{"id":1011,"date":"2025-06-20T17:28:37","date_gmt":"2025-06-20T17:28:37","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1011"},"modified":"2025-08-28T13:49:25","modified_gmt":"2025-08-28T13:49:25","slug":"polar-coordinates-and-conic-sections-background-youll-need-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/polar-coordinates-and-conic-sections-background-youll-need-3\/","title":{"raw":"Polar Coordinates and Conic Sections: Background You'll Need 3","rendered":"Polar Coordinates and Conic Sections: Background You&#8217;ll Need 3"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Apply substitution methods to find definite integrals<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Substitution for Definite Integrals<\/h2>\r\n<p id=\"fs-id1170573387843\">Substitution can be used with definite integrals. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>substitution with definite integrals<\/h3>\r\n<p id=\"fs-id1170570999523\">Let [latex]u=g(x)[\/latex] and let [latex]{g}^{\\text{\u2032}}[\/latex] be continuous over an interval [latex]\\left[a,b\\right],[\/latex] and let [latex]f[\/latex] be continuous over the range of [latex]u=g(x).[\/latex] Then,<\/p>\r\n\r\n<div id=\"fs-id1170571189407\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{a}^{b}f(g(x)){g}^{\\prime }(x)dx={\\displaystyle\\int }_{g(a)}^{g(b)}f(u)du[\/latex]<\/div>\r\n<\/div>\r\n<\/section>\r\n<p id=\"fs-id1170573502228\">Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if [latex]F(x)[\/latex] is an antiderivative of [latex]f(x),[\/latex] we have<\/p>\r\n\r\n<div id=\"fs-id1170571168095\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int f(g(x)){g}^{\\prime }(x)dx=F(g(x))+C[\/latex]<\/div>\r\n<p id=\"fs-id1170573420715\">Then<\/p>\r\n\r\n<div id=\"fs-id1170573388911\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{a}^{b}f\\left[g(x)\\right]{g}^{\\prime }(x)dx\\hfill &amp; ={F(g(x))|}_{x=a}^{x=b}\\hfill \\\\ &amp; =F(g(b))-F(g(a))\\hfill \\\\ &amp; ={F(u)|}_{u=g(a)}^{u=g(b)}\\hfill \\\\ \\\\ \\\\ &amp; ={\\displaystyle\\int }_{g(a)}^{g(b)}f(u)du,\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170571100165\">and we have the desired result.<\/p>\r\n\r\n<section class=\"textbox example\">Use substitution to evaluate [latex]{\\displaystyle\\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx.[\/latex]\r\n\r\n[reveal-answer q=\"fs-id1170572587723\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587723\"]\r\n<p id=\"fs-id1170573440252\">Let [latex]u=1+2{x}^{3},[\/latex] so [latex]du=6{x}^{2}dx.[\/latex]<\/p>\r\nSince the original function includes one factor of [latex]x^2[\/latex]\u00a0and [latex]du=6{x}^{2}dx,[\/latex] multiply both sides of the <em>du<\/em> equation by [latex]\\frac{1}{6}.[\/latex]\r\n\r\nThen,\r\n<div id=\"fs-id1170573362498\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}du\\hfill &amp; =\\hfill &amp; 6{x}^{2}dx\\hfill \\\\ \\frac{1}{6}du\\hfill &amp; =\\hfill &amp; {x}^{2}dx.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170571157085\">To adjust the limits of integration, note that when [latex]x=0,u=1+2(0)=1,[\/latex] and when [latex]x=1,u=1+2(1)=3.[\/latex]<\/p>\r\nThen,\r\n<div id=\"fs-id1170571303367\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx=\\frac{1}{6}{\\displaystyle\\int }_{1}^{3}{u}^{5}du.[\/latex]<\/div>\r\n<p id=\"fs-id1170571285285\">Evaluating this expression, we get<\/p>\r\n\r\n<div id=\"fs-id1170571285289\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\frac{1}{6}{\\displaystyle\\int }_{1}^{3}{u}^{5}du\\hfill &amp; =(\\frac{1}{6})(\\frac{{u}^{6}}{6}){|}_{1}^{3}\\hfill \\\\ &amp; =\\frac{1}{36}\\left[{(3)}^{6}-{(1)}^{6}\\right]\\hfill \\\\ &amp; =\\frac{182}{9}.\\hfill \\end{array}[\/latex]<\/div>\r\nWatch the following video to see the worked solution to this example.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Ak_y3lsBNfE?controls=0&amp;start=519&amp;end=664&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/Calculus1+Videos\/5.5Substitution519to664_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.5 Substitution\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\">Use substitution to evaluate [latex]{\\displaystyle\\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx.[\/latex]\r\n\r\n[reveal-answer q=\"fs-id1170572587724\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587724\"]\r\n<p id=\"fs-id1170571347206\">Let [latex]u=4{x}^{3}+3.[\/latex] Then, [latex]du=8xdx.[\/latex]<\/p>\r\nTo adjust the limits of integration, we note that when [latex]x=0,u=3,[\/latex] and when [latex]x=1,u=7.[\/latex]\r\n\r\nSo our substitution gives,\r\n<div id=\"fs-id1170571052992\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx\\hfill &amp; =\\frac{1}{8}{\\displaystyle\\int }_{3}^{7}{e}^{u}du\\hfill \\\\ \\\\ &amp; =\\frac{1}{8}{e}^{u}{|}_{3}^{7}\\hfill \\\\ &amp; =\\frac{{e}^{7}-{e}^{3}}{8}\\hfill \\\\ &amp; \\approx 134.568.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/section>\r\n<p id=\"fs-id1170571289709\">Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we can apply substitution. Also, we have the option of replacing the original expression for [latex]u[\/latex] after we find the antiderivative, which means that we do not have to change the limits of integration. These two approaches are shown in the following examples.<\/p>\r\n\r\n<section class=\"textbox example\">Use substitution to evaluate [latex]{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}{ \\cos }^{2}\\theta d\\theta .[\/latex]\r\n\r\n[reveal-answer q=\"fs-id1170572587725\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587725\"]\r\n<p id=\"fs-id1170571030577\">Let us first use a trigonometric identity to rewrite the integral. The trig identity [latex]{ \\cos }^{2}\\theta =\\frac{1+ \\cos 2\\theta }{2}[\/latex] allows us to rewrite the integral as<\/p>\r\n\r\n<div id=\"fs-id1170571131164\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}{ \\cos }^{2}\\theta d\\theta ={\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}\\frac{1+ \\cos 2\\theta }{2}d\\theta .[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170573351678\">Then,<\/p>\r\n\r\n<div id=\"fs-id1170573581363\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}\\left(\\frac{1+ \\cos 2\\theta }{2}\\right)d\\theta \\hfill &amp; ={\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}\\left(\\frac{1}{2}+\\frac{1}{2} \\cos 2\\theta \\right)d\\theta \\hfill \\\\ \\\\ \\\\ &amp; =\\frac{1}{2}{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}d\\theta + \\frac{1}{2}{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2} \\cos 2\\theta d\\theta .\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571374921\">We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let [latex]u=2\\theta .[\/latex] Then, [latex]du=2d\\theta ,[\/latex] or [latex]\\frac{1}{2}du=d\\theta .[\/latex] Also, when [latex]\\theta =0,u=0,[\/latex] and when [latex]\\theta =\\pi \\text{\/}2,u=\\pi .[\/latex] Expressing the second integral in terms of [latex]u[\/latex], we have<\/p>\r\n\r\n<div id=\"fs-id1170571244772\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ \\frac{1}{2}{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}d\\theta +\\frac{1}{2}{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2} \\cos 2\\theta d\\theta \\hfill &amp; =\\frac{1}{2}{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}d\\theta +\\frac{1}{2}(\\frac{1}{2}){\\displaystyle\\int }_{0}^{\\pi } \\cos udu\\hfill \\\\ &amp; =\\frac{\\theta }{2}{|}_{\\theta =0}^{\\theta =\\pi \\text{\/}2}+\\frac{1}{4} \\sin u{|}_{u=0}^{u=\\theta }\\hfill \\\\ &amp; =(\\frac{\\pi }{4}-0)+(0-0)=\\frac{\\pi }{4}.\\hfill \\end{array}[\/latex]<\/div>\r\nWatch the following video to see the worked solution to this example.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Ak_y3lsBNfE?controls=0&amp;start=774&amp;end=978&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/Calculus1+Videos\/5.5Substitution774to978_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.5 Substitution\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]211747[\/ohm_question]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Apply substitution methods to find definite integrals<\/li>\n<\/ul>\n<\/section>\n<h2>Substitution for Definite Integrals<\/h2>\n<p id=\"fs-id1170573387843\">Substitution can be used with definite integrals. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>substitution with definite integrals<\/h3>\n<p id=\"fs-id1170570999523\">Let [latex]u=g(x)[\/latex] and let [latex]{g}^{\\text{\u2032}}[\/latex] be continuous over an interval [latex]\\left[a,b\\right],[\/latex] and let [latex]f[\/latex] be continuous over the range of [latex]u=g(x).[\/latex] Then,<\/p>\n<div id=\"fs-id1170571189407\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{a}^{b}f(g(x)){g}^{\\prime }(x)dx={\\displaystyle\\int }_{g(a)}^{g(b)}f(u)du[\/latex]<\/div>\n<\/div>\n<\/section>\n<p id=\"fs-id1170573502228\">Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if [latex]F(x)[\/latex] is an antiderivative of [latex]f(x),[\/latex] we have<\/p>\n<div id=\"fs-id1170571168095\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int f(g(x)){g}^{\\prime }(x)dx=F(g(x))+C[\/latex]<\/div>\n<p id=\"fs-id1170573420715\">Then<\/p>\n<div id=\"fs-id1170573388911\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{a}^{b}f\\left[g(x)\\right]{g}^{\\prime }(x)dx\\hfill & ={F(g(x))|}_{x=a}^{x=b}\\hfill \\\\ & =F(g(b))-F(g(a))\\hfill \\\\ & ={F(u)|}_{u=g(a)}^{u=g(b)}\\hfill \\\\ \\\\ \\\\ & ={\\displaystyle\\int }_{g(a)}^{g(b)}f(u)du,\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571100165\">and we have the desired result.<\/p>\n<section class=\"textbox example\">Use substitution to evaluate [latex]{\\displaystyle\\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572587723\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572587723\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573440252\">Let [latex]u=1+2{x}^{3},[\/latex] so [latex]du=6{x}^{2}dx.[\/latex]<\/p>\n<p>Since the original function includes one factor of [latex]x^2[\/latex]\u00a0and [latex]du=6{x}^{2}dx,[\/latex] multiply both sides of the <em>du<\/em> equation by [latex]\\frac{1}{6}.[\/latex]<\/p>\n<p>Then,<\/p>\n<div id=\"fs-id1170573362498\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}du\\hfill & =\\hfill & 6{x}^{2}dx\\hfill \\\\ \\frac{1}{6}du\\hfill & =\\hfill & {x}^{2}dx.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571157085\">To adjust the limits of integration, note that when [latex]x=0,u=1+2(0)=1,[\/latex] and when [latex]x=1,u=1+2(1)=3.[\/latex]<\/p>\n<p>Then,<\/p>\n<div id=\"fs-id1170571303367\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx=\\frac{1}{6}{\\displaystyle\\int }_{1}^{3}{u}^{5}du.[\/latex]<\/div>\n<p id=\"fs-id1170571285285\">Evaluating this expression, we get<\/p>\n<div id=\"fs-id1170571285289\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\frac{1}{6}{\\displaystyle\\int }_{1}^{3}{u}^{5}du\\hfill & =(\\frac{1}{6})(\\frac{{u}^{6}}{6}){|}_{1}^{3}\\hfill \\\\ & =\\frac{1}{36}\\left[{(3)}^{6}-{(1)}^{6}\\right]\\hfill \\\\ & =\\frac{182}{9}.\\hfill \\end{array}[\/latex]<\/div>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Ak_y3lsBNfE?controls=0&amp;start=519&amp;end=664&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/Calculus1+Videos\/5.5Substitution519to664_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.5 Substitution&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Use substitution to evaluate [latex]{\\displaystyle\\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572587724\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572587724\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571347206\">Let [latex]u=4{x}^{3}+3.[\/latex] Then, [latex]du=8xdx.[\/latex]<\/p>\n<p>To adjust the limits of integration, we note that when [latex]x=0,u=3,[\/latex] and when [latex]x=1,u=7.[\/latex]<\/p>\n<p>So our substitution gives,<\/p>\n<div id=\"fs-id1170571052992\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx\\hfill & =\\frac{1}{8}{\\displaystyle\\int }_{3}^{7}{e}^{u}du\\hfill \\\\ \\\\ & =\\frac{1}{8}{e}^{u}{|}_{3}^{7}\\hfill \\\\ & =\\frac{{e}^{7}-{e}^{3}}{8}\\hfill \\\\ & \\approx 134.568.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<p id=\"fs-id1170571289709\">Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we can apply substitution. Also, we have the option of replacing the original expression for [latex]u[\/latex] after we find the antiderivative, which means that we do not have to change the limits of integration. These two approaches are shown in the following examples.<\/p>\n<section class=\"textbox example\">Use substitution to evaluate [latex]{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}{ \\cos }^{2}\\theta d\\theta .[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572587725\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572587725\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571030577\">Let us first use a trigonometric identity to rewrite the integral. The trig identity [latex]{ \\cos }^{2}\\theta =\\frac{1+ \\cos 2\\theta }{2}[\/latex] allows us to rewrite the integral as<\/p>\n<div id=\"fs-id1170571131164\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}{ \\cos }^{2}\\theta d\\theta ={\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}\\frac{1+ \\cos 2\\theta }{2}d\\theta .[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573351678\">Then,<\/p>\n<div id=\"fs-id1170573581363\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}\\left(\\frac{1+ \\cos 2\\theta }{2}\\right)d\\theta \\hfill & ={\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}\\left(\\frac{1}{2}+\\frac{1}{2} \\cos 2\\theta \\right)d\\theta \\hfill \\\\ \\\\ \\\\ & =\\frac{1}{2}{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}d\\theta + \\frac{1}{2}{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2} \\cos 2\\theta d\\theta .\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571374921\">We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let [latex]u=2\\theta .[\/latex] Then, [latex]du=2d\\theta ,[\/latex] or [latex]\\frac{1}{2}du=d\\theta .[\/latex] Also, when [latex]\\theta =0,u=0,[\/latex] and when [latex]\\theta =\\pi \\text{\/}2,u=\\pi .[\/latex] Expressing the second integral in terms of [latex]u[\/latex], we have<\/p>\n<div id=\"fs-id1170571244772\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ \\frac{1}{2}{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}d\\theta +\\frac{1}{2}{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2} \\cos 2\\theta d\\theta \\hfill & =\\frac{1}{2}{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}d\\theta +\\frac{1}{2}(\\frac{1}{2}){\\displaystyle\\int }_{0}^{\\pi } \\cos udu\\hfill \\\\ & =\\frac{\\theta }{2}{|}_{\\theta =0}^{\\theta =\\pi \\text{\/}2}+\\frac{1}{4} \\sin u{|}_{u=0}^{u=\\theta }\\hfill \\\\ & =(\\frac{\\pi }{4}-0)+(0-0)=\\frac{\\pi }{4}.\\hfill \\end{array}[\/latex]<\/div>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Ak_y3lsBNfE?controls=0&amp;start=774&amp;end=978&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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. 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