{"id":801,"date":"2021-06-02T23:29:01","date_gmt":"2021-06-02T23:29:01","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=801"},"modified":"2022-03-19T03:58:23","modified_gmt":"2022-03-19T03:58:23","slug":"integration-by-parts-formula","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/integration-by-parts-formula\/","title":{"raw":"Integration-by-Parts Formula","rendered":"Integration-by-Parts Formula"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Recognize when to use integration by parts.<\/li>\r\n \t<li>Use the integration-by-parts formula to solve integration problems.<\/li>\r\n \t<li>Use the integration-by-parts formula for definite integrals.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165041979164\">If, [latex]h\\left(x\\right)=f\\left(x\\right)g\\left(x\\right)[\/latex], then by using the product rule, we obtain [latex]{h}^{\\prime }\\left(x\\right)={f}^{\\prime }\\left(x\\right)g\\left(x\\right)+{g}^{\\prime }\\left(x\\right)f\\left(x\\right)[\/latex]. Although at first it may seem counterproductive, let\u2019s now integrate both sides of this equation: [latex]\\displaystyle\\int {h}^{\\prime }\\left(x\\right)dx=\\displaystyle\\int \\left(g\\left(x\\right){f}^{\\prime }\\left(x\\right)+f\\left(x\\right){g}^{\\prime }\\left(x\\right)\\right)dx[\/latex].<\/p>\r\n<p id=\"fs-id1165042262165\">This gives us<\/p>\r\n\r\n<div id=\"fs-id1165042041848\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]h\\left(x\\right)=f\\left(x\\right)g\\left(x\\right)=\\displaystyle\\int g\\left(x\\right){f}^{\\prime }\\left(x\\right)dx+\\displaystyle\\int f\\left(x\\right){g}^{\\prime }\\left(x\\right)dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041955261\">Now we solve for [latex]\\displaystyle\\int f\\left(x\\right){g}^{\\prime }\\left(x\\right)dx:[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165042235539\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int f\\left(x\\right){g}^{\\prime }\\left(x\\right)dx=f\\left(x\\right)g\\left(x\\right)-\\displaystyle\\int g\\left(x\\right){f}^{\\prime }\\left(x\\right)dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042135462\">By making the substitutions [latex]u=f\\left(x\\right)[\/latex] and [latex]v=g\\left(x\\right)[\/latex], which in turn make [latex]du={f}^{\\prime }\\left(x\\right)dx[\/latex] and [latex]dv={g}^{\\prime }\\left(x\\right)dx[\/latex], we have the more compact form<\/p>\r\n\r\n<div id=\"fs-id1165041813626\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int udv=uv-\\displaystyle\\int vdu[\/latex].<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1165042235552\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<div id=\"fs-id1165042235552\" class=\"theorem\" data-type=\"note\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Integration by Parts<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165042089614\">Let [latex]u=f\\left(x\\right)[\/latex] and [latex]v=g\\left(x\\right)[\/latex] be functions with continuous derivatives. Then, the integration-by-parts formula for the integral involving these two functions is:<\/p>\r\n\r\n<div id=\"fs-id1165042232372\" class=\"numbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int udv=uv-\\displaystyle\\int vdu[\/latex].<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165041841895\"><\/p>\r\n\r\n<\/div>\r\n<span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">The advantage of using the integration-by-parts formula is that we can use it to exchange one integral for another, possibly easier, integral. The following example illustrates its use.<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165041832973\" data-type=\"example\">\r\n<div id=\"fs-id1165041816966\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042126072\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Integration by Parts<\/h3>\r\n<div id=\"fs-id1165042126072\" data-type=\"problem\">\r\n<p id=\"fs-id1165041832755\">Use integration by parts with [latex]u=x[\/latex] and [latex]dv=\\sin{x}dx[\/latex] to evaluate [latex]\\displaystyle\\int x\\sin{x}dx[\/latex].<\/p>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n\r\n<strong>Solution<\/strong>\r\n<div id=\"fs-id1165041792874\" data-type=\"solution\">\r\n<p id=\"fs-id1165042089552\">By choosing [latex]u=x[\/latex], we have [latex]du=1dx[\/latex]. Since [latex]dv=\\sin{x}dx[\/latex], we get [latex]v=\\displaystyle\\int \\sin{x}dx=\\text{-}\\cos{x}[\/latex]. It is handy to keep track of these values as follows:<\/p>\r\n\r\n<div id=\"fs-id1165041792867\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\hfill u&amp; =\\hfill &amp; x\\hfill &amp; &amp; \\hfill dv&amp; =\\hfill &amp; \\sin{x}dx\\hfill \\\\ \\hfill du&amp; =\\hfill &amp; 1dx\\hfill &amp; &amp; \\hfill v&amp; =\\hfill &amp; \\displaystyle\\int \\sin{x}dx=\\text{-}\\cos{x}. \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040681626\">Applying the integration-by-parts formula results in<\/p>\r\n\r\n<div id=\"fs-id1165041985992\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccc}\\hfill {\\displaystyle\\int x\\sin{x}dx}&amp; =\\left(x\\right)\\left(\\text{-}\\cos{x}\\right)-{\\displaystyle\\int \\left(\\text{-}\\cos{x}\\right)\\left(1dx\\right)}\\hfill &amp; &amp; \\text{Substitute.}\\hfill \\\\ &amp; =\\text{-}x\\cos{x}+{\\displaystyle\\int \\cos{x}dx}\\hfill &amp; &amp; \\text{Simplify.}\\hfill \\\\ &amp; =\\text{-}x\\cos{x}+\\sin{x}+C.\\hfill &amp; &amp; \\text{Use}{\\displaystyle\\int \\cos{x}dx=\\sin{x}+C.}\\end{array}[\/latex]<\/div>\r\n<div id=\"fs-id1165041893024\" data-type=\"commentary\">\r\n<div data-type=\"title\"><\/div>\r\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\r\n<div data-type=\"title\"><\/div>\r\n<p id=\"fs-id1165041973005\">At this point, there are probably a few items that need clarification. First of all, you may be curious about what would have happened if we had chosen [latex]u=\\sin{x}[\/latex] and [latex]dv=x[\/latex]. If we had done so, then we would have [latex]du=\\cos{x}[\/latex] and [latex]v=\\frac{1}{2}{x}^{2}[\/latex]. Thus, after applying integration by parts, we have [latex]{\\displaystyle\\int }^{\\text{ }}x\\sin{x}dx=\\frac{1}{2}{x}^{2}\\sin{x}-{\\displaystyle\\int }^{\\text{ }}\\frac{1}{2}{x}^{2}\\cos{x}dx[\/latex]. Unfortunately, with the new integral, we are in no better position than before. It is important to keep in mind that when we apply integration by parts, we may need to try several choices for [latex]u[\/latex] and [latex]dv[\/latex] before finding a choice that works.<\/p>\r\n<p id=\"fs-id1165042098353\">Second, you may wonder why, when we find [latex]v={\\displaystyle\\int }^{\\text{ }}\\sin{x}dx=\\text{-}\\cos{x}[\/latex], we do not use [latex]v=\\text{-}\\cos{x}+K[\/latex]. To see that it makes no difference, we can rework the problem using [latex]v=\\text{-}\\cos{x}+K\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165042094536\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{\\displaystyle\\int}x\\sin{x}dx\\hfill &amp; =\\left(x\\right)\\left(\\text{-}\\cos{x}+K\\right)-{\\displaystyle\\int}\\left(\\text{-}\\cos{x}+K\\right)\\left(1dx\\right)\\hfill \\\\ \\hfill &amp; =\\text{-}x\\cos{x}+Kx+{\\displaystyle\\int}\\cos{x}dx-{\\displaystyle\\int}Kdx\\hfill \\\\ \\hfill &amp; =\\text{-}x\\cos{x}+Kx+\\sin{x}-Kx+C\\hfill \\\\ \\hfill &amp; =\\text{-}x\\cos{x}+\\sin{x}+C.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042136113\">As you can see, it makes no difference in the final solution.<\/p>\r\n<p id=\"fs-id1165041759254\">Last, we can check to make sure that our antiderivative is correct by differentiating [latex]\\text{-}x\\cos{x}+\\sin{x}+C\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165041952373\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\frac{d}{dx}\\left(\\text{-}x\\cos{x}+\\sin{x}+C\\right)\\hfill &amp; =\\left(-1\\right)\\cos{x}+\\left(\\text{-}x\\right)\\left(\\text{-}\\sin{x}\\right)+\\cos{x}\\hfill \\\\ \\hfill &amp; =x\\sin{x}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040665697\">Therefore, the antiderivative checks out.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Interactive<\/h3>\r\nWatch <a href=\"http:\/\/www.sosmath.com\/calculus\/integration\/byparts\/byparts.html\" target=\"_blank\" rel=\"noopener\">this video for an introduction to integration by parts<\/a> and visit <a href=\"http:\/\/www.sosmath.com\/calculus\/integration\/byparts\/byparts.html\" target=\"_blank\" rel=\"noopener\">this website for examples of integration by parts<\/a>.\r\n\r\n<\/div>\r\n<div id=\"fs-id1165042127896\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1165042127899\" data-type=\"exercise\">\r\n<div id=\"fs-id1165040747248\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<div id=\"fs-id1165040747248\" data-type=\"problem\">\r\n<p id=\"fs-id1165040747250\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}x{e}^{2x}dx[\/latex] using the integration-by-parts formula with [latex]u=x[\/latex] and [latex]dv={e}^{2x}dx[\/latex].<\/p>\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1165042051378\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165040681938\">Find [latex]du[\/latex] and [latex]v[\/latex], and use the previous example as a guide.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1165040642767\" data-type=\"solution\">\r\n<p id=\"fs-id1165043352090\">[latex]{\\displaystyle\\int }^{\\text{ }}x{e}^{2x}dx=\\frac{1}{2}x{e}^{2x}-\\frac{1}{4}{e}^{2x}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lpmf-HAmaEU?controls=0&amp;start=305&amp;end=430&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.1IntegrationByParts305to430_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.1 Integration by Parts\" here (opens in new window)<\/a>.\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">The natural question to ask at this point is: How do we know how to choose [latex]u[\/latex] and [latex]dv?[\/latex] Sometimes it is a matter of trial and error; however, the acronym LIATE can often help to take some of the guesswork out of our choices. This acronym stands for <\/span><strong style=\"font-size: 1rem; text-align: initial;\" data-effect=\"bold\">L<\/strong><span style=\"font-size: 1rem; text-align: initial;\">ogarithmic Functions, <\/span><strong style=\"font-size: 1rem; text-align: initial;\" data-effect=\"bold\">I<\/strong><span style=\"font-size: 1rem; text-align: initial;\">nverse Trigonometric Functions, <\/span><strong style=\"font-size: 1rem; text-align: initial;\" data-effect=\"bold\">A<\/strong><span style=\"font-size: 1rem; text-align: initial;\">lgebraic Functions, <\/span><strong style=\"font-size: 1rem; text-align: initial;\" data-effect=\"bold\">T<\/strong><span style=\"font-size: 1rem; text-align: initial;\">rigonometric Functions, and <\/span><strong style=\"font-size: 1rem; text-align: initial;\" data-effect=\"bold\">E<\/strong><span style=\"font-size: 1rem; text-align: initial;\">xponential Functions. This mnemonic serves as an aid in determining an appropriate choice for [latex]u[\/latex].<\/span>\r\n<p id=\"fs-id1165040744404\">The type of function in the integral that appears first in the list should be our first choice of [latex]u[\/latex]. For example, if an integral contains a <span class=\"no-emphasis\" data-type=\"term\">logarithmic function<\/span> and an <span class=\"no-emphasis\" data-type=\"term\">algebraic function<\/span>, we should choose [latex]u[\/latex] to be the logarithmic function, because L comes before A in LIATE. The integral in the previous example has a trigonometric function [latex]\\text{(}\\sin{x}\\text{)}[\/latex] and an algebraic function [latex]\\left(x\\right)[\/latex]. Because A comes before T in LIATE, we chose [latex]u[\/latex] to be the algebraic function. When we have chosen [latex]u[\/latex], [latex]dv[\/latex] is selected to be the remaining part of the function to be integrated, together with [latex]dx[\/latex].<\/p>\r\n<p id=\"fs-id1165040676764\">Why does this mnemonic work? Remember that whatever we pick to be [latex]dv[\/latex] must be something we can integrate. Since we do not have integration formulas that allow us to integrate simple logarithmic functions and inverse trigonometric functions, it makes sense that they should not be chosen as values for [latex]dv[\/latex]. Consequently, they should be at the head of the list as choices for [latex]u[\/latex]. Thus, we put LI at the beginning of the mnemonic. (We could just as easily have started with IL, since these two types of functions won\u2019t appear together in an integration-by-parts problem.) The exponential and trigonometric functions are at the end of our list because they are fairly easy to integrate and make good choices for [latex]dv[\/latex]. Thus, we have TE at the end of our mnemonic. (We could just as easily have used ET at the end, since when these types of functions appear together it usually doesn\u2019t really matter which one is [latex]u[\/latex] and which one is [latex]dv[\/latex].) Algebraic functions are generally easy both to integrate and to differentiate, and they come in the middle of the mnemonic.<\/p>\r\nTo use the by-parts technique successfully, it is helpful to first review the derivative rules of several familiar transcendental functions.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Derivative rules for transcendental functions<\/h3>\r\n<ol id=\"fs-id1170572169684\" style=\"list-style-type: decimal;\">\r\n \t<li>[latex] \\frac{d}{dx} (\\sin x) = \\cos x [\/latex]<\/li>\r\n \t<li>[latex] \\frac{d}{dx} (\\cos x) = -\\sin x [\/latex]<\/li>\r\n \t<li>[latex] \\frac{d}{dx} (\\ln x) = \\frac{1}{x} [\/latex]<\/li>\r\n \t<li>[latex] \\frac{d}{dx} (\\arcsin x) = \\frac{1}{\\sqrt{1-x^2}} [\/latex]<\/li>\r\n \t<li>[latex] \\frac{d}{dx} (\\arctan x) = \\frac{1}{1+x^2} [\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1165042015633\" data-type=\"example\">\r\n<div id=\"fs-id1165042015635\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042231844\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>example: Using integration by parts<\/h3>\r\n<div id=\"fs-id1165042231844\" data-type=\"problem\">\r\n<p id=\"fs-id1165042231849\">Evaluate [latex]\\displaystyle\\int \\frac{\\text{ln}x}{{x}^{3}}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1165042042007\" data-type=\"solution\">\r\n<p id=\"fs-id1165042042010\">Begin by rewriting the integral:<\/p>\r\n\r\n<div id=\"fs-id1165042008468\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{\\text{ln}x}{{x}^{3}}dx={\\displaystyle\\int }^{\\text{ }}{x}^{-3}\\text{ln}xdx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042127807\">Since this integral contains the algebraic function [latex]{x}^{-3}[\/latex] and the logarithmic function [latex]\\text{ln}x[\/latex], choose [latex]u=\\text{ln}x[\/latex], since L comes before A in LIATE. After we have chosen [latex]u=\\text{ln}x[\/latex], we must choose [latex]dv={x}^{-3}dx[\/latex].<\/p>\r\n<p id=\"fs-id1165042127784\">Next, since [latex]u=\\text{ln}x[\/latex], we have [latex]du=\\frac{1}{x}dx[\/latex]. Also, [latex]v={\\displaystyle\\int }^{\\text{ }}{x}^{-3}dx=-\\frac{1}{2}{x}^{-2}[\/latex]. Summarizing,<\/p>\r\n\r\n<div id=\"fs-id1165042109999\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\hfill u&amp; =\\hfill &amp; \\text{ln}x\\hfill &amp; &amp; \\hfill dv&amp; =\\hfill &amp; {x}^{-3}dx\\hfill \\\\ \\hfill du&amp; =\\hfill &amp; \\frac{1}{x}dx\\hfill &amp; &amp; \\hfill v&amp; =\\hfill &amp; {\\displaystyle\\int }^{\\text{ }}{x}^{-3}dx=-\\frac{1}{2}{x}^{-2}.\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042128802\">Substituting into the integration-by-parts formula gives<\/p>\r\n\r\n<div id=\"fs-id1165042128805\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill \\displaystyle\\int \\frac{\\text{ln}x}{{x}^{3}}dx&amp; ={\\displaystyle\\int }^{\\text{ }}{x}^{-3}\\text{ln}xdx=\\left(\\text{ln}x\\right)\\left(\\text{-}\\frac{1}{2}{x}^{-2}\\right)-{\\displaystyle\\int }^{\\text{ }}\\left(\\text{-}\\frac{1}{2}{x}^{-2}\\right)\\left(\\frac{1}{x}dx\\right)\\hfill &amp; &amp; &amp; \\\\ &amp; =-\\frac{1}{2}{x}^{-2}\\text{ln}x+{\\displaystyle\\int }^{\\text{ }}\\frac{1}{2}{x}^{-3}dx\\hfill &amp; &amp; &amp; \\text{Simplify}.\\hfill \\\\ &amp; =-\\frac{1}{2}{x}^{-2}\\text{ln}x-\\frac{1}{4}{x}^{-2}+C\\hfill &amp; &amp; &amp; \\text{Integrate}.\\hfill \\\\ &amp; =-\\frac{1}{2{x}^{2}}\\text{ln}x-\\frac{1}{4{x}^{2}}+C.\\hfill &amp; &amp; &amp; \\text{Rewrite with positive integers.}\\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042127888\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1165041899921\" data-type=\"exercise\">\r\n<div id=\"fs-id1165041899924\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1165041899924\" data-type=\"problem\">\r\n<p id=\"fs-id1165041899926\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}x\\text{ln}xdx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558895\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1165041864895\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165041864902\">Use [latex]u=\\text{ln}x[\/latex] and [latex]dv=xdx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558894\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1165040747233\" data-type=\"solution\">\r\n<p id=\"fs-id1165043331164\">[latex]\\frac{1}{2}{x}^{2}\\text{ln}x-\\frac{1}{4}{x}^{2}+C[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lpmf-HAmaEU?controls=0&amp;start=558&amp;end=647&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.1IntegrationByParts558to647_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.1 Integration by Parts\" here (opens in new window)<\/a>.\r\n<p id=\"fs-id1165041948188\">In some cases, as in the next two examples, it may be necessary to apply integration by parts more than once.<\/p>\r\n\r\n<div id=\"fs-id1165041948192\" data-type=\"example\">\r\n<div id=\"fs-id1165041948195\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042015541\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>example: Applying integration by parts more than once<\/h3>\r\n<div id=\"fs-id1165042015541\" data-type=\"problem\">\r\n<p id=\"fs-id1165042015546\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{x}^{2}{e}^{3x}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n<div id=\"fs-id1165040637308\" data-type=\"solution\">\r\n<p id=\"fs-id1165040637310\">Using LIATE, choose [latex]u={x}^{2}[\/latex] and [latex]dv={e}^{3x}dx[\/latex]. Thus, [latex]du=2xdx[\/latex] and [latex]v=\\displaystyle\\int {e}^{3x}dx=\\left(\\frac{1}{3}\\right){e}^{3x}[\/latex]. Therefore,<\/p>\r\n\r\n<div id=\"fs-id1165040740033\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\hfill u&amp; =\\hfill &amp; {x}^{2}\\hfill &amp; &amp; \\hfill dv&amp; =\\hfill &amp; {e}^{3x}dx\\hfill \\\\ \\hfill du&amp; =\\hfill &amp; 2xdx\\hfill &amp; &amp; \\hfill v&amp; =\\hfill &amp; \\displaystyle\\int {e}^{3x}dx=\\frac{1}{3}{e}^{3x}.\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042240114\">Substituting into the integration-by-parts formula\u00a0produces<\/p>\r\n\r\n<div id=\"fs-id1165042240117\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int {x}^{2}{e}^{3x}dx=\\frac{1}{3}{x}^{2}{e}^{3x}-\\displaystyle\\int \\frac{2}{3}x{e}^{3x}dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040756177\">We still cannot integrate [latex]\\displaystyle\\int \\frac{2}{3}x{e}^{3x}dx[\/latex] directly, but the integral now has a lower power on [latex]x[\/latex]. We can evaluate this new integral by using integration by parts again. To do this, choose [latex]u=x[\/latex] and [latex]dv=\\frac{2}{3}{e}^{3x}dx[\/latex]. Thus, [latex]du=dx[\/latex] and [latex]v=\\displaystyle\\int \\left(\\frac{2}{3}\\right){e}^{3x}dx=\\left(\\frac{2}{9}\\right){e}^{3x}[\/latex]. Now we have<\/p>\r\n\r\n<div id=\"fs-id1165042002748\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\hfill u&amp; =\\hfill &amp; x\\hfill &amp; &amp; \\hfill dv&amp; =\\hfill &amp; \\frac{2}{3}{e}^{3x}dx\\hfill \\\\ \\hfill du&amp; =\\hfill &amp; dx\\hfill &amp; &amp; \\hfill v&amp; =\\hfill &amp; \\displaystyle\\int \\frac{2}{3}{e}^{3x}dx=\\frac{2}{9}{e}^{3x}.\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040757531\">Substituting back into the previous equation yields<\/p>\r\n\r\n<div id=\"fs-id1165040757534\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}{x}^{2}{e}^{3x}dx=\\frac{1}{3}{x}^{2}{e}^{3x}-\\left(\\frac{2}{9}x{e}^{3x}-{\\displaystyle\\int }^{\\text{ }}\\frac{2}{9}{e}^{3x}dx\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042110524\">After evaluating the last integral and simplifying, we obtain<\/p>\r\n\r\n<div id=\"fs-id1165042110527\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int {x}^{2}{e}^{3x}dx=\\frac{1}{3}{x}^{2}{e}^{3x}-\\frac{2}{9}x{e}^{3x}+\\frac{2}{27}{e}^{3x}+C[\/latex].<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042132627\" data-type=\"example\">\r\n<div id=\"fs-id1165042132629\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042132631\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Applying Integration by Parts When LIATE Doesn\u2019t Quite Work<\/h3>\r\n<div id=\"fs-id1165042132631\" data-type=\"problem\">\r\n<p id=\"fs-id1165042132636\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{t}^{3}{e}^{{t}^{2}}dt[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n<div id=\"fs-id1165042132671\" data-type=\"solution\">\r\n<p id=\"fs-id1165042132674\">If we use a strict interpretation of the mnemonic LIATE to make our choice of [latex]u[\/latex], we end up with [latex]u={t}^{3}[\/latex] and [latex]dv={e}^{{t}^{2}}dt[\/latex]. Unfortunately, this choice won\u2019t work because we are unable to evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{e}^{{t}^{2}}dt[\/latex]. However, since we can evaluate [latex]{\\displaystyle\\int }^{\\text{ }}t{e}^{{t}^{2}}dx[\/latex], we can try choosing [latex]u={t}^{2}[\/latex] and [latex]dv=t{e}^{{t}^{2}}dt[\/latex]. With these choices we have<\/p>\r\n\r\n<div id=\"fs-id1165040715819\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\hfill u&amp; =\\hfill &amp; {t}^{2}\\hfill &amp; &amp; \\hfill dv&amp; =\\hfill &amp; t{e}^{{t}^{2}}dt\\hfill \\\\ \\hfill du&amp; =\\hfill &amp; 2tdt\\hfill &amp; &amp; \\hfill v&amp; =\\hfill &amp; {\\displaystyle\\int }^{\\text{ }}t{e}^{{t}^{2}}dt=\\frac{1}{2}{e}^{{t}^{2}}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042013342\">Thus, we obtain<\/p>\r\n\r\n<div id=\"fs-id1165042013345\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{\\displaystyle\\int {t}^{3}{e}^{{t}^{2}}dt}\\hfill &amp; =\\frac{1}{2}{t}^{2}{e}^{{t}^{2}}-{\\displaystyle\\int \\frac{1}{2}{e}^{{t}^{2}}2tdt}\\hfill \\\\ \\hfill &amp; =\\frac{1}{2}{t}^{2}{e}^{{t}^{2}}-\\frac{1}{2}{e}^{{t}^{2}}+C.\\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042035446\" data-type=\"example\">\r\n<div id=\"fs-id1165042035448\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042035450\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Applying Integration by Parts More Than Once<\/h3>\r\n<div id=\"fs-id1165042035450\" data-type=\"problem\">\r\n<p id=\"fs-id1165042035456\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558891\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n\r\n<strong>Solution<\/strong>\r\n<div id=\"fs-id1165042035489\" data-type=\"solution\">\r\n<p id=\"fs-id1165042035491\">This integral appears to have only one function\u2014namely, [latex]\\sin\\left(\\text{ln}x\\right)[\/latex] \u2014however, we can always use the constant function 1 as the other function. In this example, let\u2019s choose [latex]u=\\sin\\left(\\text{ln}x\\right)[\/latex] and [latex]dv=1dx[\/latex]. (The decision to use [latex]u=\\sin\\left(\\text{ln}x\\right)[\/latex] is easy. We can\u2019t choose [latex]dv=\\sin\\left(\\text{ln}x\\right)dx[\/latex] because if we could integrate it, we wouldn\u2019t be using integration by parts in the first place!) Consequently, [latex]du=\\frac{1}{x}\\cos\\left(\\text{ln}x\\right)dx[\/latex] and [latex]v={\\displaystyle\\int }^{\\text{ }}1dx=x[\/latex]. After applying integration by parts to the integral and simplifying, we have<\/p>\r\n\r\n<div id=\"fs-id1165042257200\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx=x\\sin\\left(\\text{ln}x\\right)-{\\displaystyle\\int }^{\\text{ }}\\cos\\left(\\text{ln}x\\right)dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042110018\">Unfortunately, this process leaves us with a new integral that is very similar to the original. However, let\u2019s see what happens when we apply integration by parts again. This time let\u2019s choose [latex]u=\\cos\\left(\\text{ln}x\\right)[\/latex] and [latex]dv=1dx[\/latex], making [latex]du=\\text{-}\\frac{1}{x}\\sin\\left(\\text{ln}x\\right)dx[\/latex] and [latex]v={\\displaystyle\\int }^{\\text{ }}1dx=x[\/latex]. Substituting, we have<\/p>\r\n\r\n<div id=\"fs-id1165042110131\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx=x\\sin\\left(\\text{ln}x\\right)-\\left(x\\cos\\left(\\text{ln}x\\right)-{\\displaystyle\\int }^{\\text{ }}-\\sin\\left(\\text{ln}x\\right)dx\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040741882\">After simplifying, we obtain<\/p>\r\n\r\n<div id=\"fs-id1165040741886\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx=x\\sin\\left(\\text{ln}x\\right)-x\\cos\\left(\\text{ln}x\\right)-{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040741980\">The last integral is now the same as the original. It may seem that we have simply gone in a circle, but now we can actually evaluate the integral. To see how to do this more clearly, substitute [latex]I={\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx[\/latex]. Thus, the equation becomes<\/p>\r\n\r\n<div id=\"fs-id1165040742019\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]I=x\\sin\\left(\\text{ln}x\\right)-x\\cos\\left(\\text{ln}x\\right)-I[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040742068\">First, add [latex]I[\/latex] to both sides of the equation to obtain<\/p>\r\n\r\n<div id=\"fs-id1165040742075\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]2I=x\\sin\\left(\\text{ln}x\\right)-x\\cos\\left(\\text{ln}x\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042266548\">Next, divide by 2:<\/p>\r\n\r\n<div id=\"fs-id1165042266551\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]I=\\frac{1}{2}x\\sin\\left(\\text{ln}x\\right)-\\frac{1}{2}x\\cos\\left(\\text{ln}x\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042266606\">Substituting [latex]I={\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx[\/latex] again, we have<\/p>\r\n\r\n<div id=\"fs-id1165042266641\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx=\\frac{1}{2}x\\sin\\left(\\text{ln}x\\right)-\\frac{1}{2}x\\cos\\left(\\text{ln}x\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042266721\">From this we see that [latex]\\frac{1}{2}x\\sin\\left(\\text{ln}x\\right)-\\frac{1}{2}x\\cos\\left(\\text{ln}x\\right)[\/latex] is an antiderivative of [latex]\\sin\\left(\\text{ln}x\\right)dx[\/latex]. For the most general antiderivative, add [latex]+C\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165042001966\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx=\\frac{1}{2}x\\sin\\left(\\text{ln}x\\right)-\\frac{1}{2}x\\cos\\left(\\text{ln}x\\right)+C[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1165042002057\" data-type=\"commentary\">\r\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\r\n<p id=\"fs-id1165042002062\">If this method feels a little strange at first, we can check the answer by differentiation:<\/p>\r\n\r\n<div id=\"fs-id1165042002066\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}\\frac{d}{dx}\\left(\\frac{1}{2}x\\sin\\left(\\text{ln}x\\right)-\\frac{1}{2}x\\cos\\left(\\text{ln}x\\right)\\right)\\hfill \\\\ \\\\ =\\frac{1}{2}\\left(\\sin\\left(\\text{ln}x\\right)\\right)+\\cos\\left(\\text{ln}x\\right)\\cdot \\frac{1}{x}\\cdot \\frac{1}{2}x-\\left(\\frac{1}{2}\\cos\\left(\\text{ln}x\\right)-\\sin\\left(\\text{ln}x\\right)\\cdot \\frac{1}{x}\\cdot \\frac{1}{2}x\\right)\\hfill \\\\ =\\sin\\left(\\text{ln}x\\right).\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042295952\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1165042295956\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042295958\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1165042295958\" data-type=\"problem\">\r\n<p id=\"fs-id1165042295960\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{x}^{2}\\sin{x}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558889\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n<div id=\"fs-id1165042231011\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165042231018\">This is similar to the previous example: Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{x}^{2}{e}^{3x}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558890\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n<div id=\"fs-id1165042230965\" data-type=\"solution\">\r\n<p id=\"fs-id1165042407711\">[latex]\\text{-}{x}^{2}\\cos{x}+2x\\sin{x}+2\\cos{x}+C[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lpmf-HAmaEU?controls=0&amp;start=1419&amp;end=1575&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.1IntegrationByParts1419to1575_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.1 Integration by Parts\" here (opens in new window)<\/a>.\r\n\r\n<section id=\"fs-id1165042231026\" data-depth=\"1\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]169241[\/ohm_question]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Integration by Parts for Definite Integrals<\/h2>\r\n<p id=\"fs-id1165042231031\">Now that we have used integration by parts successfully to evaluate <span class=\"no-emphasis\" data-type=\"term\">indefinite integrals<\/span>, we turn our attention to definite integrals. The integration technique is really the same, only we add a step to evaluate the integral at the upper and lower limits of integration.<\/p>\r\n\r\n<div id=\"fs-id1165042231041\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Integration by Parts for Definite Integrals<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165042231049\">Let [latex]u=f\\left(x\\right)[\/latex] and [latex]v=g\\left(x\\right)[\/latex] be functions with continuous derivatives on [latex]\\left[a,b\\right][\/latex]. Then<\/p>\r\n\r\n<div id=\"fs-id1165042231112\" class=\"numbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{a}^{b}udv={uv|}_{a}^{b}-{\\displaystyle\\int }_{a}^{b}vdu[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042231180\" data-type=\"example\">\r\n<div id=\"fs-id1165042231182\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042231184\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the area of a region<\/h3>\r\n<div id=\"fs-id1165042231184\" data-type=\"problem\">\r\n<p id=\"fs-id1165042237852\">Find the area of the region bounded above by the graph of [latex]y={\\tan}^{-1}x[\/latex] and below by the [latex]x[\/latex] -axis over the interval [latex]\\left[0,1\\right][\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558879\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558879\"]\r\n<div id=\"fs-id1165042237897\" data-type=\"solution\">\r\n<p id=\"fs-id1165042237899\">This region is shown in Figure 1. To find the area, we must evaluate [latex]\\underset{0}{\\overset{1}{\\displaystyle\\int }}{\\tan}^{-1}xdx[\/latex].<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_07_01_001\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"562\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233750\/CNX_Calc_Figure_07_01_001.jpg\" alt=\"This figure is the graph of the inverse tangent function. It is an increasing function that passes through the origin. In the first quadrant there is a shaded region under the graph, above the x-axis. The shaded area is bounded to the right at x = 1.\" width=\"562\" height=\"570\" data-media-type=\"image\/jpeg\" \/> Figure 1. To find the area of the shaded region, we have to use integration by parts.[\/caption]<\/figure>\r\n<p id=\"fs-id1165042237957\">For this integral, let\u2019s choose [latex]u={\\tan}^{-1}x[\/latex] and [latex]dv=dx[\/latex], thereby making [latex]du=\\frac{1}{{x}^{2}+1}dx[\/latex] and [latex]v=x[\/latex]. After applying the integration-by-parts formula we obtain<\/p>\r\n\r\n<div id=\"fs-id1165042238034\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{Area}=x{\\tan}^{-1}{x|}_{0}^{1}-\\underset{0}{\\overset{1}{\\displaystyle\\int }}\\frac{x}{{x}^{2}+1}dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042304278\">Use <em data-effect=\"italics\">u<\/em>-substitution to obtain<\/p>\r\n\r\n<div id=\"fs-id1165042304286\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{0}{\\overset{1}{\\displaystyle\\int }}\\frac{x}{{x}^{2}+1}dx=\\frac{1}{2}\\text{ln}|{x}^{2}+1{|}_{0}^{1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042304368\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1165042304371\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{Area}=x{\\tan }^{-1}x{|}_{0}^{1}-\\frac{1}{2}\\mathrm{ln}|{x}^{2}+1|{|}_{0}^{1}=\\frac{\\pi }{4}-\\frac{1}{2}\\mathrm{ln}2[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042304463\">At this point it might not be a bad idea to do a \"reality check\" on the reasonableness of our solution. Since [latex]\\frac{\\pi }{4}-\\frac{1}{2}\\text{ln}2\\approx 0.4388[\/latex], and from Figure 1 we expect our area to be slightly less than 0.5, this solution appears to be reasonable.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042058871\" data-type=\"example\">\r\n<div id=\"fs-id1165042058873\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042058875\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding a volume of revolution<\/h3>\r\n<div id=\"fs-id1165042058875\" data-type=\"problem\">\r\n<p id=\"fs-id1165042058881\">Find the volume of the solid obtained by revolving the region bounded by the graph of [latex]f\\left(x\\right)={e}^{\\text{-}x}[\/latex], the <em data-effect=\"italics\">x<\/em>-axis, the <em data-effect=\"italics\">y<\/em>-axis, and the line [latex]x=1[\/latex] about the <em data-effect=\"italics\">y<\/em>-axis.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558869\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558869\"]\r\n<div id=\"fs-id1165042058936\" data-type=\"solution\">\r\n\r\n<strong>Solution<\/strong>\r\n<p id=\"fs-id1165042058938\">The best option to solving this problem is to use the shell method. Begin by sketching the region to be revolved, along with a typical rectangle (see the following graph).<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_07_01_002\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233752\/CNX_Calc_Figure_07_01_002.jpg\" alt=\"This figure is the graph of the function e^-x. It is an increasing function on the left side of the y-axis and decreasing on the right side of the y-axis. The curve also comes to a point on the y-axis at y=1. Under the curve there is a shaded rectangle in the first quadrant. There is also a cylinder under the graph, formed by revolving the rectangle around the y-axis.\" width=\"731\" height=\"272\" data-media-type=\"image\/jpeg\" \/> Figure 2. We can use the shell method to find a volume of revolution.[\/caption]<\/figure>\r\n<p id=\"fs-id1165042058966\">To find the volume using shells, we must evaluate [latex]2\\pi {\\displaystyle\\int }_{0}^{1}x{e}^{\\text{-}x}dx[\/latex]. To do this, let [latex]u=x[\/latex] and [latex]dv={e}^{\\text{-}x}[\/latex]. These choices lead to [latex]du=dx[\/latex] and [latex]v={\\displaystyle\\int }^{\\text{ }}{e}^{\\text{-}x}=\\text{-}{e}^{\\text{-}x}[\/latex]. Substituting into the integration-by-parts for definite integrals formula, we obtain<\/p>\r\n\r\n<div id=\"fs-id1165042281368\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccc}\\hfill \\text{Volume}&amp; =2\\pi \\underset{0}{\\overset{1}{\\displaystyle\\int }}x{e}^{\\text{-}x}dx=2\\pi \\left(\\text{-}x{e}^{\\text{-}x}{|}_{0}^{1}+\\underset{0}{\\overset{1}{\\displaystyle\\int }}{e}^{\\text{-}x}dx\\right)\\hfill &amp; &amp; \\text{Use integration by parts}.\\hfill \\\\ &amp; =-2\\pi x{e}^{\\text{-}x}{|}_{0}^{1}-2\\pi {e}^{\\text{-}x}{|}_{0}^{1}\\hfill &amp; &amp; \\text{Evaluate}\\underset{0}{\\overset{1}{\\displaystyle\\int }}{e}^{\\text{-}x}dx=\\text{-}{e}^{\\text{-}x}{|}_{0}^{1}.\\hfill \\\\ &amp; =2\\pi -\\frac{4\\pi }{e}.\\hfill &amp; &amp; \\text{Evaluate and simplify}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1165042071427\" data-type=\"commentary\">\r\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\r\n<p id=\"fs-id1165042071432\">Again, it is a good idea to check the reasonableness of our solution. We observe that the solid has a volume slightly less than that of a cylinder of radius [latex]1[\/latex] and height of [latex]\\frac{1}{e}[\/latex] added to the volume of a cone of base radius [latex]1[\/latex] and height of [latex]1-\\frac{1}{3}[\/latex]. Consequently, the solid should have a volume a bit less than<\/p>\r\n\r\n<div id=\"fs-id1165042071470\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\pi {\\left(1\\right)}^{2}\\frac{1}{e}+\\left(\\frac{\\pi }{3}\\right){\\left(1\\right)}^{2}\\left(1-\\frac{1}{e}\\right)=\\frac{2\\pi }{3e}-\\frac{\\pi }{3}\\approx 1.8177[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042071569\">Since [latex]2\\pi -\\frac{4\\pi }{e}\\approx 1.6603[\/latex], we see that our calculated volume is reasonable.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042299594\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1165042299598\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042299600\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1165042299600\" data-type=\"problem\">\r\n<p id=\"fs-id1165042299602\">Evaluate [latex]{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}x\\cos{x}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558819\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558819\"]\r\n<div id=\"fs-id1165042299659\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165042299667\">Use the integration-by-parts formula with [latex]u=x[\/latex] and [latex]dv=\\cos{x}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558829\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558829\"]\r\n<div id=\"fs-id1165042299639\" data-type=\"solution\">\r\n<p id=\"fs-id1165042583725\">[latex]\\frac{\\pi }{2}-1[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lpmf-HAmaEU?controls=0&amp;start=1913&amp;end=2043&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.1IntegrationByParts1913to2043_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.1 Integration by Parts\" here (opens in new window)<\/a>.\r\n\r\n<\/section>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]169243[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Recognize when to use integration by parts.<\/li>\n<li>Use the integration-by-parts formula to solve integration problems.<\/li>\n<li>Use the integration-by-parts formula for definite integrals.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165041979164\">If, [latex]h\\left(x\\right)=f\\left(x\\right)g\\left(x\\right)[\/latex], then by using the product rule, we obtain [latex]{h}^{\\prime }\\left(x\\right)={f}^{\\prime }\\left(x\\right)g\\left(x\\right)+{g}^{\\prime }\\left(x\\right)f\\left(x\\right)[\/latex]. Although at first it may seem counterproductive, let\u2019s now integrate both sides of this equation: [latex]\\displaystyle\\int {h}^{\\prime }\\left(x\\right)dx=\\displaystyle\\int \\left(g\\left(x\\right){f}^{\\prime }\\left(x\\right)+f\\left(x\\right){g}^{\\prime }\\left(x\\right)\\right)dx[\/latex].<\/p>\n<p id=\"fs-id1165042262165\">This gives us<\/p>\n<div id=\"fs-id1165042041848\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]h\\left(x\\right)=f\\left(x\\right)g\\left(x\\right)=\\displaystyle\\int g\\left(x\\right){f}^{\\prime }\\left(x\\right)dx+\\displaystyle\\int f\\left(x\\right){g}^{\\prime }\\left(x\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041955261\">Now we solve for [latex]\\displaystyle\\int f\\left(x\\right){g}^{\\prime }\\left(x\\right)dx:[\/latex]<\/p>\n<div id=\"fs-id1165042235539\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int f\\left(x\\right){g}^{\\prime }\\left(x\\right)dx=f\\left(x\\right)g\\left(x\\right)-\\displaystyle\\int g\\left(x\\right){f}^{\\prime }\\left(x\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042135462\">By making the substitutions [latex]u=f\\left(x\\right)[\/latex] and [latex]v=g\\left(x\\right)[\/latex], which in turn make [latex]du={f}^{\\prime }\\left(x\\right)dx[\/latex] and [latex]dv={g}^{\\prime }\\left(x\\right)dx[\/latex], we have the more compact form<\/p>\n<div id=\"fs-id1165041813626\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int udv=uv-\\displaystyle\\int vdu[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1165042235552\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<div id=\"fs-id1165042235552\" class=\"theorem\" data-type=\"note\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Integration by Parts<\/h3>\n<hr \/>\n<p id=\"fs-id1165042089614\">Let [latex]u=f\\left(x\\right)[\/latex] and [latex]v=g\\left(x\\right)[\/latex] be functions with continuous derivatives. Then, the integration-by-parts formula for the integral involving these two functions is:<\/p>\n<div id=\"fs-id1165042232372\" class=\"numbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int udv=uv-\\displaystyle\\int vdu[\/latex].<\/div>\n<\/div>\n<p id=\"fs-id1165041841895\">\n<\/div>\n<p><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">The advantage of using the integration-by-parts formula is that we can use it to exchange one integral for another, possibly easier, integral. The following example illustrates its use.<\/span><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165041832973\" data-type=\"example\">\n<div id=\"fs-id1165041816966\" data-type=\"exercise\">\n<div id=\"fs-id1165042126072\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Using Integration by Parts<\/h3>\n<div id=\"fs-id1165042126072\" data-type=\"problem\">\n<p id=\"fs-id1165041832755\">Use integration by parts with [latex]u=x[\/latex] and [latex]dv=\\sin{x}dx[\/latex] to evaluate [latex]\\displaystyle\\int x\\sin{x}dx[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558899\">Show Solution<\/span><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Solution<\/strong><\/p>\n<div id=\"fs-id1165041792874\" data-type=\"solution\">\n<p id=\"fs-id1165042089552\">By choosing [latex]u=x[\/latex], we have [latex]du=1dx[\/latex]. Since [latex]dv=\\sin{x}dx[\/latex], we get [latex]v=\\displaystyle\\int \\sin{x}dx=\\text{-}\\cos{x}[\/latex]. It is handy to keep track of these values as follows:<\/p>\n<div id=\"fs-id1165041792867\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\hfill u& =\\hfill & x\\hfill & & \\hfill dv& =\\hfill & \\sin{x}dx\\hfill \\\\ \\hfill du& =\\hfill & 1dx\\hfill & & \\hfill v& =\\hfill & \\displaystyle\\int \\sin{x}dx=\\text{-}\\cos{x}. \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040681626\">Applying the integration-by-parts formula results in<\/p>\n<div id=\"fs-id1165041985992\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccc}\\hfill {\\displaystyle\\int x\\sin{x}dx}& =\\left(x\\right)\\left(\\text{-}\\cos{x}\\right)-{\\displaystyle\\int \\left(\\text{-}\\cos{x}\\right)\\left(1dx\\right)}\\hfill & & \\text{Substitute.}\\hfill \\\\ & =\\text{-}x\\cos{x}+{\\displaystyle\\int \\cos{x}dx}\\hfill & & \\text{Simplify.}\\hfill \\\\ & =\\text{-}x\\cos{x}+\\sin{x}+C.\\hfill & & \\text{Use}{\\displaystyle\\int \\cos{x}dx=\\sin{x}+C.}\\end{array}[\/latex]<\/div>\n<div id=\"fs-id1165041893024\" data-type=\"commentary\">\n<div data-type=\"title\"><\/div>\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\n<div data-type=\"title\"><\/div>\n<p id=\"fs-id1165041973005\">At this point, there are probably a few items that need clarification. First of all, you may be curious about what would have happened if we had chosen [latex]u=\\sin{x}[\/latex] and [latex]dv=x[\/latex]. If we had done so, then we would have [latex]du=\\cos{x}[\/latex] and [latex]v=\\frac{1}{2}{x}^{2}[\/latex]. Thus, after applying integration by parts, we have [latex]{\\displaystyle\\int }^{\\text{ }}x\\sin{x}dx=\\frac{1}{2}{x}^{2}\\sin{x}-{\\displaystyle\\int }^{\\text{ }}\\frac{1}{2}{x}^{2}\\cos{x}dx[\/latex]. Unfortunately, with the new integral, we are in no better position than before. It is important to keep in mind that when we apply integration by parts, we may need to try several choices for [latex]u[\/latex] and [latex]dv[\/latex] before finding a choice that works.<\/p>\n<p id=\"fs-id1165042098353\">Second, you may wonder why, when we find [latex]v={\\displaystyle\\int }^{\\text{ }}\\sin{x}dx=\\text{-}\\cos{x}[\/latex], we do not use [latex]v=\\text{-}\\cos{x}+K[\/latex]. To see that it makes no difference, we can rework the problem using [latex]v=\\text{-}\\cos{x}+K\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1165042094536\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{\\displaystyle\\int}x\\sin{x}dx\\hfill & =\\left(x\\right)\\left(\\text{-}\\cos{x}+K\\right)-{\\displaystyle\\int}\\left(\\text{-}\\cos{x}+K\\right)\\left(1dx\\right)\\hfill \\\\ \\hfill & =\\text{-}x\\cos{x}+Kx+{\\displaystyle\\int}\\cos{x}dx-{\\displaystyle\\int}Kdx\\hfill \\\\ \\hfill & =\\text{-}x\\cos{x}+Kx+\\sin{x}-Kx+C\\hfill \\\\ \\hfill & =\\text{-}x\\cos{x}+\\sin{x}+C.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042136113\">As you can see, it makes no difference in the final solution.<\/p>\n<p id=\"fs-id1165041759254\">Last, we can check to make sure that our antiderivative is correct by differentiating [latex]\\text{-}x\\cos{x}+\\sin{x}+C\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1165041952373\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\frac{d}{dx}\\left(\\text{-}x\\cos{x}+\\sin{x}+C\\right)\\hfill & =\\left(-1\\right)\\cos{x}+\\left(\\text{-}x\\right)\\left(\\text{-}\\sin{x}\\right)+\\cos{x}\\hfill \\\\ \\hfill & =x\\sin{x}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040665697\">Therefore, the antiderivative checks out.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Interactive<\/h3>\n<p>Watch <a href=\"http:\/\/www.sosmath.com\/calculus\/integration\/byparts\/byparts.html\" target=\"_blank\" rel=\"noopener\">this video for an introduction to integration by parts<\/a> and visit <a href=\"http:\/\/www.sosmath.com\/calculus\/integration\/byparts\/byparts.html\" target=\"_blank\" rel=\"noopener\">this website for examples of integration by parts<\/a>.<\/p>\n<\/div>\n<div id=\"fs-id1165042127896\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1165042127899\" data-type=\"exercise\">\n<div id=\"fs-id1165040747248\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<div id=\"fs-id1165040747248\" data-type=\"problem\">\n<p id=\"fs-id1165040747250\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}x{e}^{2x}dx[\/latex] using the integration-by-parts formula with [latex]u=x[\/latex] and [latex]dv={e}^{2x}dx[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558897\">Hint<\/span><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042051378\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165040681938\">Find [latex]du[\/latex] and [latex]v[\/latex], and use the previous example as a guide.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558898\">Show Solution<\/span><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040642767\" data-type=\"solution\">\n<p id=\"fs-id1165043352090\">[latex]{\\displaystyle\\int }^{\\text{ }}x{e}^{2x}dx=\\frac{1}{2}x{e}^{2x}-\\frac{1}{4}{e}^{2x}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lpmf-HAmaEU?controls=0&amp;start=305&amp;end=430&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.1IntegrationByParts305to430_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.1 Integration by Parts&#8221; here (opens in new window)<\/a>.<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">The natural question to ask at this point is: How do we know how to choose [latex]u[\/latex] and [latex]dv?[\/latex] Sometimes it is a matter of trial and error; however, the acronym LIATE can often help to take some of the guesswork out of our choices. This acronym stands for <\/span><strong style=\"font-size: 1rem; text-align: initial;\" data-effect=\"bold\">L<\/strong><span style=\"font-size: 1rem; text-align: initial;\">ogarithmic Functions, <\/span><strong style=\"font-size: 1rem; text-align: initial;\" data-effect=\"bold\">I<\/strong><span style=\"font-size: 1rem; text-align: initial;\">nverse Trigonometric Functions, <\/span><strong style=\"font-size: 1rem; text-align: initial;\" data-effect=\"bold\">A<\/strong><span style=\"font-size: 1rem; text-align: initial;\">lgebraic Functions, <\/span><strong style=\"font-size: 1rem; text-align: initial;\" data-effect=\"bold\">T<\/strong><span style=\"font-size: 1rem; text-align: initial;\">rigonometric Functions, and <\/span><strong style=\"font-size: 1rem; text-align: initial;\" data-effect=\"bold\">E<\/strong><span style=\"font-size: 1rem; text-align: initial;\">xponential Functions. This mnemonic serves as an aid in determining an appropriate choice for [latex]u[\/latex].<\/span><\/p>\n<p id=\"fs-id1165040744404\">The type of function in the integral that appears first in the list should be our first choice of [latex]u[\/latex]. For example, if an integral contains a <span class=\"no-emphasis\" data-type=\"term\">logarithmic function<\/span> and an <span class=\"no-emphasis\" data-type=\"term\">algebraic function<\/span>, we should choose [latex]u[\/latex] to be the logarithmic function, because L comes before A in LIATE. The integral in the previous example has a trigonometric function [latex]\\text{(}\\sin{x}\\text{)}[\/latex] and an algebraic function [latex]\\left(x\\right)[\/latex]. Because A comes before T in LIATE, we chose [latex]u[\/latex] to be the algebraic function. When we have chosen [latex]u[\/latex], [latex]dv[\/latex] is selected to be the remaining part of the function to be integrated, together with [latex]dx[\/latex].<\/p>\n<p id=\"fs-id1165040676764\">Why does this mnemonic work? Remember that whatever we pick to be [latex]dv[\/latex] must be something we can integrate. Since we do not have integration formulas that allow us to integrate simple logarithmic functions and inverse trigonometric functions, it makes sense that they should not be chosen as values for [latex]dv[\/latex]. Consequently, they should be at the head of the list as choices for [latex]u[\/latex]. Thus, we put LI at the beginning of the mnemonic. (We could just as easily have started with IL, since these two types of functions won\u2019t appear together in an integration-by-parts problem.) The exponential and trigonometric functions are at the end of our list because they are fairly easy to integrate and make good choices for [latex]dv[\/latex]. Thus, we have TE at the end of our mnemonic. (We could just as easily have used ET at the end, since when these types of functions appear together it usually doesn\u2019t really matter which one is [latex]u[\/latex] and which one is [latex]dv[\/latex].) Algebraic functions are generally easy both to integrate and to differentiate, and they come in the middle of the mnemonic.<\/p>\n<p>To use the by-parts technique successfully, it is helpful to first review the derivative rules of several familiar transcendental functions.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Derivative rules for transcendental functions<\/h3>\n<ol id=\"fs-id1170572169684\" style=\"list-style-type: decimal;\">\n<li>[latex]\\frac{d}{dx} (\\sin x) = \\cos x[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx} (\\cos x) = -\\sin x[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx} (\\ln x) = \\frac{1}{x}[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx} (\\arcsin x) = \\frac{1}{\\sqrt{1-x^2}}[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx} (\\arctan x) = \\frac{1}{1+x^2}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1165042015633\" data-type=\"example\">\n<div id=\"fs-id1165042015635\" data-type=\"exercise\">\n<div id=\"fs-id1165042231844\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>example: Using integration by parts<\/h3>\n<div id=\"fs-id1165042231844\" data-type=\"problem\">\n<p id=\"fs-id1165042231849\">Evaluate [latex]\\displaystyle\\int \\frac{\\text{ln}x}{{x}^{3}}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558896\">Show Solution<\/span><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042042007\" data-type=\"solution\">\n<p id=\"fs-id1165042042010\">Begin by rewriting the integral:<\/p>\n<div id=\"fs-id1165042008468\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{\\text{ln}x}{{x}^{3}}dx={\\displaystyle\\int }^{\\text{ }}{x}^{-3}\\text{ln}xdx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042127807\">Since this integral contains the algebraic function [latex]{x}^{-3}[\/latex] and the logarithmic function [latex]\\text{ln}x[\/latex], choose [latex]u=\\text{ln}x[\/latex], since L comes before A in LIATE. After we have chosen [latex]u=\\text{ln}x[\/latex], we must choose [latex]dv={x}^{-3}dx[\/latex].<\/p>\n<p id=\"fs-id1165042127784\">Next, since [latex]u=\\text{ln}x[\/latex], we have [latex]du=\\frac{1}{x}dx[\/latex]. Also, [latex]v={\\displaystyle\\int }^{\\text{ }}{x}^{-3}dx=-\\frac{1}{2}{x}^{-2}[\/latex]. Summarizing,<\/p>\n<div id=\"fs-id1165042109999\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\hfill u& =\\hfill & \\text{ln}x\\hfill & & \\hfill dv& =\\hfill & {x}^{-3}dx\\hfill \\\\ \\hfill du& =\\hfill & \\frac{1}{x}dx\\hfill & & \\hfill v& =\\hfill & {\\displaystyle\\int }^{\\text{ }}{x}^{-3}dx=-\\frac{1}{2}{x}^{-2}.\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042128802\">Substituting into the integration-by-parts formula gives<\/p>\n<div id=\"fs-id1165042128805\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill \\displaystyle\\int \\frac{\\text{ln}x}{{x}^{3}}dx& ={\\displaystyle\\int }^{\\text{ }}{x}^{-3}\\text{ln}xdx=\\left(\\text{ln}x\\right)\\left(\\text{-}\\frac{1}{2}{x}^{-2}\\right)-{\\displaystyle\\int }^{\\text{ }}\\left(\\text{-}\\frac{1}{2}{x}^{-2}\\right)\\left(\\frac{1}{x}dx\\right)\\hfill & & & \\\\ & =-\\frac{1}{2}{x}^{-2}\\text{ln}x+{\\displaystyle\\int }^{\\text{ }}\\frac{1}{2}{x}^{-3}dx\\hfill & & & \\text{Simplify}.\\hfill \\\\ & =-\\frac{1}{2}{x}^{-2}\\text{ln}x-\\frac{1}{4}{x}^{-2}+C\\hfill & & & \\text{Integrate}.\\hfill \\\\ & =-\\frac{1}{2{x}^{2}}\\text{ln}x-\\frac{1}{4{x}^{2}}+C.\\hfill & & & \\text{Rewrite with positive integers.}\\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042127888\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1165041899921\" data-type=\"exercise\">\n<div id=\"fs-id1165041899924\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1165041899924\" data-type=\"problem\">\n<p id=\"fs-id1165041899926\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}x\\text{ln}xdx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558895\">Hint<\/span><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041864895\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165041864902\">Use [latex]u=\\text{ln}x[\/latex] and [latex]dv=xdx[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558894\">Show Solution<\/span><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040747233\" data-type=\"solution\">\n<p id=\"fs-id1165043331164\">[latex]\\frac{1}{2}{x}^{2}\\text{ln}x-\\frac{1}{4}{x}^{2}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lpmf-HAmaEU?controls=0&amp;start=558&amp;end=647&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.1IntegrationByParts558to647_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.1 Integration by Parts&#8221; here (opens in new window)<\/a>.<\/p>\n<p id=\"fs-id1165041948188\">In some cases, as in the next two examples, it may be necessary to apply integration by parts more than once.<\/p>\n<div id=\"fs-id1165041948192\" data-type=\"example\">\n<div id=\"fs-id1165041948195\" data-type=\"exercise\">\n<div id=\"fs-id1165042015541\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>example: Applying integration by parts more than once<\/h3>\n<div id=\"fs-id1165042015541\" data-type=\"problem\">\n<p id=\"fs-id1165042015546\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{x}^{2}{e}^{3x}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558893\">Show Solution<\/span><\/p>\n<div id=\"q44558893\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040637308\" data-type=\"solution\">\n<p id=\"fs-id1165040637310\">Using LIATE, choose [latex]u={x}^{2}[\/latex] and [latex]dv={e}^{3x}dx[\/latex]. Thus, [latex]du=2xdx[\/latex] and [latex]v=\\displaystyle\\int {e}^{3x}dx=\\left(\\frac{1}{3}\\right){e}^{3x}[\/latex]. Therefore,<\/p>\n<div id=\"fs-id1165040740033\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\hfill u& =\\hfill & {x}^{2}\\hfill & & \\hfill dv& =\\hfill & {e}^{3x}dx\\hfill \\\\ \\hfill du& =\\hfill & 2xdx\\hfill & & \\hfill v& =\\hfill & \\displaystyle\\int {e}^{3x}dx=\\frac{1}{3}{e}^{3x}.\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042240114\">Substituting into the integration-by-parts formula\u00a0produces<\/p>\n<div id=\"fs-id1165042240117\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int {x}^{2}{e}^{3x}dx=\\frac{1}{3}{x}^{2}{e}^{3x}-\\displaystyle\\int \\frac{2}{3}x{e}^{3x}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040756177\">We still cannot integrate [latex]\\displaystyle\\int \\frac{2}{3}x{e}^{3x}dx[\/latex] directly, but the integral now has a lower power on [latex]x[\/latex]. We can evaluate this new integral by using integration by parts again. To do this, choose [latex]u=x[\/latex] and [latex]dv=\\frac{2}{3}{e}^{3x}dx[\/latex]. Thus, [latex]du=dx[\/latex] and [latex]v=\\displaystyle\\int \\left(\\frac{2}{3}\\right){e}^{3x}dx=\\left(\\frac{2}{9}\\right){e}^{3x}[\/latex]. Now we have<\/p>\n<div id=\"fs-id1165042002748\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\hfill u& =\\hfill & x\\hfill & & \\hfill dv& =\\hfill & \\frac{2}{3}{e}^{3x}dx\\hfill \\\\ \\hfill du& =\\hfill & dx\\hfill & & \\hfill v& =\\hfill & \\displaystyle\\int \\frac{2}{3}{e}^{3x}dx=\\frac{2}{9}{e}^{3x}.\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040757531\">Substituting back into the previous equation yields<\/p>\n<div id=\"fs-id1165040757534\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}{x}^{2}{e}^{3x}dx=\\frac{1}{3}{x}^{2}{e}^{3x}-\\left(\\frac{2}{9}x{e}^{3x}-{\\displaystyle\\int }^{\\text{ }}\\frac{2}{9}{e}^{3x}dx\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042110524\">After evaluating the last integral and simplifying, we obtain<\/p>\n<div id=\"fs-id1165042110527\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int {x}^{2}{e}^{3x}dx=\\frac{1}{3}{x}^{2}{e}^{3x}-\\frac{2}{9}x{e}^{3x}+\\frac{2}{27}{e}^{3x}+C[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042132627\" data-type=\"example\">\n<div id=\"fs-id1165042132629\" data-type=\"exercise\">\n<div id=\"fs-id1165042132631\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Applying Integration by Parts When LIATE Doesn\u2019t Quite Work<\/h3>\n<div id=\"fs-id1165042132631\" data-type=\"problem\">\n<p id=\"fs-id1165042132636\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{t}^{3}{e}^{{t}^{2}}dt[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558892\">Show Solution<\/span><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042132671\" data-type=\"solution\">\n<p id=\"fs-id1165042132674\">If we use a strict interpretation of the mnemonic LIATE to make our choice of [latex]u[\/latex], we end up with [latex]u={t}^{3}[\/latex] and [latex]dv={e}^{{t}^{2}}dt[\/latex]. Unfortunately, this choice won\u2019t work because we are unable to evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{e}^{{t}^{2}}dt[\/latex]. However, since we can evaluate [latex]{\\displaystyle\\int }^{\\text{ }}t{e}^{{t}^{2}}dx[\/latex], we can try choosing [latex]u={t}^{2}[\/latex] and [latex]dv=t{e}^{{t}^{2}}dt[\/latex]. With these choices we have<\/p>\n<div id=\"fs-id1165040715819\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\hfill u& =\\hfill & {t}^{2}\\hfill & & \\hfill dv& =\\hfill & t{e}^{{t}^{2}}dt\\hfill \\\\ \\hfill du& =\\hfill & 2tdt\\hfill & & \\hfill v& =\\hfill & {\\displaystyle\\int }^{\\text{ }}t{e}^{{t}^{2}}dt=\\frac{1}{2}{e}^{{t}^{2}}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042013342\">Thus, we obtain<\/p>\n<div id=\"fs-id1165042013345\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{\\displaystyle\\int {t}^{3}{e}^{{t}^{2}}dt}\\hfill & =\\frac{1}{2}{t}^{2}{e}^{{t}^{2}}-{\\displaystyle\\int \\frac{1}{2}{e}^{{t}^{2}}2tdt}\\hfill \\\\ \\hfill & =\\frac{1}{2}{t}^{2}{e}^{{t}^{2}}-\\frac{1}{2}{e}^{{t}^{2}}+C.\\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042035446\" data-type=\"example\">\n<div id=\"fs-id1165042035448\" data-type=\"exercise\">\n<div id=\"fs-id1165042035450\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Applying Integration by Parts More Than Once<\/h3>\n<div id=\"fs-id1165042035450\" data-type=\"problem\">\n<p id=\"fs-id1165042035456\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558891\">Show Solution<\/span><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Solution<\/strong><\/p>\n<div id=\"fs-id1165042035489\" data-type=\"solution\">\n<p id=\"fs-id1165042035491\">This integral appears to have only one function\u2014namely, [latex]\\sin\\left(\\text{ln}x\\right)[\/latex] \u2014however, we can always use the constant function 1 as the other function. In this example, let\u2019s choose [latex]u=\\sin\\left(\\text{ln}x\\right)[\/latex] and [latex]dv=1dx[\/latex]. (The decision to use [latex]u=\\sin\\left(\\text{ln}x\\right)[\/latex] is easy. We can\u2019t choose [latex]dv=\\sin\\left(\\text{ln}x\\right)dx[\/latex] because if we could integrate it, we wouldn\u2019t be using integration by parts in the first place!) Consequently, [latex]du=\\frac{1}{x}\\cos\\left(\\text{ln}x\\right)dx[\/latex] and [latex]v={\\displaystyle\\int }^{\\text{ }}1dx=x[\/latex]. After applying integration by parts to the integral and simplifying, we have<\/p>\n<div id=\"fs-id1165042257200\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx=x\\sin\\left(\\text{ln}x\\right)-{\\displaystyle\\int }^{\\text{ }}\\cos\\left(\\text{ln}x\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042110018\">Unfortunately, this process leaves us with a new integral that is very similar to the original. However, let\u2019s see what happens when we apply integration by parts again. This time let\u2019s choose [latex]u=\\cos\\left(\\text{ln}x\\right)[\/latex] and [latex]dv=1dx[\/latex], making [latex]du=\\text{-}\\frac{1}{x}\\sin\\left(\\text{ln}x\\right)dx[\/latex] and [latex]v={\\displaystyle\\int }^{\\text{ }}1dx=x[\/latex]. Substituting, we have<\/p>\n<div id=\"fs-id1165042110131\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx=x\\sin\\left(\\text{ln}x\\right)-\\left(x\\cos\\left(\\text{ln}x\\right)-{\\displaystyle\\int }^{\\text{ }}-\\sin\\left(\\text{ln}x\\right)dx\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040741882\">After simplifying, we obtain<\/p>\n<div id=\"fs-id1165040741886\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx=x\\sin\\left(\\text{ln}x\\right)-x\\cos\\left(\\text{ln}x\\right)-{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040741980\">The last integral is now the same as the original. It may seem that we have simply gone in a circle, but now we can actually evaluate the integral. To see how to do this more clearly, substitute [latex]I={\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx[\/latex]. Thus, the equation becomes<\/p>\n<div id=\"fs-id1165040742019\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]I=x\\sin\\left(\\text{ln}x\\right)-x\\cos\\left(\\text{ln}x\\right)-I[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040742068\">First, add [latex]I[\/latex] to both sides of the equation to obtain<\/p>\n<div id=\"fs-id1165040742075\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]2I=x\\sin\\left(\\text{ln}x\\right)-x\\cos\\left(\\text{ln}x\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042266548\">Next, divide by 2:<\/p>\n<div id=\"fs-id1165042266551\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]I=\\frac{1}{2}x\\sin\\left(\\text{ln}x\\right)-\\frac{1}{2}x\\cos\\left(\\text{ln}x\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042266606\">Substituting [latex]I={\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx[\/latex] again, we have<\/p>\n<div id=\"fs-id1165042266641\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx=\\frac{1}{2}x\\sin\\left(\\text{ln}x\\right)-\\frac{1}{2}x\\cos\\left(\\text{ln}x\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042266721\">From this we see that [latex]\\frac{1}{2}x\\sin\\left(\\text{ln}x\\right)-\\frac{1}{2}x\\cos\\left(\\text{ln}x\\right)[\/latex] is an antiderivative of [latex]\\sin\\left(\\text{ln}x\\right)dx[\/latex]. For the most general antiderivative, add [latex]+C\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1165042001966\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx=\\frac{1}{2}x\\sin\\left(\\text{ln}x\\right)-\\frac{1}{2}x\\cos\\left(\\text{ln}x\\right)+C[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1165042002057\" data-type=\"commentary\">\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\n<p id=\"fs-id1165042002062\">If this method feels a little strange at first, we can check the answer by differentiation:<\/p>\n<div id=\"fs-id1165042002066\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}\\frac{d}{dx}\\left(\\frac{1}{2}x\\sin\\left(\\text{ln}x\\right)-\\frac{1}{2}x\\cos\\left(\\text{ln}x\\right)\\right)\\hfill \\\\ \\\\ =\\frac{1}{2}\\left(\\sin\\left(\\text{ln}x\\right)\\right)+\\cos\\left(\\text{ln}x\\right)\\cdot \\frac{1}{x}\\cdot \\frac{1}{2}x-\\left(\\frac{1}{2}\\cos\\left(\\text{ln}x\\right)-\\sin\\left(\\text{ln}x\\right)\\cdot \\frac{1}{x}\\cdot \\frac{1}{2}x\\right)\\hfill \\\\ =\\sin\\left(\\text{ln}x\\right).\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042295952\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1165042295956\" data-type=\"exercise\">\n<div id=\"fs-id1165042295958\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1165042295958\" data-type=\"problem\">\n<p id=\"fs-id1165042295960\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{x}^{2}\\sin{x}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558889\">Hint<\/span><\/p>\n<div id=\"q44558889\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042231011\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165042231018\">This is similar to the previous example: Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{x}^{2}{e}^{3x}dx[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558890\">Show Solution<\/span><\/p>\n<div id=\"q44558890\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042230965\" data-type=\"solution\">\n<p id=\"fs-id1165042407711\">[latex]\\text{-}{x}^{2}\\cos{x}+2x\\sin{x}+2\\cos{x}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lpmf-HAmaEU?controls=0&amp;start=1419&amp;end=1575&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.1IntegrationByParts1419to1575_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.1 Integration by Parts&#8221; here (opens in new window)<\/a>.<\/p>\n<section id=\"fs-id1165042231026\" data-depth=\"1\">\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm169241\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169241&theme=oea&iframe_resize_id=ohm169241&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2 data-type=\"title\">Integration by Parts for Definite Integrals<\/h2>\n<p id=\"fs-id1165042231031\">Now that we have used integration by parts successfully to evaluate <span class=\"no-emphasis\" data-type=\"term\">indefinite integrals<\/span>, we turn our attention to definite integrals. The integration technique is really the same, only we add a step to evaluate the integral at the upper and lower limits of integration.<\/p>\n<div id=\"fs-id1165042231041\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Integration by Parts for Definite Integrals<\/h3>\n<hr \/>\n<p id=\"fs-id1165042231049\">Let [latex]u=f\\left(x\\right)[\/latex] and [latex]v=g\\left(x\\right)[\/latex] be functions with continuous derivatives on [latex]\\left[a,b\\right][\/latex]. Then<\/p>\n<div id=\"fs-id1165042231112\" class=\"numbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{a}^{b}udv={uv|}_{a}^{b}-{\\displaystyle\\int }_{a}^{b}vdu[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042231180\" data-type=\"example\">\n<div id=\"fs-id1165042231182\" data-type=\"exercise\">\n<div id=\"fs-id1165042231184\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: finding the area of a region<\/h3>\n<div id=\"fs-id1165042231184\" data-type=\"problem\">\n<p id=\"fs-id1165042237852\">Find the area of the region bounded above by the graph of [latex]y={\\tan}^{-1}x[\/latex] and below by the [latex]x[\/latex] -axis over the interval [latex]\\left[0,1\\right][\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558879\">Show Solution<\/span><\/p>\n<div id=\"q44558879\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042237897\" data-type=\"solution\">\n<p id=\"fs-id1165042237899\">This region is shown in Figure 1. To find the area, we must evaluate [latex]\\underset{0}{\\overset{1}{\\displaystyle\\int }}{\\tan}^{-1}xdx[\/latex].<\/p>\n<figure id=\"CNX_Calc_Figure_07_01_001\"><figcaption><\/figcaption><div style=\"width: 572px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233750\/CNX_Calc_Figure_07_01_001.jpg\" alt=\"This figure is the graph of the inverse tangent function. It is an increasing function that passes through the origin. In the first quadrant there is a shaded region under the graph, above the x-axis. The shaded area is bounded to the right at x = 1.\" width=\"562\" height=\"570\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. To find the area of the shaded region, we have to use integration by parts.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165042237957\">For this integral, let\u2019s choose [latex]u={\\tan}^{-1}x[\/latex] and [latex]dv=dx[\/latex], thereby making [latex]du=\\frac{1}{{x}^{2}+1}dx[\/latex] and [latex]v=x[\/latex]. After applying the integration-by-parts formula we obtain<\/p>\n<div id=\"fs-id1165042238034\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{Area}=x{\\tan}^{-1}{x|}_{0}^{1}-\\underset{0}{\\overset{1}{\\displaystyle\\int }}\\frac{x}{{x}^{2}+1}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042304278\">Use <em data-effect=\"italics\">u<\/em>-substitution to obtain<\/p>\n<div id=\"fs-id1165042304286\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{0}{\\overset{1}{\\displaystyle\\int }}\\frac{x}{{x}^{2}+1}dx=\\frac{1}{2}\\text{ln}|{x}^{2}+1{|}_{0}^{1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042304368\">Thus,<\/p>\n<div id=\"fs-id1165042304371\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{Area}=x{\\tan }^{-1}x{|}_{0}^{1}-\\frac{1}{2}\\mathrm{ln}|{x}^{2}+1|{|}_{0}^{1}=\\frac{\\pi }{4}-\\frac{1}{2}\\mathrm{ln}2[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042304463\">At this point it might not be a bad idea to do a &#8220;reality check&#8221; on the reasonableness of our solution. Since [latex]\\frac{\\pi }{4}-\\frac{1}{2}\\text{ln}2\\approx 0.4388[\/latex], and from Figure 1 we expect our area to be slightly less than 0.5, this solution appears to be reasonable.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042058871\" data-type=\"example\">\n<div id=\"fs-id1165042058873\" data-type=\"exercise\">\n<div id=\"fs-id1165042058875\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Finding a volume of revolution<\/h3>\n<div id=\"fs-id1165042058875\" data-type=\"problem\">\n<p id=\"fs-id1165042058881\">Find the volume of the solid obtained by revolving the region bounded by the graph of [latex]f\\left(x\\right)={e}^{\\text{-}x}[\/latex], the <em data-effect=\"italics\">x<\/em>-axis, the <em data-effect=\"italics\">y<\/em>-axis, and the line [latex]x=1[\/latex] about the <em data-effect=\"italics\">y<\/em>-axis.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558869\">Show Solution<\/span><\/p>\n<div id=\"q44558869\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042058936\" data-type=\"solution\">\n<p><strong>Solution<\/strong><\/p>\n<p id=\"fs-id1165042058938\">The best option to solving this problem is to use the shell method. Begin by sketching the region to be revolved, along with a typical rectangle (see the following graph).<\/p>\n<figure id=\"CNX_Calc_Figure_07_01_002\"><figcaption><\/figcaption><div style=\"width: 741px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233752\/CNX_Calc_Figure_07_01_002.jpg\" alt=\"This figure is the graph of the function e^-x. It is an increasing function on the left side of the y-axis and decreasing on the right side of the y-axis. The curve also comes to a point on the y-axis at y=1. Under the curve there is a shaded rectangle in the first quadrant. There is also a cylinder under the graph, formed by revolving the rectangle around the y-axis.\" width=\"731\" height=\"272\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. We can use the shell method to find a volume of revolution.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165042058966\">To find the volume using shells, we must evaluate [latex]2\\pi {\\displaystyle\\int }_{0}^{1}x{e}^{\\text{-}x}dx[\/latex]. To do this, let [latex]u=x[\/latex] and [latex]dv={e}^{\\text{-}x}[\/latex]. These choices lead to [latex]du=dx[\/latex] and [latex]v={\\displaystyle\\int }^{\\text{ }}{e}^{\\text{-}x}=\\text{-}{e}^{\\text{-}x}[\/latex]. Substituting into the integration-by-parts for definite integrals formula, we obtain<\/p>\n<div id=\"fs-id1165042281368\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccc}\\hfill \\text{Volume}& =2\\pi \\underset{0}{\\overset{1}{\\displaystyle\\int }}x{e}^{\\text{-}x}dx=2\\pi \\left(\\text{-}x{e}^{\\text{-}x}{|}_{0}^{1}+\\underset{0}{\\overset{1}{\\displaystyle\\int }}{e}^{\\text{-}x}dx\\right)\\hfill & & \\text{Use integration by parts}.\\hfill \\\\ & =-2\\pi x{e}^{\\text{-}x}{|}_{0}^{1}-2\\pi {e}^{\\text{-}x}{|}_{0}^{1}\\hfill & & \\text{Evaluate}\\underset{0}{\\overset{1}{\\displaystyle\\int }}{e}^{\\text{-}x}dx=\\text{-}{e}^{\\text{-}x}{|}_{0}^{1}.\\hfill \\\\ & =2\\pi -\\frac{4\\pi }{e}.\\hfill & & \\text{Evaluate and simplify}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1165042071427\" data-type=\"commentary\">\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\n<p id=\"fs-id1165042071432\">Again, it is a good idea to check the reasonableness of our solution. We observe that the solid has a volume slightly less than that of a cylinder of radius [latex]1[\/latex] and height of [latex]\\frac{1}{e}[\/latex] added to the volume of a cone of base radius [latex]1[\/latex] and height of [latex]1-\\frac{1}{3}[\/latex]. Consequently, the solid should have a volume a bit less than<\/p>\n<div id=\"fs-id1165042071470\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\pi {\\left(1\\right)}^{2}\\frac{1}{e}+\\left(\\frac{\\pi }{3}\\right){\\left(1\\right)}^{2}\\left(1-\\frac{1}{e}\\right)=\\frac{2\\pi }{3e}-\\frac{\\pi }{3}\\approx 1.8177[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042071569\">Since [latex]2\\pi -\\frac{4\\pi }{e}\\approx 1.6603[\/latex], we see that our calculated volume is reasonable.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042299594\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1165042299598\" data-type=\"exercise\">\n<div id=\"fs-id1165042299600\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1165042299600\" data-type=\"problem\">\n<p id=\"fs-id1165042299602\">Evaluate [latex]{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}x\\cos{x}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558819\">Hint<\/span><\/p>\n<div id=\"q44558819\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042299659\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165042299667\">Use the integration-by-parts formula with [latex]u=x[\/latex] and [latex]dv=\\cos{x}dx[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558829\">Show Solution<\/span><\/p>\n<div id=\"q44558829\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042299639\" data-type=\"solution\">\n<p id=\"fs-id1165042583725\">[latex]\\frac{\\pi }{2}-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lpmf-HAmaEU?controls=0&amp;start=1913&amp;end=2043&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.1IntegrationByParts1913to2043_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.1 Integration by Parts&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm169243\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169243&theme=oea&iframe_resize_id=ohm169243&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-801\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>3.1 Integration by Parts. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"3.1 Integration by Parts\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-801","chapter","type-chapter","status-publish","hentry"],"part":158,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/801","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":20,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/801\/revisions"}],"predecessor-version":[{"id":2050,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/801\/revisions\/2050"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/158"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/801\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=801"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=801"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=801"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=801"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}