{"id":1984,"date":"2021-08-19T16:06:59","date_gmt":"2021-08-19T16:06:59","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-trigonometric-integrals\/"},"modified":"2021-11-19T03:01:17","modified_gmt":"2021-11-19T03:01:17","slug":"skills-review-for-trigonometric-integrals","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-trigonometric-integrals\/","title":{"raw":"Skills Review for Trigonometric Integrals","rendered":"Skills Review for Trigonometric Integrals"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Apply the Pythagorean identities<\/li>\r\n \t<li>Apply the product-to-sum formulas<\/li>\r\n \t<li>Use substitution to evaluate indefinite integrals containing trigonometric functions<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Trigonometric Integrals section, we will learn how to evaluate integrals that contain trigonometric functions raised to powers. Here we will review trigonometric identifies and how to use substitution to evaluate trigonometric integrals.\r\n<h2>Apply the Pythagorean Identities<\/h2>\r\n<strong>Pythagorean identities <\/strong>are equations involving trigonometric functions based on the properties of a right triangle.\r\n<table id=\"Table_07_01_01\" summary=\"&quot;Pythagorean Identities&quot; with three cells. First: sin(theta)^2 + cos(theta)^2 = 1. Second: 1 + cot(theta)^2 = csc(theta)^2. Third: 1 + tan(theta)^2 = sec(theta)^2.\"><colgroup> <col \/> <col \/> <col \/><\/colgroup>\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center;\" colspan=\"3\">Pythagorean Identities<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]{\\sin }^{2}\\theta +{\\cos }^{2}\\theta =1[\/latex]<\/td>\r\n<td>[latex]1+{\\tan }^{2}\\theta ={\\sec }^{2}\\theta[\/latex]<\/td>\r\n<td>[latex]1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div>We are especially interested in the first two Pythagorean Identities for the Trigonometric Integrals section.<\/div>\r\n<div>Notice the identity\u00a0[latex]{\\sin }^{2}\\theta +{\\cos }^{2}\\theta =1[\/latex] can be rearranged into the following useful alternative forms:<\/div>\r\n<ul>\r\n \t<li>[latex]{\\cos }^{2}\\theta =1-{\\sin }^{2}\\theta [\/latex]<\/li>\r\n \t<li>[latex]{\\sin }^{2}\\theta =1 - {\\cos }^{2}\\theta[\/latex]<\/li>\r\n<\/ul>\r\n<div>Notice the identity\u00a0[latex]1+{\\tan }^{2}\\theta ={\\sec }^{2}\\theta[\/latex] can be rearranged into the following useful alternative forms:<\/div>\r\n<ul>\r\n \t<li>[latex]{\\tan }^{2}\\theta = {\\sec }^{2}\\theta - 1[\/latex]<\/li>\r\n \t<li>[latex]{\\sec }^{2}\\theta = 1+{\\tan }^{2}\\theta[\/latex] (same as the original identity but sides of the equation are swapped)<\/li>\r\n<\/ul>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Pythagorean Identities to Rewrite Expressions<\/h3>\r\nSimplify [latex]\\frac{{\\sec }^{2}\\theta -1}{{\\sec }^{2}\\theta }[\/latex].\r\n\r\n[reveal-answer q=\"565941\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"565941\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{{\\sec }^{2}\\theta -1}{{\\sec }^{2}\\theta }&amp;=\\frac{\\left({\\tan }^{2}\\theta +1\\right)-1}{{\\sec }^{2}\\theta }&amp;&amp; {\\sec}^{2}\\theta ={\\tan }^{2}\\theta +1 \\\\ &amp;=\\frac{{\\tan }^{2}\\theta }{{\\sec }^{2}\\theta } \\\\ &amp;={\\tan }^{2}\\theta \\left(\\frac{1}{{\\sec }^{2}\\theta }\\right) \\\\ &amp;={\\tan }^{2}\\theta \\left({\\cos }^{2}\\theta \\right)&amp;&amp; {\\cos }^{2}\\theta =\\frac{1}{{\\sec }^{2}\\theta } \\\\ &amp;=\\left(\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta }\\right)\\left({\\cos }^{2}\\theta \\right)&amp;&amp; {\\tan}^{2}\\theta =\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta } \\\\ &amp;=\\left(\\frac{{\\sin }^{2}\\theta }{\\cancel{{\\cos }^{2}\\theta}}\\right)\\left(\\cancel{{\\cos }^{2}\\theta} \\right) \\\\ &amp;={\\sin }^{2}\\theta \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Pythagorean Identities to Rewrite Expressions<\/h3>\r\nRewrite the expression [latex]{\\sin}\\theta {\\cos^4}\\theta[\/latex] using the identity [latex]{\\cos }^{2}\\theta =1 - {\\sin }^{2}\\theta[\/latex].\r\n\r\n.\r\n\r\n[reveal-answer q=\"565958\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"565958\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\sin}\\theta {\\cos^4}\\theta&amp;={\\sin}\\theta ({cos^2}\\theta)^2 \\\\ &amp;={\\sin}\\theta (1-{\\sin^2}\\theta)^2 &amp;&amp; {\\cos }^{2}\\theta =1 - {\\sin }^{2}\\theta \\\\ &amp;= {\\sin}\\theta (1-2{\\sin^2}\\theta +{\\sin^4}\\theta) \\\\ &amp;={\\sin}\\theta-2{\\sin^3}\\theta+{\\sin^5}\\theta \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nRewrite the expression [latex]{\\tan^4}\\theta[\/latex] using the identity [latex]{\\tan }^{2}\\theta = {\\sec }^{2}\\theta - 1[\/latex].\r\n\r\n[reveal-answer q=\"813945\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"813945\"]\r\n<p style=\"text-align: left;\">[latex]{\\sec^4}\\theta-2{\\sec^2}\\theta+1[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nRewrite the expression [latex]{\\sin^2}\\theta {\\cos}\\theta[\/latex] using the identity [latex]{\\sin }^{2}\\theta =1 - {\\cos }^{2}\\theta[\/latex].\r\n\r\n[reveal-answer q=\"307900\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"307900\"]\r\n\r\n[latex]{\\cos}\\theta-{\\cos^3}\\theta[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div>\r\n<h2>Apply the Product-to-Sum Formulas<\/h2>\r\nSometimes, we may need to express the product of cosine and sine as a sum. We can use the <strong>product-to-sum formulas<\/strong>, which express products of trigonometric functions as sums. Note the following product-to-sum formulas.\r\n<ul>\r\n \t<li>\r\n<p style=\"text-align: left;\">[latex]\\cos \\alpha \\cos \\beta =\\frac{1}{2}\\left[\\cos \\left(\\alpha -\\beta \\right)+\\cos \\left(\\alpha +\\beta \\right)\\right][\/latex]<\/p>\r\n<\/li>\r\n \t<li>\r\n<p style=\"text-align: left;\">[latex]\\sin \\alpha \\cos \\beta =\\frac{1}{2}\\left[\\sin \\left(\\alpha +\\beta \\right)+\\sin \\left(\\alpha -\\beta \\right)\\right][\/latex]<\/p>\r\n<\/li>\r\n \t<li>\r\n<p style=\"text-align: left;\">[latex]\\sin \\alpha \\sin \\beta =\\frac{1}{2}\\left[\\cos \\left(\\alpha -\\beta \\right)-\\cos \\left(\\alpha +\\beta \\right)\\right][\/latex]<\/p>\r\n<\/li>\r\n \t<li>\r\n<p style=\"text-align: left;\">[latex]\\cos \\alpha \\sin \\beta =\\frac{1}{2}\\left[\\sin \\left(\\alpha +\\beta \\right)-\\sin \\left(\\alpha -\\beta \\right)\\right][\/latex]<\/p>\r\n<\/li>\r\n<\/ul>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Expanding Using a Product-To-Sum Formula<\/h3>\r\nWrite the following product of cosines as a sum: [latex]2\\cos \\left(\\frac{7x}{2}\\right)\\cos \\left(\\frac{3x}{2}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"440362\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"440362\"]\r\n\r\nWe begin by writing the formula for the product of cosines:\r\n<p style=\"text-align: center;\">[latex]\\cos \\alpha \\cos \\beta =\\frac{1}{2}\\left[\\cos \\left(\\alpha -\\beta \\right)+\\cos \\left(\\alpha +\\beta \\right)\\right][\/latex]<\/p>\r\nWe can then substitute the given angles into the formula and simplify.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2\\cos \\left(\\frac{7x}{2}\\right)\\cos \\left(\\frac{3x}{2}\\right)&amp;=\\left(2\\right)\\left(\\frac{1}{2}\\right)\\left[\\cos \\left(\\frac{7x}{2}-\\frac{3x}{2}\\right)+\\cos \\left(\\frac{7x}{2}+\\frac{3x}{2}\\right)\\right] \\\\ &amp;=\\left[\\cos \\left(\\frac{4x}{2}\\right)+\\cos \\left(\\frac{10x}{2}\\right)\\right] \\\\ &amp;=\\cos 2x+\\cos 5x \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Expanding Using a Product-To-Sum Formula<\/h3>\r\nExpress the following product as a sum containing only sine or cosine and no products: [latex]\\sin \\left(4\\theta \\right)\\cos \\left(2\\theta \\right)[\/latex].\r\n\r\n[reveal-answer q=\"816809\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"816809\"]\r\n\r\nWrite the formula for the product of sine and cosine. Then substitute the given values into the formula and simplify.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sin \\alpha \\cos \\beta &amp;=\\frac{1}{2}\\left[\\sin \\left(\\alpha +\\beta \\right)+\\sin \\left(\\alpha -\\beta \\right)\\right] \\\\ \\sin \\left(4\\theta \\right)\\cos \\left(2\\theta \\right)&amp;=\\frac{1}{2}\\left[\\sin \\left(4\\theta +2\\theta \\right)+\\sin \\left(4\\theta -2\\theta \\right)\\right] \\\\ &amp;=\\frac{1}{2}\\left[\\sin \\left(6\\theta \\right)+\\sin \\left(2\\theta \\right)\\right] \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>\u00a0Try It<\/h3>\r\nUse the product-to-sum formula to write the product as a sum or difference: [latex]\\cos \\left(2\\theta \\right)\\cos \\left(4\\theta \\right)[\/latex].\r\n\r\n[reveal-answer q=\"293192\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"293192\"][latex]\\frac{1}{2}\\left(\\cos 6\\theta +\\cos 2\\theta \\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]4642[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]4631[\/ohm_question]\r\n\r\n<\/div>\r\n<h2><span style=\"font-size: 1.15em;\">Use Substitution to Evaluate Indefinite Integrals Containing Trigonometric Functions<\/span><\/h2>\r\n<\/div>\r\n<div id=\"fs-id1170573413133\" class=\"bc-section section\">\r\n<p id=\"fs-id1170573363034\">We can generalize substitution using the following steps:<\/p>\r\n\r\n<ol id=\"fs-id1170570997711\">\r\n \t<li>Look carefully at the integrand and select an expression [latex]g(x)[\/latex] within the integrand to set equal to [latex]u[\/latex]. Let\u2019s select [latex]g(x).[\/latex] such that [latex]{g}^{\\prime }(x)[\/latex] is also part of the integrand.<\/li>\r\n \t<li>Substitute [latex]u=g(x)[\/latex] and [latex]du={g}^{\\prime }(x)dx.[\/latex] into the integral.<\/li>\r\n \t<li>We should now be able to evaluate the integral with respect to [latex]u[\/latex]. If the integral can\u2019t be evaluated we need to go back and select a different expression to use as [latex]u[\/latex].<\/li>\r\n \t<li>Evaluate the integral in terms of [latex]u[\/latex].<\/li>\r\n \t<li>Write the result in terms of [latex]x[\/latex] and the expression [latex]g(x).[\/latex]<\/li>\r\n<\/ol>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Applying Substitution to Integrals with Trigonometric Functions<\/h3>\r\nUse substitution to evaluate the integral [latex]\\displaystyle\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt.[\/latex]\r\n\r\n[reveal-answer q=\"fs-id1170572587721\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587721\"]\r\n<div id=\"qfs-id1170573437523\" class=\"hidden-answer\">\r\n<p id=\"fs-id1170573437523\">We know the derivative of [latex] \\cos t[\/latex] is [latex]\\text{\u2212} \\sin t,[\/latex] so we set [latex]u= \\cos t.[\/latex] Then [latex]du=\\text{\u2212} \\sin tdt.[\/latex] Substituting into the integral, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt=\\text{\u2212}\\displaystyle\\int \\frac{du}{{u}^{3}}.[\/latex]<\/p>\r\n<p id=\"fs-id1170573364147\">Evaluating the integral, we get<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ \\text{\u2212}{\\displaystyle\\int \\frac{du}{{u}^{3}}}\\hfill &amp; =\\text{\u2212} {\\displaystyle\\int {u}^{-3}du}\\hfill \\\\ &amp; =\\text{\u2212}(-\\frac{1}{2}){u}^{-2}+C.\\hfill \\end{array}[\/latex]<\/p>\r\n<p id=\"fs-id1170573409376\">Putting the answer back in terms of [latex]t[\/latex], we get<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc} {\\displaystyle\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt}\\hfill &amp; =\\frac{1}{2{u}^{2}}+C\\hfill \\\\ \\\\ &amp; =\\dfrac{1}{2{ \\cos }^{2}t}+C.\\hfill \\end{array}[\/latex] [\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Applying Substitution to Integrals with Trigonometric Functions<\/h3>\r\nUse substitution to evaluate the integral [latex]\\displaystyle\\int {\\tan}(t) {\\sec^2}(t) dt.[\/latex]\r\n\r\n[reveal-answer q=\"fs-id1170572587729\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587729\"]\r\n<div id=\"qfs-id1170573437523\" class=\"hidden-answer\">\r\n<p id=\"fs-id1170573437523\">We know the derivative of [latex] \\tan t[\/latex] is [latex]{sec^2}t,[\/latex] so we set [latex]u= \\tan t.[\/latex] Then [latex]du={\\sec^2}t dt.[\/latex] Substituting into the integral, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int {\\tan}(t) {\\sec^2}(t) dt=\\displaystyle\\int u du.[\/latex]<\/p>\r\n<p id=\"fs-id1170573364147\">Evaluating the integral, we get<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int u du}\\hfill &amp; = \\dfrac{u^2}{2} + C \\hfill \\end{array}[\/latex]<\/p>\r\n<p id=\"fs-id1170573409376\">Putting the answer back in terms of [latex]t[\/latex], we get<\/p>\r\n<p style=\"text-align: center;\">[latex]\\dfrac{u^2}{2} + C=\\dfrac{{\\tan^2}t}{2} + C[\/latex] [\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573733746\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse substitution to evaluate the integral [latex]\\displaystyle\\int \\frac{ \\cos t}{{ \\sin }^{2}t}dt.[\/latex]\r\n\r\n[reveal-answer q=\"fs-id1170572628519\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572628519\"]\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">[latex]-\\dfrac{1}{ \\sin t}+C[\/latex]<\/span>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571115891\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170571087102\">Use substitution to evaluate the indefinite integral [latex]\\displaystyle\\int { \\cos }^{3}(t) \\sin (t)dt.[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572628520\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572628520\"]\r\n<div id=\"qfs-id1170571285848\" class=\"hidden-answer\">\r\n<p id=\"fs-id1170571285848\">[latex]-\\dfrac{{ \\cos }^{4}t}{4}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]16228[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Apply the Pythagorean identities<\/li>\n<li>Apply the product-to-sum formulas<\/li>\n<li>Use substitution to evaluate indefinite integrals containing trigonometric functions<\/li>\n<\/ul>\n<\/div>\n<p>In the Trigonometric Integrals section, we will learn how to evaluate integrals that contain trigonometric functions raised to powers. Here we will review trigonometric identifies and how to use substitution to evaluate trigonometric integrals.<\/p>\n<h2>Apply the Pythagorean Identities<\/h2>\n<p><strong>Pythagorean identities <\/strong>are equations involving trigonometric functions based on the properties of a right triangle.<\/p>\n<table id=\"Table_07_01_01\" summary=\"&quot;Pythagorean Identities&quot; with three cells. First: sin(theta)^2 + cos(theta)^2 = 1. Second: 1 + cot(theta)^2 = csc(theta)^2. Third: 1 + tan(theta)^2 = sec(theta)^2.\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<thead>\n<tr>\n<th style=\"text-align: center;\" colspan=\"3\">Pythagorean Identities<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]{\\sin }^{2}\\theta +{\\cos }^{2}\\theta =1[\/latex]<\/td>\n<td>[latex]1+{\\tan }^{2}\\theta ={\\sec }^{2}\\theta[\/latex]<\/td>\n<td>[latex]1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div>We are especially interested in the first two Pythagorean Identities for the Trigonometric Integrals section.<\/div>\n<div>Notice the identity\u00a0[latex]{\\sin }^{2}\\theta +{\\cos }^{2}\\theta =1[\/latex] can be rearranged into the following useful alternative forms:<\/div>\n<ul>\n<li>[latex]{\\cos }^{2}\\theta =1-{\\sin }^{2}\\theta[\/latex]<\/li>\n<li>[latex]{\\sin }^{2}\\theta =1 - {\\cos }^{2}\\theta[\/latex]<\/li>\n<\/ul>\n<div>Notice the identity\u00a0[latex]1+{\\tan }^{2}\\theta ={\\sec }^{2}\\theta[\/latex] can be rearranged into the following useful alternative forms:<\/div>\n<ul>\n<li>[latex]{\\tan }^{2}\\theta = {\\sec }^{2}\\theta - 1[\/latex]<\/li>\n<li>[latex]{\\sec }^{2}\\theta = 1+{\\tan }^{2}\\theta[\/latex] (same as the original identity but sides of the equation are swapped)<\/li>\n<\/ul>\n<div class=\"textbox exercises\">\n<h3>Example: Using Pythagorean Identities to Rewrite Expressions<\/h3>\n<p>Simplify [latex]\\frac{{\\sec }^{2}\\theta -1}{{\\sec }^{2}\\theta }[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q565941\">Show Solution<\/span><\/p>\n<div id=\"q565941\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{{\\sec }^{2}\\theta -1}{{\\sec }^{2}\\theta }&=\\frac{\\left({\\tan }^{2}\\theta +1\\right)-1}{{\\sec }^{2}\\theta }&& {\\sec}^{2}\\theta ={\\tan }^{2}\\theta +1 \\\\ &=\\frac{{\\tan }^{2}\\theta }{{\\sec }^{2}\\theta } \\\\ &={\\tan }^{2}\\theta \\left(\\frac{1}{{\\sec }^{2}\\theta }\\right) \\\\ &={\\tan }^{2}\\theta \\left({\\cos }^{2}\\theta \\right)&& {\\cos }^{2}\\theta =\\frac{1}{{\\sec }^{2}\\theta } \\\\ &=\\left(\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta }\\right)\\left({\\cos }^{2}\\theta \\right)&& {\\tan}^{2}\\theta =\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta } \\\\ &=\\left(\\frac{{\\sin }^{2}\\theta }{\\cancel{{\\cos }^{2}\\theta}}\\right)\\left(\\cancel{{\\cos }^{2}\\theta} \\right) \\\\ &={\\sin }^{2}\\theta \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using Pythagorean Identities to Rewrite Expressions<\/h3>\n<p>Rewrite the expression [latex]{\\sin}\\theta {\\cos^4}\\theta[\/latex] using the identity [latex]{\\cos }^{2}\\theta =1 - {\\sin }^{2}\\theta[\/latex].<\/p>\n<p>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q565958\">Show Solution<\/span><\/p>\n<div id=\"q565958\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\sin}\\theta {\\cos^4}\\theta&={\\sin}\\theta ({cos^2}\\theta)^2 \\\\ &={\\sin}\\theta (1-{\\sin^2}\\theta)^2 && {\\cos }^{2}\\theta =1 - {\\sin }^{2}\\theta \\\\ &= {\\sin}\\theta (1-2{\\sin^2}\\theta +{\\sin^4}\\theta) \\\\ &={\\sin}\\theta-2{\\sin^3}\\theta+{\\sin^5}\\theta \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Rewrite the expression [latex]{\\tan^4}\\theta[\/latex] using the identity [latex]{\\tan }^{2}\\theta = {\\sec }^{2}\\theta - 1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q813945\">Show Solution<\/span><\/p>\n<div id=\"q813945\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">[latex]{\\sec^4}\\theta-2{\\sec^2}\\theta+1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Rewrite the expression [latex]{\\sin^2}\\theta {\\cos}\\theta[\/latex] using the identity [latex]{\\sin }^{2}\\theta =1 - {\\cos }^{2}\\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q307900\">Show Solution<\/span><\/p>\n<div id=\"q307900\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\cos}\\theta-{\\cos^3}\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<h2>Apply the Product-to-Sum Formulas<\/h2>\n<p>Sometimes, we may need to express the product of cosine and sine as a sum. We can use the <strong>product-to-sum formulas<\/strong>, which express products of trigonometric functions as sums. Note the following product-to-sum formulas.<\/p>\n<ul>\n<li>\n<p style=\"text-align: left;\">[latex]\\cos \\alpha \\cos \\beta =\\frac{1}{2}\\left[\\cos \\left(\\alpha -\\beta \\right)+\\cos \\left(\\alpha +\\beta \\right)\\right][\/latex]<\/p>\n<\/li>\n<li>\n<p style=\"text-align: left;\">[latex]\\sin \\alpha \\cos \\beta =\\frac{1}{2}\\left[\\sin \\left(\\alpha +\\beta \\right)+\\sin \\left(\\alpha -\\beta \\right)\\right][\/latex]<\/p>\n<\/li>\n<li>\n<p style=\"text-align: left;\">[latex]\\sin \\alpha \\sin \\beta =\\frac{1}{2}\\left[\\cos \\left(\\alpha -\\beta \\right)-\\cos \\left(\\alpha +\\beta \\right)\\right][\/latex]<\/p>\n<\/li>\n<li>\n<p style=\"text-align: left;\">[latex]\\cos \\alpha \\sin \\beta =\\frac{1}{2}\\left[\\sin \\left(\\alpha +\\beta \\right)-\\sin \\left(\\alpha -\\beta \\right)\\right][\/latex]<\/p>\n<\/li>\n<\/ul>\n<div class=\"textbox exercises\">\n<h3>Example: Expanding Using a Product-To-Sum Formula<\/h3>\n<p>Write the following product of cosines as a sum: [latex]2\\cos \\left(\\frac{7x}{2}\\right)\\cos \\left(\\frac{3x}{2}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q440362\">Show Solution<\/span><\/p>\n<div id=\"q440362\" class=\"hidden-answer\" style=\"display: none\">\n<p>We begin by writing the formula for the product of cosines:<\/p>\n<p style=\"text-align: center;\">[latex]\\cos \\alpha \\cos \\beta =\\frac{1}{2}\\left[\\cos \\left(\\alpha -\\beta \\right)+\\cos \\left(\\alpha +\\beta \\right)\\right][\/latex]<\/p>\n<p>We can then substitute the given angles into the formula and simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2\\cos \\left(\\frac{7x}{2}\\right)\\cos \\left(\\frac{3x}{2}\\right)&=\\left(2\\right)\\left(\\frac{1}{2}\\right)\\left[\\cos \\left(\\frac{7x}{2}-\\frac{3x}{2}\\right)+\\cos \\left(\\frac{7x}{2}+\\frac{3x}{2}\\right)\\right] \\\\ &=\\left[\\cos \\left(\\frac{4x}{2}\\right)+\\cos \\left(\\frac{10x}{2}\\right)\\right] \\\\ &=\\cos 2x+\\cos 5x \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Expanding Using a Product-To-Sum Formula<\/h3>\n<p>Express the following product as a sum containing only sine or cosine and no products: [latex]\\sin \\left(4\\theta \\right)\\cos \\left(2\\theta \\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q816809\">Show Solution<\/span><\/p>\n<div id=\"q816809\" class=\"hidden-answer\" style=\"display: none\">\n<p>Write the formula for the product of sine and cosine. Then substitute the given values into the formula and simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sin \\alpha \\cos \\beta &=\\frac{1}{2}\\left[\\sin \\left(\\alpha +\\beta \\right)+\\sin \\left(\\alpha -\\beta \\right)\\right] \\\\ \\sin \\left(4\\theta \\right)\\cos \\left(2\\theta \\right)&=\\frac{1}{2}\\left[\\sin \\left(4\\theta +2\\theta \\right)+\\sin \\left(4\\theta -2\\theta \\right)\\right] \\\\ &=\\frac{1}{2}\\left[\\sin \\left(6\\theta \\right)+\\sin \\left(2\\theta \\right)\\right] \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>\u00a0Try It<\/h3>\n<p>Use the product-to-sum formula to write the product as a sum or difference: [latex]\\cos \\left(2\\theta \\right)\\cos \\left(4\\theta \\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q293192\">Show Solution<\/span><\/p>\n<div id=\"q293192\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac{1}{2}\\left(\\cos 6\\theta +\\cos 2\\theta \\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm4642\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=4642&theme=oea&iframe_resize_id=ohm4642\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm4631\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=4631&theme=oea&iframe_resize_id=ohm4631\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2><span style=\"font-size: 1.15em;\">Use Substitution to Evaluate Indefinite Integrals Containing Trigonometric Functions<\/span><\/h2>\n<\/div>\n<div id=\"fs-id1170573413133\" class=\"bc-section section\">\n<p id=\"fs-id1170573363034\">We can generalize substitution using the following steps:<\/p>\n<ol id=\"fs-id1170570997711\">\n<li>Look carefully at the integrand and select an expression [latex]g(x)[\/latex] within the integrand to set equal to [latex]u[\/latex]. Let\u2019s select [latex]g(x).[\/latex] such that [latex]{g}^{\\prime }(x)[\/latex] is also part of the integrand.<\/li>\n<li>Substitute [latex]u=g(x)[\/latex] and [latex]du={g}^{\\prime }(x)dx.[\/latex] into the integral.<\/li>\n<li>We should now be able to evaluate the integral with respect to [latex]u[\/latex]. If the integral can\u2019t be evaluated we need to go back and select a different expression to use as [latex]u[\/latex].<\/li>\n<li>Evaluate the integral in terms of [latex]u[\/latex].<\/li>\n<li>Write the result in terms of [latex]x[\/latex] and the expression [latex]g(x).[\/latex]<\/li>\n<\/ol>\n<div class=\"textbox exercises\">\n<h3>Example: Applying Substitution to Integrals with Trigonometric Functions<\/h3>\n<p>Use substitution to evaluate the integral [latex]\\displaystyle\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572587721\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572587721\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"qfs-id1170573437523\" class=\"hidden-answer\">\n<p id=\"fs-id1170573437523\">We know the derivative of [latex]\\cos t[\/latex] is [latex]\\text{\u2212} \\sin t,[\/latex] so we set [latex]u= \\cos t.[\/latex] Then [latex]du=\\text{\u2212} \\sin tdt.[\/latex] Substituting into the integral, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt=\\text{\u2212}\\displaystyle\\int \\frac{du}{{u}^{3}}.[\/latex]<\/p>\n<p id=\"fs-id1170573364147\">Evaluating the integral, we get<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ \\text{\u2212}{\\displaystyle\\int \\frac{du}{{u}^{3}}}\\hfill & =\\text{\u2212} {\\displaystyle\\int {u}^{-3}du}\\hfill \\\\ & =\\text{\u2212}(-\\frac{1}{2}){u}^{-2}+C.\\hfill \\end{array}[\/latex]<\/p>\n<p id=\"fs-id1170573409376\">Putting the answer back in terms of [latex]t[\/latex], we get<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc} {\\displaystyle\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt}\\hfill & =\\frac{1}{2{u}^{2}}+C\\hfill \\\\ \\\\ & =\\dfrac{1}{2{ \\cos }^{2}t}+C.\\hfill \\end{array}[\/latex] <\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Applying Substitution to Integrals with Trigonometric Functions<\/h3>\n<p>Use substitution to evaluate the integral [latex]\\displaystyle\\int {\\tan}(t) {\\sec^2}(t) dt.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572587729\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572587729\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"qfs-id1170573437523\" class=\"hidden-answer\">\n<p id=\"fs-id1170573437523\">We know the derivative of [latex]\\tan t[\/latex] is [latex]{sec^2}t,[\/latex] so we set [latex]u= \\tan t.[\/latex] Then [latex]du={\\sec^2}t dt.[\/latex] Substituting into the integral, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int {\\tan}(t) {\\sec^2}(t) dt=\\displaystyle\\int u du.[\/latex]<\/p>\n<p id=\"fs-id1170573364147\">Evaluating the integral, we get<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int u du}\\hfill & = \\dfrac{u^2}{2} + C \\hfill \\end{array}[\/latex]<\/p>\n<p id=\"fs-id1170573409376\">Putting the answer back in terms of [latex]t[\/latex], we get<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{u^2}{2} + C=\\dfrac{{\\tan^2}t}{2} + C[\/latex] <\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573733746\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use substitution to evaluate the integral [latex]\\displaystyle\\int \\frac{ \\cos t}{{ \\sin }^{2}t}dt.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572628519\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572628519\" class=\"hidden-answer\" style=\"display: none\">\n<p><span style=\"font-size: 1rem; text-align: initial;\">[latex]-\\dfrac{1}{ \\sin t}+C[\/latex]<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571115891\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170571087102\">Use substitution to evaluate the indefinite integral [latex]\\displaystyle\\int { \\cos }^{3}(t) \\sin (t)dt.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572628520\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572628520\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"qfs-id1170571285848\" class=\"hidden-answer\">\n<p id=\"fs-id1170571285848\">[latex]-\\dfrac{{ \\cos }^{4}t}{4}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm16228\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=16228&theme=oea&iframe_resize_id=ohm16228&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-1984\">\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>Modification and Revision. <strong>Provided by<\/strong>: Lumen Learning. <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>College Algebra Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/a>. <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>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Modification and Revision\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra 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https:\/\/courses.lumenlearning.com\/precalculus\/\",\"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-1984","chapter","type-chapter","status-publish","hentry"],"part":535,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1984","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\/349141"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1984\/revisions"}],"predecessor-version":[{"id":2534,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1984\/revisions\/2534"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/535"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1984\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1984"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1984"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1984"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1984"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}