{"id":701,"date":"2025-06-20T17:06:09","date_gmt":"2025-06-20T17:06:09","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=701"},"modified":"2025-09-10T14:09:57","modified_gmt":"2025-09-10T14:09:57","slug":"trigonometric-integrals-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/trigonometric-integrals-learn-it-1\/","title":{"raw":"Trigonometric Integrals: Learn It 1","rendered":"Trigonometric Integrals: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Integrate expressions containing products and powers of sine and cosine<\/li>\r\n \t<li>Integrate expressions containing products and powers of tangent and secant<\/li>\r\n \t<li>Use reduction formulas to simplify and solve trigonometric integrals<\/li>\r\n \t<li>Integrate expressions containing square roots of sums or differences of squares<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Integrating Products and Powers of [latex]\\sin{x}[\/latex] and [latex]\\cos{x}[\/latex]<\/h2>\r\nIn this section, you'll learn how to integrate products of trigonometric functions\u2014skills that will be essential for advanced techniques like trigonometric substitution and coordinate systems you'll encounter later. We'll start with products of [latex]\\sin{x}[\/latex] and [latex]\\cos{x}[\/latex] and build from there.\r\n<p id=\"fs-id1165042469843\">A key idea behind the strategy used to integrate combinations of products and powers of [latex]\\sin{x}[\/latex] and [latex]\\cos{x}[\/latex] involves rewriting these expressions as sums and differences of integrals of the form [latex]\\displaystyle\\int\\sin^{j}x\\cos{x}dx[\/latex] or [latex]{\\displaystyle\\int}{\\cos}^{j}x\\sin{x}dx[\/latex]. After rewriting these integrals, we evaluate them using <em data-effect=\"italics\">u<\/em>-substitution.<\/p>\r\nBefore describing the general process in detail, let\u2019s take a look at a couple of examples.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165043250980\" data-type=\"example\">\r\n<div id=\"fs-id1165043100234\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042449637\" data-type=\"problem\">\r\n<p id=\"fs-id1165042320064\">Evaluate [latex]{\\displaystyle\\int}{\\cos}^{3}x\\sin{x}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1165043081174\" data-type=\"solution\">\r\n<p id=\"fs-id1165042358050\">Use [latex]u[\/latex] -substitution and let [latex]u=\\cos{x}[\/latex]. In this case, [latex]du=\\text{-}\\sin{x}dx[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1165043093877\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{\\displaystyle\\int}{\\cos}^{3}x\\sin{x}dx\\hfill &amp; =\\text{-}{\\displaystyle\\int}{u}^{3}du\\hfill \\\\ \\hfill &amp; =-\\frac{1}{4}{u}^{4}+C\\hfill \\\\ \\hfill &amp; =-\\frac{1}{4}{\\cos}^{4}x+C.\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165043250980\" data-type=\"example\">\r\n<div id=\"fs-id1165043100234\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042449637\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n\r\n<span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">Evaluate [latex]{\\displaystyle\\int}{\\sin}^{4}x\\cos{x}dx[\/latex].<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042964834\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1165043354058\" data-type=\"exercise\">\r\n\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1165043428113\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<div data-type=\"title\"><\/div>\r\n<p id=\"fs-id1165042364488\">Let [latex]u=\\sin{x}[\/latex].<\/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-id1165043051946\" data-type=\"solution\">\r\n<p id=\"fs-id1165040774446\" style=\"text-align: center;\">[latex]\\frac{1}{5}{\\sin}^{5}x+C[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/OW-JQPR36co?controls=0&amp;start=118&amp;end=159&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/3.2TrigonometricIntegrals118to159_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.2 Trigonometric Integrals\" here (opens in new window)<\/a>.\r\n\r\n<\/section>In addition to the technique of [latex]u-[\/latex] substitution, the problems in this section and the next make frequent use of the Pythagorean Identity and its implications for how to rewrite trigonometric functions in terms of other trigonometric functions. Here's a quick review of these essential relationships.\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p id=\"fs-id1170572169681\">For any angle [latex] x [\/latex]:<\/p>\r\n\r\n<center>[latex] \\sin^2 x + \\cos^2 x = 1 [\/latex]<\/center>Subtracting by [latex] \\sin^2 x [\/latex] allows a square power of cosine in terms of sine:\r\n\r\n<center>[latex] \\cos^2 x = 1-\\sin^2 x [\/latex]<\/center>Subtracting instead by [latex] \\cos^2 x [\/latex] allows a square power of sine to be written in terms of cosine:\r\n\r\n<center>[latex] \\sin^2 x = 1 -\\cos^2 x [\/latex]<\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165043078643\" data-type=\"problem\">\r\n<p id=\"fs-id1165042873576\">Evaluate [latex]{\\displaystyle\\int}{\\cos}^{2}x{\\sin}^{3}xdx[\/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-id1165042621331\" data-type=\"solution\">\r\n<p id=\"fs-id1165043035592\">To convert this integral to integrals of the form [latex]{\\displaystyle\\int}{\\cos}^{j}x\\sin{x}dx[\/latex], rewrite [latex]{\\sin}^{3}x={\\sin}^{2}x\\sin{x}[\/latex] and make the substitution [latex]{\\sin}^{2}x=1-{\\cos}^{2}x[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1165042880012\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}{\\displaystyle\\int}{\\cos}^{2}x{\\sin}^{3}xdx\\hfill &amp; ={\\displaystyle\\int}{\\cos}^{2}x\\left(1-{\\cos}^{2}x\\right)\\sin{x}dx\\hfill &amp; \\text{Let }u=\\cos{x};\\text{then }du=\\text{-}\\sin{x}dx.\\hfill \\\\ \\hfill &amp; =\\text{-}{\\displaystyle\\int}{u}^{2}\\left(1-{u}^{2}\\right)du\\hfill &amp; \\hfill \\\\ \\hfill &amp; ={\\displaystyle\\int}\\left({u}^{4}-{u}^{2}\\right)du\\hfill &amp; \\hfill \\\\ \\hfill &amp; =\\frac{1}{5}{u}^{5}-\\frac{1}{3}{u}^{3}+C\\hfill &amp; \\hfill \\\\ \\hfill &amp; =\\frac{1}{5}{\\cos}^{5}x-\\frac{1}{3}{\\cos}^{3}x+C.\\hfill &amp; \\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>\r\n<div id=\"fs-id1165042449637\" data-type=\"problem\">\r\n<div id=\"fs-id1165043281403\" data-type=\"example\">\r\n<div id=\"fs-id1165043257290\" data-type=\"exercise\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\r\n<div id=\"fs-id1165042449637\" data-type=\"problem\">\r\n<div id=\"fs-id1165043281403\" data-type=\"example\">\r\n<div id=\"fs-id1165043257290\" data-type=\"exercise\">\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">In the next example, we see the strategy that must be applied when there are only even powers of [latex]\\sin{x}[\/latex] and [latex]\\cos{x}[\/latex]. For integrals of this type, the identities<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165043066788\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\sin}^{2}x=\\frac{1}{2}-\\frac{1}{2}\\cos\\left(2x\\right)=\\frac{1-\\cos\\left(2x\\right)}{2}[\/latex]<\/div>\r\n<p id=\"fs-id1165042530463\">and<\/p>\r\n\r\n<div id=\"fs-id1165043331511\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\cos}^{2}x=\\frac{1}{2}+\\frac{1}{2}\\cos\\left(2x\\right)=\\frac{1+\\cos\\left(2x\\right)}{2}[\/latex]<\/div>\r\n<p id=\"fs-id1165042368394\">are invaluable. These identities are sometimes known as <span class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">power-reducing identities<\/em><\/span> and they may be derived from the double-angle identity [latex]\\cos\\left(2x\\right)={\\cos}^{2}x-{\\sin}^{2}x[\/latex] and the Pythagorean identity [latex]{\\cos}^{2}x+{\\sin}^{2}x=1[\/latex].<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165042632772\" data-type=\"problem\">\r\n<p id=\"fs-id1165042647060\">Evaluate [latex]{\\displaystyle\\int}{\\sin}^{2}xdx[\/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-id1165042965476\" data-type=\"solution\">\r\n<p id=\"fs-id1165043057271\">To evaluate this integral, let\u2019s use the trigonometric identity [latex]{\\sin}^{2}x=\\frac{1}{2}-\\frac{1}{2}\\cos\\left(2x\\right)[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1165043194450\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{\\displaystyle\\int}{\\sin}^{2}xdx\\hfill &amp; ={\\displaystyle\\int}\\left(\\frac{1}{2}-\\frac{1}{2}\\cos\\left(2x\\right)\\right)dx\\hfill \\\\ \\hfill &amp; =\\frac{1}{2}x-\\frac{1}{4}\\sin\\left(2x\\right)+C.\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]311293[\/ohm_question]<\/section><span style=\"font-size: 1rem; text-align: initial;\">The general process for integrating products of powers of [latex]\\sin{x}[\/latex] and [latex]\\cos{x}[\/latex] is summarized in the following set of guidelines.<\/span>\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>Problem-Solving Strategy: Integrating Products and Powers of [latex]\\sin{x}[\/latex] and [latex]\\cos{x}[\/latex]<\/strong>\r\n<ol id=\"fs-id1165043034337\" type=\"1\">\r\n \t<li>If [latex]k[\/latex] is odd, rewrite [latex]{\\sin}^{k}x={\\sin}^{k - 1}x\\sin{x}[\/latex] and use the identity [latex]{\\sin}^{2}x=1-{\\cos}^{2}x[\/latex] to rewrite [latex]{\\sin}^{k - 1}x[\/latex] in terms of [latex]\\cos{x}[\/latex]. Integrate using the substitution [latex]u=\\cos{x}[\/latex]. This substitution makes [latex]du=\\text{-}\\sin{x}dx[\/latex].<\/li>\r\n \t<li>If [latex]j[\/latex] is odd, rewrite [latex]{\\cos}^{j}x={\\cos}^{j - 1}x\\cos{x}[\/latex] and use the identity [latex]{\\cos}^{2}x=1-{\\sin}^{2}x[\/latex] to rewrite [latex]{\\cos}^{j - 1}x[\/latex] in terms of [latex]\\sin{x}[\/latex]. Integrate using the substitution [latex]u=\\sin{x}[\/latex]. This substitution makes [latex]du=\\cos{x}dx[\/latex]. (<em data-effect=\"italics\">Note<\/em>: If both [latex]j[\/latex] and [latex]k[\/latex] are odd, either strategy 1 or strategy 2 may be used.)<\/li>\r\n \t<li>If both [latex]j[\/latex] and [latex]k[\/latex] are even, use [latex]{\\sin}^{2}x=\\frac{1}{2}-\\frac{1}{2}\\cos\\left(2x\\right)[\/latex] and [latex]{\\cos}^{2}x=\\frac{1}{2}+\\frac{1}{2}\\cos\\left(2x\\right)[\/latex]. After applying these formulas, simplify and reapply strategies 1 through 3 as appropriate.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165043393170\" data-type=\"problem\">\r\n<p id=\"fs-id1165042907373\">Evaluate [latex]{\\displaystyle\\int}{\\cos}^{8}x{\\sin}^{5}xdx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558889\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n<div id=\"fs-id1165043350731\" data-type=\"solution\">\r\n<p id=\"fs-id1165042505411\">Since the power on [latex]\\sin{x}[\/latex] is odd, use strategy 1. Thus,<\/p>\r\n\r\n<div id=\"fs-id1165042349814\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccc}\\hfill {\\displaystyle\\int}{\\cos}^{8}x{\\sin}^{5}xdx&amp; ={\\displaystyle\\int}{\\cos}^{8}x{\\sin}^{4}x\\sin{x}dx\\hfill &amp; &amp; \\text{Break off }\\sin{x}.\\hfill \\\\ &amp; ={\\displaystyle\\int}{\\cos}^{8}x{\\left({\\sin}^{2}x\\right)}^{2}\\sin{x}dx\\hfill &amp; &amp; \\text{Rewrite }{\\sin}^{4}x={\\left({\\sin}^{2}x\\right)}^{2}.\\hfill \\\\ &amp; ={\\displaystyle\\int}{\\cos}^{8}x{\\left(1-{\\cos}^{2}x\\right)}^{2}\\sin{x}dx\\hfill &amp; &amp; \\text{Substitute }{\\sin}^{2}x=1-{\\cos}^{2}x.\\hfill \\\\ &amp; ={\\displaystyle\\int}{u}^{8}{\\left(1-{u}^{2}\\right)}^{2}\\left(\\text{-}du\\right)\\hfill &amp; &amp; \\text{Let }u=\\cos{x}\\text{ and }du=\\text{-}\\sin{x}dx.\\hfill \\\\ &amp; ={\\displaystyle\\int}\\left(\\text{-}{u}^{8}+2{u}^{10}-{u}^{12}\\right)du\\hfill &amp; &amp; \\text{Expand}.\\hfill \\\\ &amp; =-\\frac{1}{9}{u}^{9}+\\frac{2}{11}{u}^{11}-\\frac{1}{13}{u}^{13}+C\\hfill &amp; &amp; \\text{Evaluate the integral}.\\hfill \\\\ &amp; =-\\frac{1}{9}{\\cos}^{9}x+\\frac{2}{11}{\\cos}^{11}x-\\frac{1}{13}{\\cos}^{13}x+C.\\hfill &amp; &amp; \\text{Substitute }u=\\cos{x}.\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165042329519\" data-type=\"problem\">\r\n<p id=\"fs-id1165043179875\">Evaluate [latex]{\\displaystyle\\int}{\\sin}^{4}xdx[\/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-id1165042832305\" data-type=\"solution\">\r\n<p id=\"fs-id1165042832307\">Since the power on [latex]\\sin{x}[\/latex] is even [latex]\\left(k=4\\right)[\/latex] and the power on [latex]\\cos{x}[\/latex] is even [latex]\\left(j=0\\right)[\/latex], we must use strategy 3. Thus,<\/p>\r\n\r\n<div id=\"fs-id1165043089888\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccc}\\hfill {\\displaystyle\\int}{\\sin}^{4}xdx&amp; ={\\displaystyle\\int}{\\left({\\sin}^{2}x\\right)}^{2}dx\\hfill &amp; &amp; \\text{Rewrite }{\\sin}^{4}x={\\left({\\sin}^{2}x\\right)}^{2}.\\hfill \\\\ &amp; ={\\displaystyle\\int}{\\left(\\frac{1}{2}-\\frac{1}{2}\\cos\\left(2x\\right)\\right)}^{2}dx\\hfill &amp; &amp; \\text{Substitute }{\\sin}^{2}x=\\frac{1}{2}-\\frac{1}{2}\\cos\\left(2x\\right).\\hfill \\\\ &amp; ={\\displaystyle\\int}\\left(\\frac{1}{4}-\\frac{1}{2}\\cos\\left(2x\\right)+\\frac{1}{4}{\\cos}^{2}\\left(2x\\right)\\right)dx\\hfill &amp; &amp; \\text{Expand }{\\left(\\frac{1}{2}-\\frac{1}{2}\\cos\\left(2x\\right)\\right)}^{2}.\\hfill \\\\ &amp; ={\\displaystyle\\int}\\left(\\frac{1}{4}-\\frac{1}{2}\\cos\\left(2x\\right)+\\frac{1}{4}\\left(\\frac{1}{2}+\\frac{1}{2}\\cos\\left(4x\\right)\\right)\\right)dx.\\hfill &amp; &amp; \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043074297\">Since [latex]{\\cos}^{2}\\left(2x\\right)[\/latex] has an even power, substitute [latex]{\\cos}^{2}\\left(2x\\right)=\\frac{1}{2}+\\frac{1}{2}\\cos\\left(4x\\right)\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165042508466\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}={\\displaystyle\\int}\\left(\\frac{3}{8}-\\frac{1}{2}\\cos\\left(2x\\right)+\\frac{1}{8}\\cos\\left(4x\\right)\\right)dx\\hfill &amp; \\text{Simplify}.\\hfill \\\\ =\\frac{3}{8}x-\\frac{1}{4}\\sin\\left(2x\\right)+\\frac{1}{32}\\sin\\left(4x\\right)+C\\hfill &amp; \\text{Evaluate the integral}.\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>In some areas of physics, such as quantum mechanics, signal processing, and the computation of Fourier series, it is often necessary to integrate products that include [latex]\\sin\\left(ax\\right)[\/latex], [latex]\\sin\\left(bx\\right)[\/latex], [latex]\\cos\\left(ax\\right)[\/latex], and [latex]\\cos\\left(bx\\right)[\/latex]. These integrals are evaluated by applying trigonometric identities, as outlined in the following rule.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<div>\r\n<h3>integrating products of aines and cosines of different angles<\/h3>\r\nTo integrate products involving [latex]\\sin\\left(ax\\right)[\/latex], [latex]\\sin\\left(bx\\right)[\/latex], [latex]\\cos\\left(ax\\right)[\/latex], and [latex]\\cos\\left(bx\\right)[\/latex], use the substitutions\r\n\r\n<center>[latex]\\sin\\left(ax\\right)\\sin\\left(bx\\right)=\\frac{1}{2}\\cos\\left(\\left(a-b\\right)x\\right)-\\frac{1}{2}\\cos\\left(\\left(a+b\\right)x\\right)[\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex]\\sin\\left(ax\\right)\\cos\\left(bx\\right)=\\frac{1}{2}\\sin\\left(\\left(a-b\\right)x\\right)+\\frac{1}{2}\\sin\\left(\\left(a+b\\right)x\\right)[\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex]\\cos\\left(ax\\right)\\cos\\left(bx\\right)=\\frac{1}{2}\\cos\\left(\\left(a-b\\right)x\\right)+\\frac{1}{2}\\cos\\left(\\left(a+b\\right)x\\right)[\/latex]<\/center><\/div>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165043321359\" data-type=\"problem\">\r\n<p id=\"fs-id1165043272050\">Evaluate [latex]{\\displaystyle\\int}\\sin\\left(5x\\right)\\cos\\left(3x\\right)dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558799\"]\r\n<div id=\"fs-id1165043181525\" data-type=\"solution\">\r\n<p id=\"fs-id1165043181527\">Apply the identity [latex]\\sin\\left(5x\\right)\\cos\\left(3x\\right)=\\frac{1}{2}\\sin\\left(2x\\right)-\\frac{1}{2}\\cos\\left(8x\\right)[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1165042349275\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{\\displaystyle\\int}\\sin\\left(5x\\right)\\cos\\left(3x\\right)dx\\hfill &amp; ={\\displaystyle\\int}\\frac{1}{2}\\sin\\left(2x\\right)-\\frac{1}{2}\\cos\\left(8x\\right)dx\\hfill \\\\ \\hfill &amp; =-\\frac{1}{4}\\cos\\left(2x\\right)-\\frac{1}{16}\\sin\\left(8x\\right)+C.\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Integrate expressions containing products and powers of sine and cosine<\/li>\n<li>Integrate expressions containing products and powers of tangent and secant<\/li>\n<li>Use reduction formulas to simplify and solve trigonometric integrals<\/li>\n<li>Integrate expressions containing square roots of sums or differences of squares<\/li>\n<\/ul>\n<\/section>\n<h2>Integrating Products and Powers of [latex]\\sin{x}[\/latex] and [latex]\\cos{x}[\/latex]<\/h2>\n<p>In this section, you&#8217;ll learn how to integrate products of trigonometric functions\u2014skills that will be essential for advanced techniques like trigonometric substitution and coordinate systems you&#8217;ll encounter later. We&#8217;ll start with products of [latex]\\sin{x}[\/latex] and [latex]\\cos{x}[\/latex] and build from there.<\/p>\n<p id=\"fs-id1165042469843\">A key idea behind the strategy used to integrate combinations of products and powers of [latex]\\sin{x}[\/latex] and [latex]\\cos{x}[\/latex] involves rewriting these expressions as sums and differences of integrals of the form [latex]\\displaystyle\\int\\sin^{j}x\\cos{x}dx[\/latex] or [latex]{\\displaystyle\\int}{\\cos}^{j}x\\sin{x}dx[\/latex]. After rewriting these integrals, we evaluate them using <em data-effect=\"italics\">u<\/em>-substitution.<\/p>\n<p>Before describing the general process in detail, let\u2019s take a look at a couple of examples.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165043250980\" data-type=\"example\">\n<div id=\"fs-id1165043100234\" data-type=\"exercise\">\n<div id=\"fs-id1165042449637\" data-type=\"problem\">\n<p id=\"fs-id1165042320064\">Evaluate [latex]{\\displaystyle\\int}{\\cos}^{3}x\\sin{x}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558899\">Show Solution<\/button><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165043081174\" data-type=\"solution\">\n<p id=\"fs-id1165042358050\">Use [latex]u[\/latex] -substitution and let [latex]u=\\cos{x}[\/latex]. In this case, [latex]du=\\text{-}\\sin{x}dx[\/latex]. Thus,<\/p>\n<div id=\"fs-id1165043093877\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{\\displaystyle\\int}{\\cos}^{3}x\\sin{x}dx\\hfill & =\\text{-}{\\displaystyle\\int}{u}^{3}du\\hfill \\\\ \\hfill & =-\\frac{1}{4}{u}^{4}+C\\hfill \\\\ \\hfill & =-\\frac{1}{4}{\\cos}^{4}x+C.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div data-type=\"example\">\n<div data-type=\"exercise\">\n<div data-type=\"problem\">\n<div data-type=\"title\">\n<p><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">Evaluate [latex]{\\displaystyle\\int}{\\sin}^{4}x\\cos{x}dx[\/latex].<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042964834\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1165043354058\" data-type=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558897\">Hint<\/button><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165043428113\" data-type=\"commentary\" data-element-type=\"hint\">\n<div data-type=\"title\"><\/div>\n<p id=\"fs-id1165042364488\">Let [latex]u=\\sin{x}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558898\">Show Solution<\/button><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165043051946\" data-type=\"solution\">\n<p id=\"fs-id1165040774446\" style=\"text-align: center;\">[latex]\\frac{1}{5}{\\sin}^{5}x+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/OW-JQPR36co?controls=0&amp;start=118&amp;end=159&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/3.2TrigonometricIntegrals118to159_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.2 Trigonometric Integrals&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<p>In addition to the technique of [latex]u-[\/latex] substitution, the problems in this section and the next make frequent use of the Pythagorean Identity and its implications for how to rewrite trigonometric functions in terms of other trigonometric functions. Here&#8217;s a quick review of these essential relationships.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">\n<p id=\"fs-id1170572169681\">For any angle [latex]x[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\sin^2 x + \\cos^2 x = 1[\/latex]<\/div>\n<p>Subtracting by [latex]\\sin^2 x[\/latex] allows a square power of cosine in terms of sine:<\/p>\n<div style=\"text-align: center;\">[latex]\\cos^2 x = 1-\\sin^2 x[\/latex]<\/div>\n<p>Subtracting instead by [latex]\\cos^2 x[\/latex] allows a square power of sine to be written in terms of cosine:<\/p>\n<div style=\"text-align: center;\">[latex]\\sin^2 x = 1 -\\cos^2 x[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165043078643\" data-type=\"problem\">\n<p id=\"fs-id1165042873576\">Evaluate [latex]{\\displaystyle\\int}{\\cos}^{2}x{\\sin}^{3}xdx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558896\">Show Solution<\/button><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042621331\" data-type=\"solution\">\n<p id=\"fs-id1165043035592\">To convert this integral to integrals of the form [latex]{\\displaystyle\\int}{\\cos}^{j}x\\sin{x}dx[\/latex], rewrite [latex]{\\sin}^{3}x={\\sin}^{2}x\\sin{x}[\/latex] and make the substitution [latex]{\\sin}^{2}x=1-{\\cos}^{2}x[\/latex]. Thus,<\/p>\n<div id=\"fs-id1165042880012\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}{\\displaystyle\\int}{\\cos}^{2}x{\\sin}^{3}xdx\\hfill & ={\\displaystyle\\int}{\\cos}^{2}x\\left(1-{\\cos}^{2}x\\right)\\sin{x}dx\\hfill & \\text{Let }u=\\cos{x};\\text{then }du=\\text{-}\\sin{x}dx.\\hfill \\\\ \\hfill & =\\text{-}{\\displaystyle\\int}{u}^{2}\\left(1-{u}^{2}\\right)du\\hfill & \\hfill \\\\ \\hfill & ={\\displaystyle\\int}\\left({u}^{4}-{u}^{2}\\right)du\\hfill & \\hfill \\\\ \\hfill & =\\frac{1}{5}{u}^{5}-\\frac{1}{3}{u}^{3}+C\\hfill & \\hfill \\\\ \\hfill & =\\frac{1}{5}{\\cos}^{5}x-\\frac{1}{3}{\\cos}^{3}x+C.\\hfill & \\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div data-type=\"problem\">\n<div id=\"fs-id1165043281403\" data-type=\"example\">\n<div id=\"fs-id1165043257290\" data-type=\"exercise\"><\/div>\n<\/div>\n<\/div>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\n<div data-type=\"problem\">\n<div data-type=\"example\">\n<div data-type=\"exercise\">\n<p><span style=\"font-size: 1rem; text-align: initial;\">In the next example, we see the strategy that must be applied when there are only even powers of [latex]\\sin{x}[\/latex] and [latex]\\cos{x}[\/latex]. For integrals of this type, the identities<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043066788\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\sin}^{2}x=\\frac{1}{2}-\\frac{1}{2}\\cos\\left(2x\\right)=\\frac{1-\\cos\\left(2x\\right)}{2}[\/latex]<\/div>\n<p id=\"fs-id1165042530463\">and<\/p>\n<div id=\"fs-id1165043331511\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\cos}^{2}x=\\frac{1}{2}+\\frac{1}{2}\\cos\\left(2x\\right)=\\frac{1+\\cos\\left(2x\\right)}{2}[\/latex]<\/div>\n<p id=\"fs-id1165042368394\">are invaluable. These identities are sometimes known as <span class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">power-reducing identities<\/em><\/span> and they may be derived from the double-angle identity [latex]\\cos\\left(2x\\right)={\\cos}^{2}x-{\\sin}^{2}x[\/latex] and the Pythagorean identity [latex]{\\cos}^{2}x+{\\sin}^{2}x=1[\/latex].<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165042632772\" data-type=\"problem\">\n<p id=\"fs-id1165042647060\">Evaluate [latex]{\\displaystyle\\int}{\\sin}^{2}xdx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558892\">Show Solution<\/button><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042965476\" data-type=\"solution\">\n<p id=\"fs-id1165043057271\">To evaluate this integral, let\u2019s use the trigonometric identity [latex]{\\sin}^{2}x=\\frac{1}{2}-\\frac{1}{2}\\cos\\left(2x\\right)[\/latex]. Thus,<\/p>\n<div id=\"fs-id1165043194450\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{\\displaystyle\\int}{\\sin}^{2}xdx\\hfill & ={\\displaystyle\\int}\\left(\\frac{1}{2}-\\frac{1}{2}\\cos\\left(2x\\right)\\right)dx\\hfill \\\\ \\hfill & =\\frac{1}{2}x-\\frac{1}{4}\\sin\\left(2x\\right)+C.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm311293\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=311293&theme=lumen&iframe_resize_id=ohm311293&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<p><span style=\"font-size: 1rem; text-align: initial;\">The general process for integrating products of powers of [latex]\\sin{x}[\/latex] and [latex]\\cos{x}[\/latex] is summarized in the following set of guidelines.<\/span><\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>Problem-Solving Strategy: Integrating Products and Powers of [latex]\\sin{x}[\/latex] and [latex]\\cos{x}[\/latex]<\/strong><\/p>\n<ol id=\"fs-id1165043034337\" type=\"1\">\n<li>If [latex]k[\/latex] is odd, rewrite [latex]{\\sin}^{k}x={\\sin}^{k - 1}x\\sin{x}[\/latex] and use the identity [latex]{\\sin}^{2}x=1-{\\cos}^{2}x[\/latex] to rewrite [latex]{\\sin}^{k - 1}x[\/latex] in terms of [latex]\\cos{x}[\/latex]. Integrate using the substitution [latex]u=\\cos{x}[\/latex]. This substitution makes [latex]du=\\text{-}\\sin{x}dx[\/latex].<\/li>\n<li>If [latex]j[\/latex] is odd, rewrite [latex]{\\cos}^{j}x={\\cos}^{j - 1}x\\cos{x}[\/latex] and use the identity [latex]{\\cos}^{2}x=1-{\\sin}^{2}x[\/latex] to rewrite [latex]{\\cos}^{j - 1}x[\/latex] in terms of [latex]\\sin{x}[\/latex]. Integrate using the substitution [latex]u=\\sin{x}[\/latex]. This substitution makes [latex]du=\\cos{x}dx[\/latex]. (<em data-effect=\"italics\">Note<\/em>: If both [latex]j[\/latex] and [latex]k[\/latex] are odd, either strategy 1 or strategy 2 may be used.)<\/li>\n<li>If both [latex]j[\/latex] and [latex]k[\/latex] are even, use [latex]{\\sin}^{2}x=\\frac{1}{2}-\\frac{1}{2}\\cos\\left(2x\\right)[\/latex] and [latex]{\\cos}^{2}x=\\frac{1}{2}+\\frac{1}{2}\\cos\\left(2x\\right)[\/latex]. After applying these formulas, simplify and reapply strategies 1 through 3 as appropriate.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165043393170\" data-type=\"problem\">\n<p id=\"fs-id1165042907373\">Evaluate [latex]{\\displaystyle\\int}{\\cos}^{8}x{\\sin}^{5}xdx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558889\">Show Solution<\/button><\/p>\n<div id=\"q44558889\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165043350731\" data-type=\"solution\">\n<p id=\"fs-id1165042505411\">Since the power on [latex]\\sin{x}[\/latex] is odd, use strategy 1. Thus,<\/p>\n<div id=\"fs-id1165042349814\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccc}\\hfill {\\displaystyle\\int}{\\cos}^{8}x{\\sin}^{5}xdx& ={\\displaystyle\\int}{\\cos}^{8}x{\\sin}^{4}x\\sin{x}dx\\hfill & & \\text{Break off }\\sin{x}.\\hfill \\\\ & ={\\displaystyle\\int}{\\cos}^{8}x{\\left({\\sin}^{2}x\\right)}^{2}\\sin{x}dx\\hfill & & \\text{Rewrite }{\\sin}^{4}x={\\left({\\sin}^{2}x\\right)}^{2}.\\hfill \\\\ & ={\\displaystyle\\int}{\\cos}^{8}x{\\left(1-{\\cos}^{2}x\\right)}^{2}\\sin{x}dx\\hfill & & \\text{Substitute }{\\sin}^{2}x=1-{\\cos}^{2}x.\\hfill \\\\ & ={\\displaystyle\\int}{u}^{8}{\\left(1-{u}^{2}\\right)}^{2}\\left(\\text{-}du\\right)\\hfill & & \\text{Let }u=\\cos{x}\\text{ and }du=\\text{-}\\sin{x}dx.\\hfill \\\\ & ={\\displaystyle\\int}\\left(\\text{-}{u}^{8}+2{u}^{10}-{u}^{12}\\right)du\\hfill & & \\text{Expand}.\\hfill \\\\ & =-\\frac{1}{9}{u}^{9}+\\frac{2}{11}{u}^{11}-\\frac{1}{13}{u}^{13}+C\\hfill & & \\text{Evaluate the integral}.\\hfill \\\\ & =-\\frac{1}{9}{\\cos}^{9}x+\\frac{2}{11}{\\cos}^{11}x-\\frac{1}{13}{\\cos}^{13}x+C.\\hfill & & \\text{Substitute }u=\\cos{x}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165042329519\" data-type=\"problem\">\n<p id=\"fs-id1165043179875\">Evaluate [latex]{\\displaystyle\\int}{\\sin}^{4}xdx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558879\">Show Solution<\/button><\/p>\n<div id=\"q44558879\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042832305\" data-type=\"solution\">\n<p id=\"fs-id1165042832307\">Since the power on [latex]\\sin{x}[\/latex] is even [latex]\\left(k=4\\right)[\/latex] and the power on [latex]\\cos{x}[\/latex] is even [latex]\\left(j=0\\right)[\/latex], we must use strategy 3. Thus,<\/p>\n<div id=\"fs-id1165043089888\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccc}\\hfill {\\displaystyle\\int}{\\sin}^{4}xdx& ={\\displaystyle\\int}{\\left({\\sin}^{2}x\\right)}^{2}dx\\hfill & & \\text{Rewrite }{\\sin}^{4}x={\\left({\\sin}^{2}x\\right)}^{2}.\\hfill \\\\ & ={\\displaystyle\\int}{\\left(\\frac{1}{2}-\\frac{1}{2}\\cos\\left(2x\\right)\\right)}^{2}dx\\hfill & & \\text{Substitute }{\\sin}^{2}x=\\frac{1}{2}-\\frac{1}{2}\\cos\\left(2x\\right).\\hfill \\\\ & ={\\displaystyle\\int}\\left(\\frac{1}{4}-\\frac{1}{2}\\cos\\left(2x\\right)+\\frac{1}{4}{\\cos}^{2}\\left(2x\\right)\\right)dx\\hfill & & \\text{Expand }{\\left(\\frac{1}{2}-\\frac{1}{2}\\cos\\left(2x\\right)\\right)}^{2}.\\hfill \\\\ & ={\\displaystyle\\int}\\left(\\frac{1}{4}-\\frac{1}{2}\\cos\\left(2x\\right)+\\frac{1}{4}\\left(\\frac{1}{2}+\\frac{1}{2}\\cos\\left(4x\\right)\\right)\\right)dx.\\hfill & & \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043074297\">Since [latex]{\\cos}^{2}\\left(2x\\right)[\/latex] has an even power, substitute [latex]{\\cos}^{2}\\left(2x\\right)=\\frac{1}{2}+\\frac{1}{2}\\cos\\left(4x\\right)\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1165042508466\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}={\\displaystyle\\int}\\left(\\frac{3}{8}-\\frac{1}{2}\\cos\\left(2x\\right)+\\frac{1}{8}\\cos\\left(4x\\right)\\right)dx\\hfill & \\text{Simplify}.\\hfill \\\\ =\\frac{3}{8}x-\\frac{1}{4}\\sin\\left(2x\\right)+\\frac{1}{32}\\sin\\left(4x\\right)+C\\hfill & \\text{Evaluate the integral}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<p>In some areas of physics, such as quantum mechanics, signal processing, and the computation of Fourier series, it is often necessary to integrate products that include [latex]\\sin\\left(ax\\right)[\/latex], [latex]\\sin\\left(bx\\right)[\/latex], [latex]\\cos\\left(ax\\right)[\/latex], and [latex]\\cos\\left(bx\\right)[\/latex]. These integrals are evaluated by applying trigonometric identities, as outlined in the following rule.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<div>\n<h3>integrating products of aines and cosines of different angles<\/h3>\n<p>To integrate products involving [latex]\\sin\\left(ax\\right)[\/latex], [latex]\\sin\\left(bx\\right)[\/latex], [latex]\\cos\\left(ax\\right)[\/latex], and [latex]\\cos\\left(bx\\right)[\/latex], use the substitutions<\/p>\n<div style=\"text-align: center;\">[latex]\\sin\\left(ax\\right)\\sin\\left(bx\\right)=\\frac{1}{2}\\cos\\left(\\left(a-b\\right)x\\right)-\\frac{1}{2}\\cos\\left(\\left(a+b\\right)x\\right)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\sin\\left(ax\\right)\\cos\\left(bx\\right)=\\frac{1}{2}\\sin\\left(\\left(a-b\\right)x\\right)+\\frac{1}{2}\\sin\\left(\\left(a+b\\right)x\\right)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\cos\\left(ax\\right)\\cos\\left(bx\\right)=\\frac{1}{2}\\cos\\left(\\left(a-b\\right)x\\right)+\\frac{1}{2}\\cos\\left(\\left(a+b\\right)x\\right)[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165043321359\" data-type=\"problem\">\n<p id=\"fs-id1165043272050\">Evaluate [latex]{\\displaystyle\\int}\\sin\\left(5x\\right)\\cos\\left(3x\\right)dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558799\">Show Solution<\/button><\/p>\n<div id=\"q44558799\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165043181525\" data-type=\"solution\">\n<p id=\"fs-id1165043181527\">Apply the identity [latex]\\sin\\left(5x\\right)\\cos\\left(3x\\right)=\\frac{1}{2}\\sin\\left(2x\\right)-\\frac{1}{2}\\cos\\left(8x\\right)[\/latex]. Thus,<\/p>\n<div id=\"fs-id1165042349275\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{\\displaystyle\\int}\\sin\\left(5x\\right)\\cos\\left(3x\\right)dx\\hfill & ={\\displaystyle\\int}\\frac{1}{2}\\sin\\left(2x\\right)-\\frac{1}{2}\\cos\\left(8x\\right)dx\\hfill \\\\ \\hfill & =-\\frac{1}{4}\\cos\\left(2x\\right)-\\frac{1}{16}\\sin\\left(8x\\right)+C.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":12,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":541,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/701"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":7,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/701\/revisions"}],"predecessor-version":[{"id":2285,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/701\/revisions\/2285"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/541"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/701\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=701"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=701"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=701"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=701"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}