1170573405086\">[latex]\\int 3{x}^{2}{({x}^{3}-3)}^{2}dx=\\frac{1}{3}{({x}^{3}-3)}^{3}+C[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1170571728784\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1170573396896\">Let [latex]u={x}^{3}-3.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571121618\">Sometimes we need to adjust the constants in our integral if they don\u2019t match up exactly with the expressions we are substituting.<\/p>\n<div id=\"fs-id1170571334083\" class=\"textbox examples\">\n<h3>Using Substitution with Alteration<\/h3>\n<div id=\"fs-id1170573255224\" class=\"exercise\">\n<div id=\"fs-id1170571027516\" class=\"textbox\">\n<p id=\"fs-id1170573384093\">Use substitution to find the antiderivative of [latex]\\int z\\sqrt{{z}^{2}-5}dz.[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573391210\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573391210\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573391210\">Rewrite the integral as [latex]\\int z{({z}^{2}-5)}^{1\\text{\/}2}dz.[\/latex] Let [latex]u={z}^{2}-5[\/latex] and [latex]du=2zdz.[\/latex] Now we have a problem because [latex]du=2zdz[\/latex] and the original expression has only [latex]zdz.[\/latex] We have to alter our expression for <em>du<\/em> or the integral in [latex]u[\/latex] will be twice as large as it should be. If we multiply both sides of the <em>du<\/em> equation by [latex]\\frac{1}{2}.[\/latex] we can solve this problem. Thus,<\/p>\n<div id=\"fs-id1170573430996\" class=\"equation unnumbered\">[latex]\\begin{array}{}\\\\ \\hfill u& ={z}^{2}-5\\hfill \\\\ \\hfill du& =2zdz\\hfill \\\\ \\hfill \\frac{1}{2}du& =\\frac{1}{2}(2z)dz=zdz.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170570976225\">Write the integral in terms of [latex]u[\/latex], but pull the [latex]\\frac{1}{2}[\/latex] outside the integration symbol:<\/p>\n<div id=\"fs-id1170573533836\" class=\"equation unnumbered\">[latex]\\int z{({z}^{2}-5)}^{1\\text{\/}2}dz=\\frac{1}{2}\\int {u}^{1\\text{\/}2}du.[\/latex]<\/div>\n<p id=\"fs-id1170573586341\">Integrate the expression in [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170570994415\" class=\"equation unnumbered\">[latex]\\begin{array}{}\\\\ \\frac{1}{2}\\int {u}^{1\\text{\/}2}du\\hfill & =(\\frac{1}{2})\\frac{{u}^{3\\text{\/}2}}{\\frac{3}{2}}+C\\hfill \\\\ \\\\ & =(\\frac{1}{2})(\\frac{2}{3}){u}^{3\\text{\/}2}+C\\hfill \\\\ & =\\frac{1}{3}{u}^{3\\text{\/}2}+C\\hfill \\\\ & =\\frac{1}{3}{({z}^{2}-5)}^{3\\text{\/}2}+C.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573362318\" class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1170571254579\" class=\"exercise\">\n<div id=\"fs-id1170571381835\" class=\"textbox\">\n<p id=\"fs-id1170573759886\">Use substitution to find the antiderivative of [latex]\\int {x}^{2}{({x}^{3}+5)}^{9}dx.[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573361491\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573361491\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573361491\">[latex]\\frac{{({x}^{3}+5)}^{10}}{30}+C[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1170572231096\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1170571138635\">Multiply the <em>du<\/em> equation by [latex]\\frac{1}{3}.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573573975\" class=\"textbox examples\">\n<h3>Using Substitution with Integrals of Trigonometric Functions<\/h3>\n<div class=\"exercise\">\n<div id=\"fs-id1170573333636\" class=\"textbox\">\n<p id=\"fs-id1170573385625\">Use substitution to evaluate the integral [latex]\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt.[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573437523\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573437523\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573437523\">We know the derivative of [latex]\\cos t[\/latex] is [latex]\\text{\u2212} \\sin t,[\/latex] so we set [latex]u= \\cos t.[\/latex] Then [latex]du=\\text{\u2212} \\sin tdt.[\/latex] Substituting into the integral, we have<\/p>\n<div id=\"fs-id1170573337947\" class=\"equation unnumbered\">[latex]\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt=\\text{\u2212}\\int \\frac{du}{{u}^{3}}.[\/latex]<\/div>\n<p id=\"fs-id1170573364147\">Evaluating the integral, we get<\/p>\n<div id=\"fs-id1170571136365\" class=\"equation unnumbered\">[latex]\\begin{array}{}\\\\ \\\\ \\text{\u2212}\\int \\frac{du}{{u}^{3}}\\hfill & =\\text{\u2212}\\int {u}^{-3}du\\hfill \\\\ & =\\text{\u2212}(-\\frac{1}{2}){u}^{-2}+C.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170573409376\">Putting the answer back in terms of [latex]t[\/latex], we get<\/p>\n<div id=\"fs-id1170573587515\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt\\hfill & =\\frac{1}{2{u}^{2}}+C\\hfill \\\\ \\\\ & =\\frac{1}{2{ \\cos }^{2}t}+C.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573733746\" class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1170573387947\" class=\"exercise\">\n<div id=\"fs-id1170573717717\" class=\"textbox\">\n<p id=\"fs-id1170571198076\">Use substitution to evaluate the integral [latex]\\int \\frac{ \\cos t}{{ \\sin }^{2}t}dt.[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571171177\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571171177\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571171177\">[latex]-\\frac{1}{ \\sin t}+C[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1170572333217\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1170571258771\">Use the process from <a class=\"autogenerated-content\" href=\"#fs-id1170573573975\">(Figure)<\/a> to solve the problem.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170573413569\">Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a constant. We need to eliminate all the expressions within the integrand that are in terms of the original variable. When we are done, [latex]u[\/latex] should be the only variable in the integrand. In some cases, this means solving for the original variable in terms of [latex]u[\/latex]. This technique should become clear in the next example.<\/p>\n<div id=\"fs-id1170573426799\" class=\"textbox examples\">\n<h3>Finding an Antiderivative Using [latex]u[\/latex]-Substitution<\/h3>\n<div id=\"fs-id1170573417818\" class=\"exercise\">\n<div id=\"fs-id1170573533908\" class=\"textbox\">\n<p id=\"fs-id1170573328644\">Use substitution to find the antiderivative of [latex]\\int \\frac{x}{\\sqrt{x-1}}dx.[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573435928\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573435928\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573435928\">If we let [latex]u=x-1,[\/latex] then [latex]du=dx.[\/latex] But this does not account for the [latex]x[\/latex] in the numerator of the integrand. We need to express [latex]x[\/latex] in terms of [latex]u[\/latex]. If [latex]u=x-1,[\/latex] then [latex]x=u+1.[\/latex] Now we can rewrite the integral in terms of [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170573333822\" class=\"equation unnumbered\">[latex]\\begin{array}{ll}\\int \\frac{x}{\\sqrt{x-1}}dx\\hfill & =\\int \\frac{u+1}{\\sqrt{u}}du\\hfill \\\\ \\\\ & =\\int \\sqrt{u}+\\frac{1}{\\sqrt{u}}du\\hfill \\\\ & =\\int ({u}^{1\\text{\/}2}+{u}^{-1\\text{\/}2})du.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170573544033\">Then we integrate in the usual way, replace [latex]u[\/latex] with the original expression, and factor and simplify the result. Thus,<\/p>\n<div id=\"fs-id1170570997027\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}\\int ({u}^{1\\text{\/}2}+{u}^{-1\\text{\/}2})du\\hfill & =\\frac{2}{3}{u}^{3\\text{\/}2}+2{u}^{1\\text{\/}2}+C\\hfill \\\\ \\\\ & =\\frac{2}{3}{(x-1)}^{3\\text{\/}2}+2{(x-1)}^{1\\text{\/}2}+C\\hfill \\\\ & ={(x-1)}^{1\\text{\/}2}\\left[\\frac{2}{3}(x-1)+2\\right]+C\\hfill \\\\ & ={(x-1)}^{1\\text{\/}2}(\\frac{2}{3}x-\\frac{2}{3}+\\frac{6}{3})\\hfill \\\\ & ={(x-1)}^{1\\text{\/}2}(\\frac{2}{3}x+\\frac{4}{3})\\hfill \\\\ & =\\frac{2}{3}{(x-1)}^{1\\text{\/}2}(x+2)+C.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571115891\" class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1170571283562\" class=\"exercise\">\n<div id=\"fs-id1170571189256\" class=\"textbox\">\n<p id=\"fs-id1170571087102\">Use substitution to evaluate the indefinite integral [latex]\\int { \\cos }^{3}t \\sin tdt.[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571285848\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571285848\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571285848\">[latex]-\\frac{{ \\cos }^{4}t}{4}+C[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1170572295331\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1170573501914\">Use the process from <a class=\"autogenerated-content\" href=\"#fs-id1170573426799\">(Figure)<\/a> to solve the problem.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573399047\" class=\"bc-section section\">\n<h1>Substitution for Definite Integrals<\/h1>\n<p id=\"fs-id1170573387843\">Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.<\/p>\n<div id=\"fs-id1170573497309\" class=\"textbox key-takeaways theorem\">\n<h3>Substitution with Definite Integrals<\/h3>\n<p id=\"fs-id1170570999523\">Let [latex]u=g(x)[\/latex] and let [latex]{g}^{\\text{\u2032}}[\/latex] be continuous over an interval [latex]\\left[a,b\\right],[\/latex] and let [latex]f[\/latex] be continuous over the range of [latex]u=g(x).[\/latex] Then,<\/p>\n<div id=\"fs-id1170571189407\" class=\"equation unnumbered\">[latex]{\\int }_{a}^{b}f(g(x)){g}^{\\prime }(x)dx={\\int }_{g(a)}^{g(b)}f(u)du.[\/latex]<\/div>\n<\/div>\n<p id=\"fs-id1170573502228\">Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if [latex]F(x)[\/latex] is an antiderivative of [latex]f(x),[\/latex] we have<\/p>\n<div id=\"fs-id1170571168095\" class=\"equation unnumbered\">[latex]\\int f(g(x)){g}^{\\prime }(x)dx=F(g(x))+C.[\/latex]<\/div>\n<p id=\"fs-id1170573420715\">Then<\/p>\n<div id=\"fs-id1170573388911\" class=\"equation\">[latex]\\begin{array}{cc}{\\int }_{a}^{b}f\\left[g(x)\\right]{g}^{\\prime }(x)dx\\hfill & ={F(g(x))|}_{x=a}^{x=b}\\hfill \\\\ & =F(g(b))-F(g(a))\\hfill \\\\ & ={F(u)|}_{u=g(a)}^{u=g(b)}\\hfill \\\\ \\\\ \\\\ & ={\\int }_{g(a)}^{g(b)}f(u)du,\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571100165\">and we have the desired result.<\/p>\n<div id=\"fs-id1170571215481\" class=\"textbox examples\">\n<h3>Using Substitution to Evaluate a Definite Integral<\/h3>\n<div id=\"fs-id1170571215483\" class=\"exercise\">\n<div id=\"fs-id1170573750246\" class=\"textbox\">\n<p id=\"fs-id1170571382554\">Use substitution to evaluate [latex]{\\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx.[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573440252\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573440252\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573440252\">Let [latex]u=1+2{x}^{3},[\/latex] so [latex]du=6{x}^{2}dx.[\/latex] Since the original function includes one factor of [latex]x[\/latex]<sup>2<\/sup> and [latex]du=6{x}^{2}dx,[\/latex] multiply both sides of the <em>du<\/em> equation by [latex]1\\text{\/}6.[\/latex] Then,<\/p>\n<div id=\"fs-id1170573362498\" class=\"equation unnumbered\">[latex]\\begin{array}{ccc}du\\hfill & =\\hfill & 6{x}^{2}dx\\hfill \\\\ \\frac{1}{6}du\\hfill & =\\hfill & {x}^{2}dx.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571157085\">To adjust the limits of integration, note that when [latex]x=0,u=1+2(0)=1,[\/latex] and when [latex]x=1,u=1+2(1)=3.[\/latex] Then<\/p>\n<div id=\"fs-id1170571303367\" class=\"equation unnumbered\">[latex]{\\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx=\\frac{1}{6}{\\int }_{1}^{3}{u}^{5}du.[\/latex]<\/div>\n<p id=\"fs-id1170571285285\">Evaluating this expression, we get<\/p>\n<div id=\"fs-id1170571285289\" class=\"equation unnumbered\">[latex]\\begin{array}{}\\\\ \\\\ \\frac{1}{6}{\\int }_{1}^{3}{u}^{5}du\\hfill & =(\\frac{1}{6})(\\frac{{u}^{6}}{6}){|}_{1}^{3}\\hfill \\\\ & =\\frac{1}{36}\\left[{(3)}^{6}-{(1)}^{6}\\right]\\hfill \\\\ & =\\frac{182}{9}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571101513\" class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1170571061496\" class=\"exercise\">\n<div id=\"fs-id1170571061498\" class=\"textbox\">\n<p id=\"fs-id1170571061500\">Use substitution to evaluate the definite integral [latex]{\\int }_{-1}^{0}y{(2{y}^{2}-3)}^{5}dy.[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573518898\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573518898\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573518898\">[latex]\\frac{91}{3}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1170571619018\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1170573593922\">Use the steps from <a class=\"autogenerated-content\" href=\"#fs-id1170571215481\">(Figure)<\/a> to solve the problem.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573593895\" class=\"textbox examples\">\n<h3>Using Substitution with an Exponential Function<\/h3>\n<div id=\"fs-id1170573593897\" class=\"exercise\">\n<div id=\"fs-id1170571158819\" class=\"textbox\">\n<p id=\"fs-id1170571158824\">Use substitution to evaluate [latex]{\\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx.[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571347206\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571347206\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571347206\">Let [latex]u=4{x}^{3}+3.[\/latex] Then, [latex]du=8xdx.[\/latex] To adjust the limits of integration, we note that when [latex]x=0,u=3,[\/latex] and when [latex]x=1,u=7.[\/latex] So our substitution gives<\/p>\n<div id=\"fs-id1170571052992\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}{\\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx\\hfill & =\\frac{1}{8}{\\int }_{3}^{7}{e}^{u}du\\hfill \\\\ \\\\ & =\\frac{1}{8}{e}^{u}{|}_{3}^{7}\\hfill \\\\ & =\\frac{{e}^{7}-{e}^{3}}{8}\\hfill \\\\ & \\approx 134.568.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170570976160\" class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1170571099763\" class=\"exercise\">\n<div id=\"fs-id1170571099765\" class=\"textbox\">\n<p id=\"fs-id1170571099767\">Use substitution to evaluate [latex]{\\int }_{0}^{1}{x}^{2} \\cos (\\frac{\\pi }{2}{x}^{3})dx.[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573438316\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573438316\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573438316\">[latex]\\frac{2}{3\\pi }\\approx 0.2122[\/latex]<\/p>\n<\/div>\n<div class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1170573592042\">Use the process from <a class=\"autogenerated-content\" href=\"#fs-id1170573593895\">(Figure)<\/a> to solve the problem.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571289709\">Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we can apply substitution. Also, we have the option of replacing the original expression for [latex]u[\/latex] after we find the antiderivative, which means that we do not have to change the limits of integration. These two approaches are shown in <a class=\"autogenerated-content\" href=\"#fs-id1170570994085\">(Figure)<\/a>.<\/p>\n<div id=\"fs-id1170570994085\" class=\"textbox examples\">\n<h3>Using Substitution to Evaluate a Trigonometric Integral<\/h3>\n<div id=\"fs-id1170571213222\" class=\"exercise\">\n<div id=\"fs-id1170571213224\" class=\"textbox\">\n<p id=\"fs-id1170571219853\">Use substitution to evaluate [latex]{\\int }_{0}^{\\pi \\text{\/}2}{ \\cos }^{2}\\theta d\\theta .[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571030577\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571030577\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571030577\">Let us first use a trigonometric identity to rewrite the integral. The trig identity [latex]{ \\cos }^{2}\\theta =\\frac{1+ \\cos 2\\theta }{2}[\/latex] allows us to rewrite the integral as<\/p>\n<div id=\"fs-id1170571131164\" class=\"equation unnumbered\">[latex]{\\int }_{0}^{\\pi \\text{\/}2}{ \\cos }^{2}\\theta d\\theta ={\\int }_{0}^{\\pi \\text{\/}2}\\frac{1+ \\cos 2\\theta }{2}d\\theta .[\/latex]<\/div>\n<p id=\"fs-id1170573351678\">Then,<\/p>\n<div id=\"fs-id1170573581363\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}{\\int }_{0}^{\\pi \\text{\/}2}(\\frac{1+ \\cos 2\\theta }{2})d\\theta \\hfill & ={\\int }_{0}^{\\pi \\text{\/}2}(\\frac{1}{2}+\\frac{1}{2} \\cos 2\\theta )d\\theta \\hfill \\\\ \\\\ \\\\ & =\\frac{1}{2}{\\int }_{0}^{\\pi \\text{\/}2}d\\theta +{\\int }_{0}^{\\pi \\text{\/}2} \\cos 2\\theta d\\theta .\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571374921\">We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let [latex]u=2\\theta .[\/latex] Then, [latex]du=2d\\theta ,[\/latex] or [latex]\\frac{1}{2}du=d\\theta .[\/latex] Also, when [latex]\\theta =0,u=0,[\/latex] and when [latex]\\theta =\\pi \\text{\/}2,u=\\pi .[\/latex] Expressing the second integral in terms of [latex]u[\/latex], we have<\/p>\n<div id=\"fs-id1170571244772\" class=\"equation unnumbered\">[latex]\\begin{array}{}\\\\ \\\\ \\frac{1}{2}{\\int }_{0}^{\\pi \\text{\/}2}d\\theta +\\frac{1}{2}{\\int }_{0}^{\\pi \\text{\/}2} \\cos 2\\theta d\\theta \\hfill & =\\frac{1}{2}{\\int }_{0}^{\\pi \\text{\/}2}d\\theta +\\frac{1}{2}(\\frac{1}{2}){\\int }_{0}^{\\pi } \\cos udu\\hfill \\\\ & =\\frac{\\theta }{2}{|}_{\\theta =0}^{\\theta =\\pi \\text{\/}2}+\\frac{1}{4} \\sin u{|}_{u=0}^{u=\\theta }\\hfill \\\\ & =(\\frac{\\pi }{4}-0)+(0-0)=\\frac{\\pi }{4}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571220845\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\n<ul id=\"fs-id1170571098317\">\n<li>Substitution is a technique that simplifies the integration of functions that are the result of a chain-rule derivative. The term \u2018substitution\u2019 refers to changing variables or substituting the variable [latex]u[\/latex] and <em>du<\/em> for appropriate expressions in the integrand.<\/li>\n<li>When using substitution for a definite integral, we also have to change the limits of integration.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1170573581296\" class=\"key-equations\">\n<h1>Key Equations<\/h1>\n<ul id=\"fs-id1170573390237\">\n<li><strong>Substitution with Indefinite Integrals<\/strong><br \/>\n[latex]\\int f\\left[g(x)\\right]{g}^{\\prime }(x)dx=\\int f(u)du=F(u)+C=F(g(x))+C[\/latex]<\/li>\n<li><strong>Substitution with Definite Integrals<\/strong><br \/>\n[latex]{\\int }_{a}^{b}f(g(x))g\\text{'}(x)dx={\\int }_{g(a)}^{g(b)}f(u)du[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1170571209484\" class=\"textbox exercises\">\n<div id=\"fs-id1170573549514\" class=\"exercise\">\n<div id=\"fs-id1170573549516\" class=\"textbox\">\n<p id=\"fs-id1170573549518\">Why is [latex]u[\/latex]-substitution referred to as <em>change of variable<\/em>?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573575165\" class=\"exercise\">\n<div id=\"fs-id1170573575167\" class=\"textbox\">\n<p id=\"fs-id1170573575169\">2. If [latex]f=g\\circ h,[\/latex] when reversing the chain rule, [latex]\\frac{d}{dx}(g\\circ h)(x)={g}^{\\prime }(h(x)){h}^{\\prime }(x),[\/latex] should you take [latex]u=g(x)[\/latex] or [latex]u=h(x)?[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573501482\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573501482\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573501482\">[latex]u=h(x)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571118049\">In the following exercises, verify each identity using differentiation. Then, using the indicated [latex]u[\/latex]-substitution, identify [latex]f[\/latex] such that the integral takes the form [latex]\\int f(u)du.[\/latex]<\/p>\n<div id=\"fs-id1170571033648\" class=\"exercise\">\n<div id=\"fs-id1170571033650\" class=\"textbox\">\n<p id=\"fs-id1170571033652\">[latex]\\int x\\sqrt{x+1}dx=\\frac{2}{15}{(x+1)}^{3\\text{\/}2}(3x-2)+C;u=x+1[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573718856\" class=\"exercise\">\n<div id=\"fs-id1170573718859\" class=\"textbox\">\n<p id=\"fs-id1170573750112\">[latex]\\int \\frac{{x}^{2}}{\\sqrt{x-1}}dx(x>1)=\\frac{2}{15}\\sqrt{x-1}(3{x}^{2}+4x+8)+C;u=x-1[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573398685\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573398685\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573398685\">[latex]f(u)=\\frac{{(u+1)}^{2}}{\\sqrt{u}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571303033\" class=\"exercise\">\n<div id=\"fs-id1170571303035\" class=\"textbox\">\n<p id=\"fs-id1170571248540\">[latex]\\int x\\sqrt{4{x}^{2}+9}dx=\\frac{1}{12}{(4{x}^{2}+9)}^{3\\text{\/}2}+C;u=4{x}^{2}+9[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571206963\" class=\"exercise\">\n<div id=\"fs-id1170571305052\" class=\"textbox\">\n<p id=\"fs-id1170571305054\">[latex]\\int \\frac{x}{\\sqrt{4{x}^{2}+9}}dx=\\frac{1}{4}\\sqrt{4{x}^{2}+9}+C;u=4{x}^{2}+9[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573436276\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573436276\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573436276\">[latex]du=8xdx;f(u)=\\frac{1}{8\\sqrt{u}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573759850\" class=\"exercise\">\n<div id=\"fs-id1170573759852\" class=\"textbox\">\n<p id=\"fs-id1170573759854\">[latex]\\int \\frac{x}{{(4{x}^{2}+9)}^{2}}dx=-\\frac{1}{8(4{x}^{2}+9)};u=4{x}^{2}+9[\/latex]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1170570992528\">In the following exercises, find the antiderivative using the indicated substitution.<\/p>\n<div id=\"fs-id1170571119993\" class=\"exercise\">\n<div id=\"fs-id1170571119995\" class=\"textbox\">\n<p id=\"fs-id1170571119997\">[latex]\\int {(x+1)}^{4}dx;u=x+1[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573385475\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573385475\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573385475\">[latex]\\frac{1}{5}{(x+1)}^{5}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170570996054\" class=\"exercise\">\n<div id=\"fs-id1170570996056\" class=\"textbox\">\n<p id=\"fs-id1170570996058\">[latex]\\int {(x-1)}^{5}dx;u=x-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571056458\" class=\"exercise\">\n<div id=\"fs-id1170571240500\" class=\"textbox\">\n<p id=\"fs-id1170571240502\">[latex]\\int {(2x-3)}^{-7}dx;u=2x-3[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573518184\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573518184\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573518184\">[latex]-\\frac{1}{12{(3-2x)}^{6}}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573760680\" class=\"exercise\">\n<div id=\"fs-id1170573760682\" class=\"textbox\">\n<p id=\"fs-id1170573760685\">[latex]\\int {(3x-2)}^{-11}dx;u=3x-2[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573417176\" class=\"exercise\">\n<div id=\"fs-id1170573417178\" class=\"textbox\">\n<p id=\"fs-id1170573417180\">[latex]\\int \\frac{x}{\\sqrt{{x}^{2}+1}}dx;u={x}^{2}+1[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573535993\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573535993\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573535993\">[latex]\\sqrt{{x}^{2}+1}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573542517\" class=\"exercise\">\n<div id=\"fs-id1170573542519\" class=\"textbox\">\n<p id=\"fs-id1170573542521\">[latex]\\int \\frac{x}{\\sqrt{1-{x}^{2}}}dx;u=1-{x}^{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571123643\" class=\"exercise\">\n<div id=\"fs-id1170573546495\" class=\"textbox\">\n<p id=\"fs-id1170573546497\">[latex]\\int (x-1){({x}^{2}-2x)}^{3}dx;u={x}^{2}-2x[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573726592\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573726592\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573726592\">[latex]\\frac{1}{8}{({x}^{2}-2x)}^{4}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571145634\" class=\"exercise\">\n<div id=\"fs-id1170571145636\" class=\"textbox\">\n<p id=\"fs-id1170571145638\">[latex]\\int ({x}^{2}-2x){({x}^{3}-3{x}^{2})}^{2}dx;u={x}^{3}=3{x}^{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573623732\" class=\"exercise\">\n<div id=\"fs-id1170573623734\" class=\"textbox\">\n<p id=\"fs-id1170573623736\">[latex]\\int { \\cos }^{3}\\theta d\\theta ;u= \\sin \\theta[\/latex]([latex]Hint\\text{:}{ \\cos }^{2}\\theta =1-{ \\sin }^{2}\\theta[\/latex])<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573618433\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573618433\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573618433\">[latex]\\sin \\theta -\\frac{{ \\sin }^{3}\\theta }{3}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571227343\" class=\"exercise\">\n<div id=\"fs-id1170571227346\" class=\"textbox\">\n<p id=\"fs-id1170571227348\">[latex]\\int { \\sin }^{3}\\theta d\\theta ;u= \\cos \\theta[\/latex][latex](Hint\\text{:}{ \\sin }^{2}\\theta =1-{ \\cos }^{2}\\theta)[\/latex]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571136653\">In the following exercises, use a suitable change of variables to determine the indefinite integral.<\/p>\n<div id=\"fs-id1170571136656\" class=\"exercise\">\n<div id=\"fs-id1170571136658\" class=\"textbox\">\n<p id=\"fs-id1170571136660\">[latex]\\int x{(1-x)}^{99}dx[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571098395\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571098395\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571098395\">[latex]\\frac{{(1-x)}^{101}}{101}-\\frac{{(1-x)}^{100}}{100}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571304981\" class=\"exercise\">\n<div id=\"fs-id1170571304983\" class=\"textbox\">\n<p id=\"fs-id1170571304985\">[latex]\\int t{(1-{t}^{2})}^{10}dt[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571275277\" class=\"exercise\">\n<div id=\"fs-id1170571275279\" class=\"textbox\">\n<p id=\"fs-id1170571275281\">[latex]\\int {(11x-7)}^{-3}dx[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573732555\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573732555\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573732555\">[latex]-\\frac{1}{22(7-11{x}^{2})}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573545918\" class=\"exercise\">\n<div id=\"fs-id1170573422123\" class=\"textbox\">\n<p id=\"fs-id1170573422126\">[latex]\\int {(7x-11)}^{4}dx[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573727306\" class=\"exercise\">\n<div id=\"fs-id1170573727308\" class=\"textbox\">\n<p id=\"fs-id1170573727310\">[latex]\\int { \\cos }^{3}\\theta \\sin \\theta d\\theta[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573404920\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573404920\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573404920\">[latex]-\\frac{{ \\cos }^{4}\\theta }{4}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571094884\" class=\"exercise\">\n<div id=\"fs-id1170571094886\" class=\"textbox\">\n<p id=\"fs-id1170571094888\">[latex]\\int { \\sin }^{7}\\theta \\cos \\theta d\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571158870\" class=\"exercise\">\n<div id=\"fs-id1170573726094\" class=\"textbox\">\n<p id=\"fs-id1170573726096\">[latex]\\int { \\cos }^{2}(\\pi t) \\sin (\\pi t)dt[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571227213\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571227213\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571227213\">[latex]-\\frac{{ \\cos }^{3}(\\pi t)}{3\\pi }+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573759909\" class=\"exercise\">\n<div id=\"fs-id1170573759911\" class=\"textbox\">\n<p id=\"fs-id1170573759913\">[latex]\\int { \\sin }^{2}x{ \\cos }^{3}xdx[\/latex][latex](Hint\\text{:}{ \\sin }^{2}x+{ \\cos }^{2}x=1)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571340505\" class=\"exercise\">\n<div id=\"fs-id1170571340508\" class=\"textbox\">\n<p id=\"fs-id1170571340510\">[latex]\\int t \\sin ({t}^{2}) \\cos ({t}^{2})dt[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573727339\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573727339\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573727339\">[latex]-\\frac{1}{4}\\phantom{\\rule{0.05em}{0ex}}{ \\cos }^{2}({t}^{2})+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571340409\" class=\"exercise\">\n<div id=\"fs-id1170571340411\" class=\"textbox\">\n<p id=\"fs-id1170571340413\">[latex]\\int {t}^{2}{ \\cos }^{2}({t}^{3}) \\sin ({t}^{3})dt[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571232630\" class=\"exercise\">\n<div id=\"fs-id1170571232632\" class=\"textbox\">\n<p id=\"fs-id1170571232634\">[latex]\\int \\frac{{x}^{2}}{{({x}^{3}-3)}^{2}}dx[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571158899\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571158899\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571158899\">[latex]-\\frac{1}{3({x}^{3}-3)}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170570999552\" class=\"exercise\">\n<div id=\"fs-id1170570999554\" class=\"textbox\">\n<p id=\"fs-id1170570999556\">[latex]\\int \\frac{{x}^{3}}{\\sqrt{1-{x}^{2}}}dx[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571057347\" class=\"exercise\">\n<div id=\"fs-id1170571057349\" class=\"textbox\">\n<p id=\"fs-id1170571057351\">[latex]\\int \\frac{{y}^{5}}{{(1-{y}^{3})}^{3\\text{\/}2}}dy[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573525700\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573525700\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573525700\">[latex]-\\frac{2({y}^{3}-2)}{3\\sqrt{1-{y}^{3}}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571269414\" class=\"exercise\">\n<div id=\"fs-id1170571269416\" class=\"textbox\">\n<p id=\"fs-id1170571269419\">[latex]{\\int \\cos \\theta (1- \\cos \\theta )}^{99} \\sin \\theta d\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571160788\" class=\"exercise\">\n<div id=\"fs-id1170571160790\" class=\"textbox\">\n<p id=\"fs-id1170571160792\">[latex]{\\int (1-{ \\cos }^{3}\\theta )}^{10}{ \\cos }^{2}\\theta \\sin \\theta d\\theta[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571198030\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571198030\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571198030\">[latex]\\frac{1}{33}{(1-{ \\cos }^{3}\\theta )}^{11}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571379754\" class=\"exercise\">\n<div id=\"fs-id1170571379756\" class=\"textbox\">\n<p id=\"fs-id1170571379758\">[latex]\\int ( \\cos \\theta -1){({ \\cos }^{2}\\theta -2 \\cos \\theta )}^{3} \\sin \\theta d\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571030683\" class=\"exercise\">\n<div id=\"fs-id1170571030685\" class=\"textbox\">\n<p id=\"fs-id1170571030687\">[latex]\\int ({ \\sin }^{2}\\theta -2 \\sin \\theta ){({ \\sin }^{3}\\theta -3{ \\sin }^{2}\\theta )}^{3} \\cos \\theta d\\theta[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573633894\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573633894\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573633894\">[latex]\\frac{1}{12}{({ \\sin }^{3}\\theta -3{ \\sin }^{2}\\theta )}^{4}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571114997\">In the following exercises, use a calculator to estimate the area under the curve using left Riemann sums with 50 terms, then use substitution to solve for the exact answer.<\/p>\n<div id=\"fs-id1170571115002\" class=\"exercise\">\n<div id=\"fs-id1170571115004\" class=\"textbox\">\n<p id=\"fs-id1170571115006\"><strong>[T]<\/strong>[latex]y=3{(1-x)}^{2}[\/latex] over [latex]\\left[0,2\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571158789\" class=\"exercise\">\n<div id=\"fs-id1170571158791\" class=\"textbox\">\n<p id=\"fs-id1170571158793\"><strong>[T]<\/strong>[latex]y=x{(1-{x}^{2})}^{3}[\/latex] over [latex]\\left[-1,2\\right][\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571269340\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571269340\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571269340\">[latex]{L}_{50}=-8.5779.[\/latex] The exact area is [latex]\\frac{-81}{8}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571337335\" class=\"exercise\">\n<div id=\"fs-id1170571337337\" class=\"textbox\">\n<p id=\"fs-id1170571337339\"><strong>[T]<\/strong>[latex]y= \\sin x{(1- \\cos x)}^{2}[\/latex] over [latex]\\left[0,\\pi \\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571159912\" class=\"exercise\">\n<div id=\"fs-id1170571159914\" class=\"textbox\">\n<p id=\"fs-id1170571159917\"><strong>[T]<\/strong>[latex]y=\\frac{x}{{({x}^{2}+1)}^{2}}[\/latex] over [latex]\\left[-1,1\\right][\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571057417\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571057417\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571057417\">[latex]{L}_{50}=-0.006399[\/latex] \u2026 The exact area is 0.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571057435\">In the following exercises, use a change of variables to evaluate the definite integral.<\/p>\n<div id=\"fs-id1170571057439\" class=\"exercise\">\n<div id=\"fs-id1170571057441\" class=\"textbox\">\n<p id=\"fs-id1170571057443\">[latex]{\\int }_{0}^{1}x\\sqrt{1-{x}^{2}}dx[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571026374\" class=\"exercise\">\n<div id=\"fs-id1170571026376\" class=\"textbox\">\n<p id=\"fs-id1170571026378\">[latex]{\\int }_{0}^{1}\\frac{x}{\\sqrt{1+{x}^{2}}}dx[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573618361\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573618361\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573618361\">[latex]u=1+{x}^{2},du=2xdx,\\frac{1}{2}{\\int }_{1}^{2}{u}^{-1\\text{\/}2}du=\\sqrt{2}-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571037384\" class=\"exercise\">\n<div id=\"fs-id1170571037387\" class=\"textbox\">\n<p id=\"fs-id1170571037389\">[latex]{\\int }_{0}^{2}\\frac{t}{\\sqrt{5+{t}^{2}}}dt[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571329548\" class=\"exercise\">\n<div id=\"fs-id1170571329550\" class=\"textbox\">\n<p id=\"fs-id1170571150593\">[latex]{\\int }_{0}^{1}\\frac{t}{\\sqrt{1+{t}^{3}}}dt[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571150634\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571150634\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571150634\">[latex]u=1+{t}^{3},du=3{t}^{2},\\frac{1}{3}{\\int }_{1}^{2}{u}^{-1\\text{\/}2}du=\\frac{2}{3}(\\sqrt{2}-1)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571160643\" class=\"exercise\">\n<div id=\"fs-id1170571160646\" class=\"textbox\">\n<p id=\"fs-id1170571160648\">[latex]{\\int }_{0}^{\\pi \\text{\/}4}{ \\sec }^{2}\\theta \\tan \\theta d\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571078760\" class=\"exercise\">\n<div id=\"fs-id1170571078762\" class=\"textbox\">\n<p id=\"fs-id1170571078764\">[latex]{\\int }_{0}^{\\pi \\text{\/}4}\\frac{ \\sin \\theta }{{ \\cos }^{4}\\theta }d\\theta[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573704614\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573704614\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573704614\">[latex]u= \\cos \\theta ,du=\\text{\u2212} \\sin \\theta d\\theta ,{\\int }_{1\\text{\/}\\sqrt{2}}^{1}{u}^{-4}du=\\frac{1}{3}(2\\sqrt{2}-1)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571377045\">In the following exercises, evaluate the indefinite integral [latex]\\int f(x)dx[\/latex] with constant [latex]C=0[\/latex] using [latex]u[\/latex]-substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of <em>C<\/em> that would need to be added to the antiderivative to make it equal to the definite integral [latex]F(x)={\\int }_{a}^{x}f(t)dt,[\/latex] with [latex]a[\/latex] the left endpoint of the given interval.<\/p>\n<div id=\"fs-id1170571303693\" class=\"exercise\">\n<div id=\"fs-id1170571303695\" class=\"textbox\">\n<p id=\"fs-id1170571303698\"><strong>[T]<\/strong>[latex]\\int (2x+1){e}^{{x}^{2}+x-6}dx[\/latex] over [latex]\\left[-3,2\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573762457\" class=\"exercise\">\n<div id=\"fs-id1170573762459\" class=\"textbox\">\n<p id=\"fs-id1170573762461\"><strong>[T]<\/strong>[latex]\\int \\frac{ \\cos (\\text{ln}(2x))}{x}dx[\/latex] on [latex]\\left[0,2\\right][\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573762526\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573762526\" class=\"hidden-answer\" style=\"display: none\"><span id=\"fs-id1170573762529\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204236\/CNX_Calc_Figure_05_05_204.jpg\" alt=\"Two graphs. The first shows the function f(x) = cos(ln(2x)) \/ x, which increases sharply over the approximate interval (0,.25) and then decreases gradually to the x axis. The second shows the function f(x) = sin(ln(2x)), which decreases sharply on the approximate interval (0, .25) and then increases in a gently curve into the first quadrant.\" \/><\/span><br \/>\nThe antiderivative is [latex]y= \\sin (\\text{ln}(2x)).[\/latex] Since the antiderivative is not continuous at [latex]x=0,[\/latex] one cannot find a value of <em>C<\/em> that would make [latex]y= \\sin (\\text{ln}(2x))-C[\/latex] work as a definite integral.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571021797\" class=\"exercise\">\n<div id=\"fs-id1170571021799\" class=\"textbox\">\n<p id=\"fs-id1170571021802\"><strong>[T]<\/strong>[latex]\\int \\frac{3{x}^{2}+2x+1}{\\sqrt{{x}^{3}+{x}^{2}+x+4}}dx[\/latex] over [latex]\\left[-1,2\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573542457\" class=\"exercise\">\n<div id=\"fs-id1170573542459\" class=\"textbox\">\n<p id=\"fs-id1170573542461\"><strong>[T]<\/strong>[latex]\\int \\frac{ \\sin x}{{ \\cos }^{3}x}dx[\/latex] over [latex]\\left[-\\frac{\\pi }{3},\\frac{\\pi }{3}\\right][\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571340048\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571340048\" class=\"hidden-answer\" style=\"display: none\"><span id=\"fs-id1170571340054\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204239\/CNX_Calc_Figure_05_05_206.jpg\" alt=\"Two graphs. The first is the function f(x) = sin(x) \/ cos(x)^3 over &#091;-5pi\/16, 5pi\/16&#093;. It is an increasing concave down function for values less than zero and an increasing concave up function for values greater than zero. The second is the fuction f(x) = \u00bd sec(x)^2 over the same interval. It is a wide, concave up curve which decreases for values less than zero and increases for values greater than zero.\" \/><\/span><br \/>\nThe antiderivative is [latex]y=\\frac{1}{2}\\phantom{\\rule{0.05em}{0ex}}{ \\sec }^{2}x.[\/latex] You should take [latex]C=-2[\/latex] so that [latex]F(-\\frac{\\pi }{3})=0.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571340126\" class=\"exercise\">\n<div id=\"fs-id1170571340128\" class=\"textbox\">\n<p id=\"fs-id1170571340131\"><strong>[T]<\/strong>[latex]\\int (x+2){e}^{\\text{\u2212}{x}^{2}-4x+3}dx[\/latex] over [latex]\\left[-5,1\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571098887\" class=\"exercise\">\n<div id=\"fs-id1170571098889\" class=\"textbox\">\n<p id=\"fs-id1170571098892\"><strong>[T]<\/strong>[latex]\\int 3{x}^{2}\\sqrt{2{x}^{3}+1}dx[\/latex] over [latex]\\left[0,1\\right][\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571098951\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571098951\" class=\"hidden-answer\" style=\"display: none\"><span id=\"fs-id1170571098958\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204243\/CNX_Calc_Figure_05_05_208.jpg\" alt=\"Two graphs. The first shows the function f(x) = 3x^2 * sqrt(2x^3 + 1). It is an increasing concave up curve starting at the origin. The second shows the function f(x) = 1\/3 * (2x^3 + 1)^(1\/3). It is an increasing concave up curve starting at about 0.3.\" \/><\/span><br \/>\nThe antiderivative is [latex]y=\\frac{1}{3}{(2{x}^{3}+1)}^{3\\text{\/}2}.[\/latex] One should take [latex]C=-\\frac{1}{3}.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571160883\" class=\"exercise\">\n<div id=\"fs-id1170571160885\" class=\"textbox\">\n<p id=\"fs-id1170571160887\">If [latex]h(a)=h(b)[\/latex] in [latex]{\\int }_{a}^{b}g\\text{'}(h(x))h(x)dx,[\/latex] what can you say about the value of the integral?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571160975\" class=\"exercise\">\n<div id=\"fs-id1170571160977\" class=\"textbox\">\n<p id=\"fs-id1170571160979\">Is the substitution [latex]u=1-{x}^{2}[\/latex] in the definite integral [latex]{\\int }_{0}^{2}\\frac{x}{1-{x}^{2}}dx[\/latex] okay? If not, why not?<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571291124\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571291124\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571291124\">No, because the integrand is discontinuous at [latex]x=1.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571291140\">In the following exercises, use a change of variables to show that each definite integral is equal to zero.<\/p>\n<div id=\"fs-id1170571291144\" class=\"exercise\">\n<div id=\"fs-id1170571291146\" class=\"textbox\">\n<p id=\"fs-id1170571291148\">[latex]{\\int }_{0}^{\\pi }{ \\cos }^{2}(2\\theta ) \\sin (2\\theta )d\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571158577\" class=\"exercise\">\n<div id=\"fs-id1170571158579\" class=\"textbox\">\n<p id=\"fs-id1170571158582\">[latex]{\\int }_{0}^{\\sqrt{\\pi }}t \\cos ({t}^{2}) \\sin ({t}^{2})dt[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571158639\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571158639\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571158639\">[latex]u= \\sin ({t}^{2});[\/latex] the integral becomes [latex]\\frac{1}{2}{\\int }_{0}^{0}udu.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571046470\" class=\"exercise\">\n<div id=\"fs-id1170571046472\" class=\"textbox\">\n<p id=\"fs-id1170571046474\">[latex]{\\int }_{0}^{1}(1-2t)dt[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571046569\" class=\"exercise\">\n<div id=\"fs-id1170571046572\" class=\"textbox\">\n<p id=\"fs-id1170571046574\">[latex]{\\int }_{0}^{1}\\frac{1-2t}{(1+{(t-\\frac{1}{2})}^{2})}dt[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571327399\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571327399\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571327399\">[latex]u=(1+{(t-\\frac{1}{2})}^{2});[\/latex] the integral becomes [latex]\\text{\u2212}{\\int }_{5\\text{\/}4}^{5\\text{\/}4}\\frac{1}{u}du.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571327483\" class=\"exercise\">\n<div id=\"fs-id1170571327486\" class=\"textbox\">\n<p id=\"fs-id1170571327488\">[latex]{\\int }_{0}^{\\pi } \\sin ({(t-\\frac{\\pi }{2})}^{3}) \\cos (t-\\frac{\\pi }{2})dt[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571371143\" class=\"exercise\">\n<div id=\"fs-id1170571371145\" class=\"textbox\">\n<p id=\"fs-id1170571371147\">[latex]{\\int }_{0}^{2}(1-t) \\cos (\\pi t)dt[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571050111\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571050111\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571050111\">[latex]u=1-t;[\/latex] the integral becomes<\/p>\n<p>[latex]\\begin{array}{l}{\\int }_{1}^{-1}u \\cos (\\pi (1-u))du\\hfill \\\\ ={\\int }_{1}^{-1}u\\left[ \\cos \\pi \\cos u- \\sin \\pi \\sin u\\right]du\\hfill \\\\ =\\text{\u2212}{\\int }_{1}^{-1}u \\cos udu\\hfill \\\\ ={\\int }_{-1}^{1}u \\cos udu=0\\hfill \\end{array}[\/latex]<br \/>\nsince the integrand is odd.<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571134790\" class=\"exercise\">\n<div id=\"fs-id1170571134792\" class=\"textbox\">\n<p id=\"fs-id1170571134794\">[latex]{\\int }_{\\pi \\text{\/}4}^{3\\pi \\text{\/}4}{ \\sin }^{2}t \\cos tdt[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571276371\" class=\"exercise\">\n<div id=\"fs-id1170571276373\" class=\"textbox\">\n<p id=\"fs-id1170571276375\">Show that the average value of [latex]f(x)[\/latex] over an interval [latex]\\left[a,b\\right][\/latex] is the same as the average value of [latex]f(cx)[\/latex] over the interval [latex]\\left[\\frac{a}{c},\\frac{b}{c}\\right][\/latex] for [latex]c>0.[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571276459\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571276459\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571276459\">Setting [latex]u=cx[\/latex] and [latex]du=cdx[\/latex] gets you [latex]\\frac{1}{\\frac{b}{c}-\\frac{a}{c}}{\\int }_{a\\text{\/}c}^{b\\text{\/}c}f(cx)dx=\\frac{c}{b-a}{\\int }_{u=a}^{u=b}f(u)\\frac{du}{c}=\\frac{1}{b-a}{\\int }_{a}^{b}f(u)du.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571273653\" class=\"exercise\">\n<div id=\"fs-id1170571273655\" class=\"textbox\">\n<p id=\"fs-id1170571273657\">Find the area under the graph of [latex]f(t)=\\frac{t}{{(1+{t}^{2})}^{a}}[\/latex] between [latex]t=0[\/latex] and [latex]t=x[\/latex] where [latex]a>0[\/latex] and [latex]a\\ne 1[\/latex] is fixed, and evaluate the limit as [latex]x\\to \\infty .[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571132654\" class=\"exercise\">\n<div id=\"fs-id1170571132656\" class=\"textbox\">\n<p id=\"fs-id1170571132658\">Find the area under the graph of [latex]g(t)=\\frac{t}{{(1-{t}^{2})}^{a}}[\/latex] between [latex]t=0[\/latex] and [latex]t=x,[\/latex] where [latex]0<x<1[\/latex] and [latex]a>0[\/latex] is fixed. Evaluate the limit as [latex]x\\to 1.[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571245712\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571245712\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571245712\">[latex]{\\int }_{0}^{x}g(t)dt=\\frac{1}{2}{\\int }_{u=1-{x}^{2}}^{1}\\frac{du}{{u}^{a}}=\\frac{1}{2(1-a)}{u}^{1-a}{|}_{u=1-{x}^{2}}^{1}=\\frac{1}{2(1-a)}(1-{(1-{x}^{2})}^{1-a}).[\/latex] As [latex]x\\to 1[\/latex] the limit is [latex]\\frac{1}{2(1-a)}[\/latex] if [latex]a<1,[\/latex] and the limit diverges to +\u221e if [latex]a>1.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573713777\" class=\"exercise\">\n<div id=\"fs-id1170573713779\" class=\"textbox\">\n<p id=\"fs-id1170573713781\">The area of a semicircle of radius 1 can be expressed as [latex]{\\int }_{-1}^{1}\\sqrt{1-{x}^{2}}dx.[\/latex] Use the substitution [latex]x= \\cos t[\/latex] to express the area of a semicircle as the integral of a trigonometric function. You do not need to compute the integral.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571169773\" class=\"exercise\">\n<div id=\"fs-id1170571169775\" class=\"textbox\">\n<p id=\"fs-id1170571169777\">The area of the top half of an ellipse with a major axis that is the [latex]x[\/latex]-axis from [latex]x=-1[\/latex] to [latex]a[\/latex] and with a minor axis that is the [latex]y[\/latex]-axis from [latex]y=\\text{\u2212}b[\/latex] to [latex]b[\/latex] can be written as [latex]{\\int }_{\\text{\u2212}a}^{a}b\\sqrt{1-\\frac{{x}^{2}}{{a}^{2}}}dx.[\/latex] Use the substitution [latex]x=a \\cos t[\/latex] to express this area in terms of an integral of a trigonometric function. You do not need to compute the integral.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571169897\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571169897\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571169897\">[latex]{\\int }_{t=\\pi }^{0}b\\sqrt{1-{ \\cos }^{2}t}\u00d7(\\text{\u2212}a \\sin t)dt={\\int }_{t=0}^{\\pi }ab{ \\sin }^{2}tdt[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571068045\" class=\"exercise\">\n<div id=\"fs-id1170571068047\" class=\"textbox\">\n<p id=\"fs-id1170571068049\"><strong>[T]<\/strong> The following graph is of a function of the form [latex]f(t)=a \\sin (nt)+b \\sin (mt).[\/latex] Estimate the coefficients [latex]a[\/latex] and [latex]b[\/latex], and the frequency parameters [latex]n[\/latex] and [latex]m[\/latex]. Use these estimates to approximate [latex]{\\int }_{0}^{\\pi }f(t)dt.[\/latex]<\/p>\n<p><span id=\"fs-id1170571060823\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204247\/CNX_Calc_Figure_05_05_201.jpg\" alt=\"A graph of a function of the given form over [0, 2pi], which has six turning points. They are located at just before pi\/4, just after pi\/2, between 3pi\/4 and pi, between pi and 5pi\/4, just before 3pi\/2, and just after 7pi\/4 at about 3, -2, 1, -1, 2, and -3. It begins at the origin and ends at (2pi, 0). It crosses the x axis between pi\/4 and pi\/2, just before 3pi\/4, pi, just after 5pi\/4, and between 3pi\/2 and 4pi\/4.\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571249463\" class=\"exercise\">\n<div id=\"fs-id1170571249465\" class=\"textbox\">\n<p id=\"fs-id1170571249467\"><strong>[T]<\/strong> The following graph is of a function of the form [latex]f(x)=a \\cos (nt)+b \\cos (mt).[\/latex] Estimate the coefficients [latex]a[\/latex] and [latex]b[\/latex] and the frequency parameters [latex]n[\/latex] and [latex]m[\/latex]. Use these estimates to approximate [latex]{\\int }_{0}^{\\pi }f(t)dt.[\/latex]<\/p>\n<p><span id=\"fs-id1170571249581\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204249\/CNX_Calc_Figure_05_05_202.jpg\" alt=\"The graph of a function of the given form over [0, 2pi]. It begins at (0,1) and ends at (2pi, 1). It has five turning points, located just after pi\/4, between pi\/2 and 3pi\/4, pi, between 5pi\/4 and 3pi\/2, and just before 7pi\/4 at about -1.5, 2.5, -3, 2.5, and -1. It crosses the x axis between 0 and pi\/4, just before pi\/2, just after 3pi\/4, just before 5pi\/4, just after 3pi\/2, and between 7pi\/4 and 2pi.\" \/><\/span><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573497145\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573497145\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573497145\">[latex]f(t)=2 \\cos (3t)- \\cos (2t);{\\int }_{0}^{\\pi \\text{\/}2}(2 \\cos (3t)- \\cos (2t))=-\\frac{2}{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1170573497269\" class=\"definition\">\n<dt>change of variables<\/dt>\n<dd id=\"fs-id1170573497274\">the substitution of a variable, such as [latex]u[\/latex], for an expression in the integrand<\/dd>\n<\/dl>\n<dl id=\"fs-id1170573497284\" class=\"definition\">\n<dt>integration by substitution<\/dt>\n<dd id=\"fs-id1170573497289\">a technique for integration that allows integration of functions that are the result of a chain-rule derivative<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":311,"menu_order":6,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1764","chapter","type-chapter","status-publish","hentry"],"part":1684,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapters\/1764","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapters\/1764\/revisions"}],"predecessor-version":[{"id":2453,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapters\/1764\/revisions\/2453"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/parts\/1684"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapters\/1764\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/wp\/v2\/media?parent=1764"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapter-type?post=1764"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/wp\/v2\/contributor?post=1764"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/wp\/v2\/license?post=1764"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}