{"id":813,"date":"2021-06-02T23:38:35","date_gmt":"2021-06-02T23:38:35","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=813"},"modified":"2022-03-19T04:02:53","modified_gmt":"2022-03-19T04:02:53","slug":"integrating-square-roots-of-binomials","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/integrating-square-roots-of-binomials\/","title":{"raw":"Integrating Square Roots of Binomials","rendered":"Integrating Square Roots of Binomials"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve integration problems involving the square root of a sum or difference of two squares<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1165041917030\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Integrals Involving [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex]<\/h2>\r\n<p id=\"fs-id1165041987630\">Before developing a general strategy for integrals containing [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex], consider the integral [latex]\\displaystyle\\int \\sqrt{9-{x}^{2}}dx[\/latex]. This integral cannot be evaluated using any of the techniques we have discussed so far. However, if we make the substitution [latex]x=3\\sin\\theta [\/latex], we have [latex]dx=3\\cos\\theta d\\theta [\/latex]. After substituting into the integral, we have<\/p>\r\n\r\n<div id=\"fs-id1165042199492\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\sqrt{9-{x}^{2}}dx={\\displaystyle\\int }^{\\text{ }}\\sqrt{9-{\\left(3\\sin\\theta \\right)}^{2}}3\\cos\\theta d\\theta [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042090183\">After simplifying, we have<\/p>\r\n\r\n<div id=\"fs-id1165042101997\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}\\sqrt{9-{x}^{2}}dx={\\displaystyle\\int }^{\\text{ }}9\\sqrt{1-{\\sin}^{2}\\theta }\\cos\\theta d\\theta [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041909368\">Letting [latex]1-{\\sin}^{2}\\theta ={\\cos}^{2}\\theta [\/latex], we now have<\/p>\r\n\r\n<div id=\"fs-id1165040752105\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}\\sqrt{9-{x}^{2}}dx={\\displaystyle\\int }^{\\text{ }}9\\sqrt{{\\cos}^{2}\\theta }\\cos\\theta d\\theta [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041781797\">Assuming that [latex]\\cos\\theta \\ge 0[\/latex], we have<\/p>\r\n\r\n<div id=\"fs-id1165042007513\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}\\sqrt{9-{x}^{2}}dx={\\displaystyle\\int }^{\\text{ }}9{\\cos}^{2}\\theta d\\theta [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042126138\">At this point, we can evaluate the integral using the techniques developed for integrating powers and products of trigonometric functions. Before completing this example, let\u2019s take a look at the general theory behind this idea.<\/p>\r\n<p id=\"fs-id1165042128862\">To evaluate integrals involving [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex], we make the substitution [latex]x=a\\sin\\theta [\/latex] and [latex]dx=a\\cos\\theta [\/latex]. To see that this actually makes sense, consider the following argument: The domain of [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex] is [latex]\\left[\\text{-}a,a\\right][\/latex]. Thus, [latex]\\text{-}a\\le x\\le a[\/latex]. Consequently, [latex]-1\\le \\frac{x}{a}\\le 1[\/latex]. Since the range of [latex]\\sin{x}[\/latex] over [latex]\\left[\\text{-}\\frac{\\pi}{2},\\frac{\\pi}{2}\\right][\/latex] is [latex]\\left[-1,1\\right][\/latex], there is a unique angle [latex]\\theta [\/latex] satisfying [latex]\\text{-}\\frac{\\pi}{2}\\le \\theta \\le \\frac{\\pi}{2}[\/latex] so that [latex]\\sin\\theta =x\\text{\/}a[\/latex], or equivalently, so that [latex]x=a\\sin\\theta [\/latex]. If we substitute [latex]x=a\\sin\\theta [\/latex] into [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex], we get<\/p>\r\n\r\n<div id=\"fs-id1165041762738\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill \\sqrt{{a}^{2}-{x}^{2}}&amp; =\\sqrt{{a}^{2}-{\\left(a\\sin\\theta \\right)}^{2}}\\hfill &amp; &amp; &amp; \\text{Let }x=a\\sin\\theta \\text{ where }-\\frac{\\pi }{2}\\le \\theta \\le \\frac{\\pi }{2}.\\text{Simplify.}\\hfill \\\\ &amp; =\\sqrt{{a}^{2}-{a}^{2}{\\sin}^{2}\\theta }\\hfill &amp; &amp; &amp; \\text{Factor out }{a}^{2}.\\hfill \\\\ &amp; =\\sqrt{{a}^{2}\\left(1-{\\sin}^{2}\\theta \\right)}\\hfill &amp; &amp; &amp; \\text{Substitute }1-{\\sin}^{2}x={\\cos}^{2}x.\\hfill \\\\ &amp; =\\sqrt{{a}^{2}{\\cos}^{2}\\theta }\\hfill &amp; &amp; &amp; \\text{Take the square root.}\\hfill \\\\ &amp; =|a\\cos\\theta |\\hfill &amp; &amp; &amp; \\\\ &amp; =a\\cos\\theta .\\hfill &amp; &amp; &amp; \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041814740\">Since [latex]\\cos\\theta \\ge 0[\/latex] on [latex]-\\frac{\\pi }{2}\\le \\theta \\le \\frac{\\pi }{2}[\/latex] and [latex]a&gt;0[\/latex], [latex]|a\\cos\\theta |=a\\cos\\theta [\/latex]. We can see, from this discussion, that by making the substitution [latex]x=a\\sin\\theta [\/latex], we are able to convert an integral involving a radical into an integral involving trigonometric functions. After we evaluate the integral, we can convert the solution back to an expression involving [latex]x[\/latex]. To see how to do this, let\u2019s begin by assuming that [latex]0&lt;x&lt;a[\/latex]. In this case, [latex]0&lt;\\theta &lt;\\frac{\\pi }{2}[\/latex]. Since [latex]\\sin\\theta =\\frac{x}{a}[\/latex], we can draw the reference triangle in Figure 1 to assist in expressing the values of [latex]\\cos\\theta [\/latex], [latex]\\tan\\theta [\/latex], and the remaining trigonometric functions in terms of [latex]x[\/latex].\u00a0 Before drawing the reference triangle, we recall the definition of trigonometric functions in terms of ratios of side lengths below.<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Right Triangle Trigonometry Relationships<\/h3>\r\nGiven a right triangle with an acute angle of [latex]\\theta[\/latex],\r\n\r\n[latex]\\begin{align}&amp;\\sin \\left(\\theta \\right)=\\frac{\\text{opposite}}{\\text{hypotenuse}} \\\\ &amp;\\cos \\left(\\theta \\right)=\\frac{\\text{adjacent}}{\\text{hypotenuse}} \\\\ &amp;\\tan \\left(\\theta \\right)=\\frac{\\text{opposite}}{\\text{adjacent}} \\end{align}[\/latex]\r\n\r\nA common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of \"Sine is opposite over hypotenuse, Cosine is adjacent over hypotenuse, Tangent is opposite over adjacent.\"\r\n\r\n<\/div>\r\nIt can be shown that this triangle actually produces the correct values of the trigonometric functions evaluated at [latex]\\theta [\/latex] for all [latex]\\theta [\/latex] satisfying [latex]-\\frac{\\pi }{2}\\le \\theta \\le \\frac{\\pi }{2}[\/latex]. It is useful to observe that the expression [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex] actually appears as the length of one side of the triangle. Last, should [latex]\\theta [\/latex] appear by itself, we use [latex]\\theta ={\\sin}^{-1}\\left(\\frac{x}{a}\\right)[\/latex].\r\n<figure id=\"CNX_Calc_Figure_07_03_001\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233757\/CNX_Calc_Figure_07_03_001.jpg\" alt=\"This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled a, the vertical leg is labeled x, and the horizontal leg is labeled as the square root of (a^2 \u2013 x^2). To the left of the triangle is the equation sin(theta) = x\/a.\" width=\"487\" height=\"181\" data-media-type=\"image\/jpeg\" \/> Figure 1. A reference triangle can help express the trigonometric functions evaluated at [latex]\\theta [\/latex] in terms of [latex]x[\/latex].[\/caption]<\/figure>\r\n<p id=\"fs-id1165040764019\">The essential part of this discussion is summarized in the following problem-solving strategy.<\/p>\r\n\r\n<div id=\"fs-id1165042314803\" class=\"problem-solving\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox examples\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Problem-Solving Strategy: Integrating Expressions Involving [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex]<\/h3>\r\n\r\n<hr \/>\r\n\r\n<ol id=\"fs-id1165040638957\" type=\"1\">\r\n \t<li>It is a good idea to make sure the integral cannot be evaluated easily in another way. For example, although this method can be applied to integrals of the form [latex]\\displaystyle\\int \\frac{1}{\\sqrt{{a}^{2}-{x}^{2}}}dx[\/latex], [latex]\\displaystyle\\int \\frac{x}{\\sqrt{{a}^{2}-{x}^{2}}}dx[\/latex], and [latex]\\displaystyle\\int x\\sqrt{{a}^{2}-{x}^{2}}dx[\/latex], they can each be integrated directly either by formula or by a simple <em data-effect=\"italics\">u<\/em>-substitution.<\/li>\r\n \t<li>Make the substitution [latex]x=a\\sin\\theta [\/latex] and [latex]dx=a\\cos\\theta d\\theta [\/latex]. <em data-effect=\"italics\">Note<\/em>: This substitution yields [latex]\\sqrt{{a}^{2}-{x}^{2}}=a\\cos\\theta[\/latex].<\/li>\r\n \t<li>Simplify the expression.<\/li>\r\n \t<li>Evaluate the integral using techniques from the section on trigonometric integrals.<\/li>\r\n \t<li>Use the reference triangle from Figure 1 to rewrite the result in terms of [latex]x[\/latex]. You may also need to use some trigonometric identities and the relationship [latex]\\theta ={\\sin}^{-1}\\left(\\frac{x}{a}\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165041837235\" data-type=\"example\">\r\n<div id=\"fs-id1165042128156\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042093676\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Integrating an Expression Involving [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex]<\/h3>\r\n<div id=\"fs-id1165041837235\" data-type=\"example\">\r\n<div id=\"fs-id1165042128156\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042093676\" data-type=\"problem\">\r\n<p id=\"fs-id1165042297418\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}\\sqrt{9-{x}^{2}}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-id1165041913990\" data-type=\"solution\">\r\n<p id=\"fs-id1165041893409\">Begin by making the substitutions [latex]x=3\\sin\\theta [\/latex] and [latex]dx=3\\cos\\theta d\\theta [\/latex]. Since [latex]\\sin\\theta =\\frac{x}{3}[\/latex], we can construct the reference triangle shown in the following figure.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_07_03_002\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233800\/CNX_Calc_Figure_07_03_002.jpg\" alt=\"This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled 3, the vertical leg is labeled x, and the horizontal leg is labeled as the square root of (9 \u2013 x^2). To the left of the triangle is the equation sin(theta) = x\/3.\" width=\"487\" height=\"181\" data-media-type=\"image\/jpeg\" \/> Figure 2. A reference triangle can be constructed for this example.[\/caption]<\/figure>\r\n<p id=\"fs-id1165040796218\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1165041949238\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }^{\\text{ }}\\sqrt{9-{x}^{2}}dx&amp; ={\\displaystyle\\int }^{\\text{ }}\\sqrt{9-{\\left(3\\sin\\theta \\right)}^{2}}3\\cos\\theta d\\theta \\hfill &amp; &amp; &amp; \\text{Substitute }x=3\\sin\\theta \\text{ and }dx=3\\cos\\theta d\\theta .\\hfill \\\\ &amp; ={\\displaystyle\\int }^{\\text{ }}\\sqrt{9\\left(1-{\\sin}^{2}\\theta \\right)}3\\cos\\theta d\\theta \\hfill &amp; &amp; &amp; \\text{Simplify.}\\hfill \\\\ &amp; ={\\displaystyle\\int }^{\\text{ }}\\sqrt{9{\\cos}^{2}\\theta }3\\cos\\theta d\\theta \\hfill &amp; &amp; &amp; \\text{Substitute }{\\cos}^{2}\\theta =1-{\\sin}^{2}\\theta .\\hfill \\\\ &amp; ={\\displaystyle\\int }^{\\text{ }}3|\\cos\\theta |3\\cos\\theta d\\theta \\hfill &amp; &amp; &amp; \\text{Take the square root.}\\hfill \\\\ &amp; ={\\displaystyle\\int }^{\\text{ }}9{\\cos}^{2}\\theta d\\theta \\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Simplify. Since}-\\frac{\\pi }{2}\\le \\theta \\le \\frac{\\pi }{2},\\cos\\theta \\ge 0\\text{ and }\\hfill \\\\ |\\cos\\theta |=\\cos\\theta .\\hfill \\end{array}\\hfill \\\\ &amp; ={\\displaystyle\\int }^{\\text{ }}9\\left(\\frac{1}{2}+\\frac{1}{2}\\cos\\left(2\\theta \\right)\\right)d\\theta \\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Use the strategy for integrating an even power}\\hfill \\\\ \\text{of }\\cos\\theta .\\hfill \\end{array}\\hfill \\\\ &amp; =\\frac{9}{2}\\theta +\\frac{9}{4}\\sin\\left(2\\theta \\right)+C\\hfill &amp; &amp; &amp; \\text{Evaluate the integral.}\\hfill \\\\ &amp; =\\frac{9}{2}\\theta +\\frac{9}{4}\\left(2\\sin\\theta \\cos\\theta \\right)+C\\hfill &amp; &amp; &amp; \\text{Substitute }\\sin\\left(2\\theta \\right)=2\\sin\\theta \\cos\\theta .\\hfill \\\\ &amp; =\\frac{9}{2}{\\sin}^{-1}\\left(\\frac{x}{3}\\right)+\\frac{9}{2}\\cdot \\frac{x}{3}\\cdot \\frac{\\sqrt{9-{x}^{2}}}{3}+C\\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Substitute }{\\sin}^{-1}\\left(\\frac{x}{3}\\right)=\\theta \\text{ and }\\sin\\theta =\\frac{x}{3}.\\text{Use}\\hfill \\\\ \\text{the reference triangle to see that}\\hfill \\\\ \\cos\\theta =\\frac{\\sqrt{9-{x}^{2}}}{3}\\text{and make this substitution.}\\hfill \\end{array}\\hfill \\\\ &amp; =\\frac{9}{2}{\\sin}^{-1}\\left(\\frac{x}{3}\\right)+\\frac{x\\sqrt{9-{x}^{2}}}{2}+C.\\hfill &amp; &amp; &amp; \\text{Simplify.}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042229562\" data-type=\"example\">\r\n<div id=\"fs-id1165042073728\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042073730\" data-type=\"problem\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Integrating an Expression Involving [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex]<\/h3>\r\n<div id=\"fs-id1165042073730\" data-type=\"problem\">\r\n<p id=\"fs-id1165042004532\">Evaluate [latex]\\displaystyle\\int \\frac{\\sqrt{4-{x}^{2}}}{x}dx[\/latex].<\/p>\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-id1165040797018\" data-type=\"solution\">\r\n<p id=\"fs-id1165041759427\">First make the substitutions [latex]x=2\\sin\\theta [\/latex] and [latex]dx=2\\cos\\theta d\\theta [\/latex]. Since [latex]\\sin\\theta =\\frac{x}{2}[\/latex], we can construct the reference triangle shown in the following figure.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_07_03_003\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"380\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233803\/CNX_Calc_Figure_07_03_003.jpg\" alt=\"This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The vertical leg is labeled x, and the horizontal leg is labeled as the square root of (4 \u2013 x^2). To the left of the triangle is the equation sin(theta) = x\/2.\" width=\"380\" height=\"177\" data-media-type=\"image\/jpeg\" \/> Figure 3. A reference triangle can be constructed for this example.[\/caption]<\/figure>\r\n<p id=\"fs-id1165040739936\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1165040697663\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int \\frac{\\sqrt{4-{x}^{2}}}{x}dx}&amp; ={\\displaystyle\\int \\frac{\\sqrt{4-{\\left(2\\sin\\theta \\right)}^{2}}}{2\\sin\\theta }2\\cos\\theta d\\theta} \\hfill &amp; &amp; &amp; \\text{Substitute}x=2\\sin\\theta \\text{and}=2\\cos\\theta d\\theta .\\hfill \\\\ &amp; ={\\displaystyle\\int \\frac{2{\\cos}^{2}\\theta }{\\sin\\theta }d\\theta} \\hfill &amp; &amp; &amp; \\text{Substitute}{\\cos}^{2}\\theta =1-{\\sin}^{2}\\theta \\text{and simplify.}\\hfill \\\\ &amp; ={\\displaystyle\\int \\frac{2\\left(1-{\\sin}^{2}\\theta \\right)}{\\sin\\theta }d\\theta} \\hfill &amp; &amp; &amp; \\text{Substitute}{\\sin}^{2}\\theta =1-{\\cos}^{2}\\theta .\\hfill \\\\ &amp; ={\\displaystyle\\int \\left(2\\csc\\theta -2\\sin\\theta \\right)d\\theta} \\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Separate the numerator, simplify, and use}\\hfill \\\\ \\csc\\theta =\\frac{1}{\\sin\\theta }.\\hfill \\end{array}\\hfill \\\\ &amp; =2\\text{ln}|\\csc\\theta -\\cot\\theta |+2\\cos\\theta +C\\hfill &amp; &amp; &amp; \\text{Evaluate the integral.}\\hfill \\\\ &amp; =2\\text{ln}|\\frac{2}{x}-\\frac{\\sqrt{4-{x}^{2}}}{x}|+\\sqrt{4-{x}^{2}}+C.\\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Use the reference triangle to rewrite the}\\hfill \\\\ \\text{expression in terms of}x\\text{and simplify.}\\hfill \\end{array}\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165042092095\">In the next example, we see that we sometimes have a choice of methods.<\/p>\r\n\r\n<div id=\"fs-id1165042092098\" data-type=\"example\">\r\n<div id=\"fs-id1165040642691\" data-type=\"exercise\">\r\n<div id=\"fs-id1165040642694\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Integrating an Expression Involving [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex] Two Ways<\/h3>\r\n<div id=\"fs-id1165040642694\" data-type=\"problem\">\r\n<p id=\"fs-id1165040744148\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{x}^{3}\\sqrt{1-{x}^{2}}dx[\/latex] two ways: first by using the substitution [latex]u=1-{x}^{2}[\/latex] and then by using a trigonometric substitution.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1165041958780\" data-type=\"solution\">\r\n<p id=\"fs-id1167794138271\"><strong data-effect=\"bold\">Method 1<\/strong><\/p>\r\n<p id=\"fs-id1165041958783\">Let [latex]u=1-{x}^{2}[\/latex] and hence [latex]{x}^{2}=1-u[\/latex]. Thus, [latex]du=-2xdx[\/latex]. In this case, the integral becomes<\/p>\r\n\r\n<div id=\"fs-id1165042301844\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int ^{\\text{ }}{x}^{3}\\sqrt{1-{x}^{2}}dx}&amp; =-\\frac{1}{2}{\\displaystyle\\int ^{\\text{ }}{x}^{2}\\sqrt{1-{x}^{2}}\\left(-2xdx\\right)}\\hfill &amp; &amp; &amp; \\text{Make the substitution.}\\hfill \\\\ &amp; =-\\frac{1}{2}{\\displaystyle\\int ^{\\text{ }}\\left(1-u\\right)\\sqrt{u}du}\\hfill &amp; &amp; &amp; \\text{Expand the expression.}\\hfill \\\\ &amp; =-\\frac{1}{2}{\\displaystyle\\int \\left({u}^{1\\text{\/}2}-{u}^{3\\text{\/}2}\\right)du}\\hfill &amp; &amp; &amp; \\text{Evaluate the integral.}\\hfill \\\\ &amp; =-\\frac{1}{2}\\left(\\frac{2}{3}{u}^{3\\text{\/}2}-\\frac{2}{5}{u}^{5\\text{\/}2}\\right)+C\\hfill &amp; &amp; &amp; \\text{Rewrite in terms of}x.\\hfill \\\\ &amp; =-\\frac{1}{3}{\\left(1-{x}^{2}\\right)}^{3\\text{\/}2}+\\frac{1}{5}{\\left(1-{x}^{2}\\right)}^{5\\text{\/}2}+C.\\hfill &amp; &amp; &amp; \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793929406\"><strong data-effect=\"bold\">Method 2<\/strong><\/p>\r\n<p id=\"fs-id1165041766752\">Let [latex]x=\\sin\\theta [\/latex]. In this case, [latex]dx=\\cos\\theta d\\theta [\/latex]. Using this substitution, we have<\/p>\r\n\r\n<div id=\"fs-id1165040793998\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }^{\\text{ }}{x}^{3}\\sqrt{1-{x}^{2}}dx&amp; ={\\displaystyle\\int }^{\\text{ }}{\\sin}^{3}\\theta {\\cos}^{2}\\theta d\\theta \\hfill &amp; &amp; &amp; \\\\ &amp; ={\\displaystyle\\int }^{\\text{ }}\\left(1-{\\cos}^{2}\\theta \\right){\\cos}^{2}\\theta \\sin\\theta d\\theta \\hfill &amp; &amp; &amp; \\text{Let }u=\\cos\\theta .\\text{Thus, }du=\\text{-}\\sin\\theta d\\theta .\\hfill \\\\ &amp; ={\\displaystyle\\int }^{\\text{ }}\\left({u}^{4}-{u}^{2}\\right)du\\hfill &amp; &amp; &amp; \\\\ &amp; =\\frac{1}{5}{u}^{5}-\\frac{1}{3}{u}^{3}+C\\hfill &amp; &amp; &amp; \\text{Substitute }\\cos\\theta =u.\\hfill \\\\ &amp; =\\frac{1}{5}{\\cos}^{5}\\theta -\\frac{1}{3}{\\cos}^{3}\\theta +C\\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Use a reference triangle to see that}\\hfill \\\\ \\cos\\theta =\\sqrt{1-{x}^{2}}.\\hfill \\end{array}\\hfill \\\\ &amp; =\\frac{1}{5}{\\left(1-{x}^{2}\\right)}^{5\\text{\/}2}-\\frac{1}{3}{\\left(1-{x}^{2}\\right)}^{3\\text{\/}2}+C.\\hfill &amp; &amp; &amp; \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042094001\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1165042094004\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042033160\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1165042033160\" data-type=\"problem\">\r\n<p id=\"fs-id1165042033163\">Rewrite the integral [latex]\\displaystyle\\int \\frac{{x}^{3}}{\\sqrt{25-{x}^{2}}}dx[\/latex] using the appropriate trigonometric substitution (do not evaluate the integral).<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558895\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1165041841595\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165041951969\">Substitute [latex]x=5\\sin\\theta [\/latex] and [latex]dx=5\\cos\\theta d\\theta [\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1165042237758\" data-type=\"solution\">\r\n<p id=\"fs-id1165040640415\">[latex]{\\displaystyle\\int }^{\\text{ }}125{\\sin}^{3}\\theta d\\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/bCIrhw0sjlU?controls=0&amp;start=1071&amp;end=1182&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.3TrigonometricSubstitution1071to1182_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.3 Trigonometric Substitution\" here (opens in new window)<\/a>.\r\n\r\n<\/section><section id=\"fs-id1165041813590\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Integrating Expressions Involving [latex]\\sqrt{{a}^{2}+{x}^{2}}[\/latex]<\/h2>\r\n<p id=\"fs-id1165042278317\">For integrals containing [latex]\\sqrt{{a}^{2}+{x}^{2},}[\/latex] let\u2019s first consider the domain of this expression. Since [latex]\\sqrt{{a}^{2}+{x}^{2}}[\/latex] is defined for all real values of [latex]x[\/latex], we restrict our choice to those trigonometric functions that have a range of all real numbers. Thus, our choice is restricted to selecting either [latex]x=a\\tan\\theta [\/latex] or [latex]x=a\\cot\\theta [\/latex]. Either of these substitutions would actually work, but the standard substitution is [latex]x=a\\tan\\theta [\/latex] or, equivalently, [latex]\\tan\\theta =\\frac{x}{a}[\/latex]. With this substitution, we make the assumption that [latex]\\text{-}\\frac{\\pi}{2}&lt;\\theta &lt;\\frac{\\pi}{2}[\/latex], so that we also have [latex]\\theta ={\\tan}^{-1}\\left(x\\text{\/}a\\right)[\/latex]. The procedure for using this substitution is outlined in the following problem-solving strategy.<\/p>\r\n\r\n<div id=\"fs-id1165042110570\" class=\"problem-solving\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox examples\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Problem-Solving Strategy: Integrating Expressions Involving [latex]\\sqrt{{a}^{2}+{x}^{2}}[\/latex]<\/h3>\r\n\r\n<hr \/>\r\n\r\n<ol id=\"fs-id1165041979050\" type=\"1\">\r\n \t<li>Check to see whether the integral can be evaluated easily by using another method. In some cases, it is more convenient to use an alternative method.<\/li>\r\n \t<li>Substitute [latex]x=a\\tan\\theta [\/latex] and [latex]dx=a{\\sec}^{2}\\theta d\\theta [\/latex]. This substitution yields<span data-type=\"newline\">\r\n<\/span>\r\n[latex]\\sqrt{{a}^{2}+{x}^{2}}=\\sqrt{{a}^{2}+{\\left(a\\tan\\theta \\right)}^{2}}=\\sqrt{{a}^{2}\\left(1+{\\tan}^{2}\\theta \\right)}=\\sqrt{{a}^{2}{\\sec}^{2}\\theta }=|a\\sec\\theta |=a\\sec\\theta [\/latex]. (Since [latex]-\\frac{\\pi }{2}&lt;\\theta &lt;\\frac{\\pi }{2}[\/latex] and [latex]\\sec\\theta &gt;0[\/latex] over this interval, [latex]|a\\sec\\theta |=a\\sec\\theta[\/latex] )<\/li>\r\n \t<li>Simplify the expression.<\/li>\r\n \t<li>Evaluate the integral using techniques from the section on trigonometric integrals.<\/li>\r\n \t<li>Use the reference triangle from Figure 4 to rewrite the result in terms of [latex]x[\/latex]. You may also need to use some trigonometric identities and the relationship [latex]\\theta ={\\tan}^{-1}\\left(\\frac{x}{a}\\right)[\/latex]. (<em data-effect=\"italics\">Note<\/em>: The reference triangle is based on the assumption that [latex]x&gt;0[\/latex]; however, the trigonometric ratios produced from the reference triangle are the same as the ratios for which [latex]x \\le{0}[\/latex] )<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<figure id=\"CNX_Calc_Figure_07_03_004\">[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233806\/CNX_Calc_Figure_07_03_004.jpg\" alt=\"This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled the square root of (a^2+x^2), the vertical leg is labeled x, and the horizontal leg is labeled a. To the left of the triangle is the equation tan(theta) = x\/a.\" width=\"487\" height=\"173\" data-media-type=\"image\/jpeg\" \/> Figure 4. A reference triangle can be constructed to express the trigonometric functions evaluated at [latex]\\theta [\/latex] in terms of [latex]x[\/latex].[\/caption]<\/figure>\r\n<div id=\"fs-id1165042089656\" data-type=\"example\">\r\n<div id=\"fs-id1165042089658\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042089660\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Integrating an Expression Involving [latex]\\sqrt{{a}^{2}+{x}^{2}}[\/latex]<\/h3>\r\n<div id=\"fs-id1165042089660\" data-type=\"problem\">\r\n<p id=\"fs-id1165040773006\">Evaluate [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}[\/latex] and check the solution by differentiating.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1165041887608\" data-type=\"solution\">\r\n<p id=\"fs-id1165041887610\">Begin with the substitution [latex]x=\\tan\\theta [\/latex] and [latex]dx={\\sec}^{2}\\theta d\\theta [\/latex]. Since [latex]\\tan\\theta =x[\/latex], draw the reference triangle in the following figure.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_07_03_005\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233810\/CNX_Calc_Figure_07_03_005.jpg\" alt=\"This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled the square root of (1+x^2), the vertical leg is labeled x, and the horizontal leg is labeled 1. To the left of the triangle is the equation tan(theta) = x\/1.\" width=\"487\" height=\"173\" data-media-type=\"image\/jpeg\" \/> Figure 5. The reference triangle for this example.[\/caption]<\/figure>\r\n<p id=\"fs-id1165040750681\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1165040750684\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}}&amp; ={\\displaystyle\\int \\frac{{\\sec}^{2}\\theta }{\\sec\\theta }d\\theta }\\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Substitute}x=\\tan\\theta \\text{and}dx={\\sec}^{2}\\theta d\\theta .\\text{This}\\hfill \\\\ \\text{substitution makes}\\sqrt{1+{x}^{2}}=\\sec\\theta .\\text{Simplify.}\\hfill \\end{array}\\hfill \\\\ &amp; ={\\displaystyle\\int ^{\\text{ }}\\sec\\theta d\\theta }\\hfill &amp; &amp; &amp; \\text{Evaluate the integral.}\\hfill \\\\ &amp; =\\text{ln}|\\sec\\theta +\\tan\\theta |+C\\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Use the reference triangle to express the result}\\hfill \\\\ \\text{in terms of}x.\\hfill \\end{array}\\hfill \\\\ &amp; =\\text{ln}|\\sqrt{1+{x}^{2}}+x|+C.\\hfill &amp; &amp; &amp; \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040744515\">To check the solution, differentiate:<\/p>\r\n\r\n<div id=\"fs-id1165040744518\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{d}{dx}\\left(\\text{ln}|\\sqrt{1+{x}^{2}}+x|\\right)&amp; =\\frac{1}{\\sqrt{1+{x}^{2}}+x}\\cdot \\left(\\frac{x}{\\sqrt{1+{x}^{2}}}+1\\right)\\hfill \\\\ &amp; =\\frac{1}{\\sqrt{1+{x}^{2}}+x}\\cdot \\frac{x+\\sqrt{1+{x}^{2}}}{\\sqrt{1+{x}^{2}}}\\hfill \\\\ &amp; =\\frac{1}{\\sqrt{1+{x}^{2}}}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041980958\">Since [latex]\\sqrt{1+{x}^{2}}+x&gt;0[\/latex] for all values of [latex]x[\/latex], we could rewrite [latex]\\text{ln}|\\sqrt{1+{x}^{2}}+x|+C=\\text{ln}\\left(\\sqrt{1+{x}^{2}}+x\\right)+C[\/latex], if desired.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042301696\" data-type=\"example\">\r\n<div id=\"fs-id1165042301698\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042301700\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n\r\nIn the example below, we explore how to use hyperbolic trigonometric functions as an alternative substitution method. First, we briefly review the definition of these functions and their derivatives.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Definition and Derivatives of Hyperbolic Trigonometric Functions<\/h3>\r\nFor any real number [latex] x [\/latex], the hyperbolic sine and hyperbolic cosine are defined as:\r\n<p style=\"padding-left: 30px;\">[latex] \\sinh x = \\frac{e^x - e^{-x}}{2} \\: \\text{and}\\:\\cosh x = \\frac{e^x + e^{-x}}{2} [\/latex]<\/p>\r\nTheir derivatives are given by:\r\n<p style=\"padding-left: 30px;\">[latex] \\frac{d}{dx} \\left( \\sinh x \\right) = \\cosh x \\:\\text{and}\\:\\frac{d}{dx} \\left( \\cosh x \\right) = \\sinh x [\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Evaluating [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}[\/latex] Using a Different Substitution<\/h3>\r\n<div id=\"fs-id1165042301700\" data-type=\"problem\">\r\n<p id=\"fs-id1165042272858\">Use the substitution [latex]x=\\text{sinh}\\theta [\/latex] to evaluate [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n<div id=\"fs-id1165041828644\" data-type=\"solution\">\r\n<p id=\"fs-id1165041828646\">Because [latex]\\text{sinh}\\theta [\/latex] has a range of all real numbers, and [latex]1+{\\text{sinh}}^{2}\\theta ={\\text{cosh}}^{2}\\theta [\/latex], we may also use the substitution [latex]x=\\text{sinh}\\theta [\/latex] to evaluate this integral. In this case, [latex]dx=\\text{cosh}\\theta d\\theta [\/latex]. Consequently,<\/p>\r\n\r\n<div id=\"fs-id1165042004330\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}}&amp; ={\\displaystyle\\int \\frac{\\text{cosh}\\theta }{\\sqrt{1+{\\text{sinh}}^{2}\\theta }}d\\theta }\\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Substitute}x=\\text{sinh}\\theta \\text{and}dx=\\text{cosh}\\theta d\\theta .\\hfill \\\\ \\text{Substitute}1+{\\text{sinh}}^{2}\\theta ={\\text{cosh}}^{2}\\theta .\\hfill \\end{array}\\hfill \\\\ &amp; ={\\displaystyle\\int \\frac{\\text{cosh}\\theta }{\\sqrt{{\\text{cosh}}^{2}\\theta }}d\\theta }\\hfill &amp; &amp; &amp; \\sqrt{{\\text{cosh}}^{2}\\theta }=|\\text{cosh}\\theta |\\hfill \\\\ &amp; ={\\displaystyle\\int \\frac{\\text{cosh}\\theta }{|\\text{cosh}\\theta |}d\\theta }\\hfill &amp; &amp; &amp; |\\text{cosh}\\theta |=\\text{cosh}\\theta \\text{since}\\text{cosh}\\theta &gt;0\\text{for all}\\theta .\\hfill \\\\ &amp; ={\\displaystyle\\int \\frac{\\text{cosh}\\theta }{\\text{cosh}\\theta }d\\theta }\\hfill &amp; &amp; &amp; \\text{Simplify.}\\hfill \\\\ &amp; ={\\displaystyle\\int ^{\\text{ }}1d\\theta }\\hfill &amp; &amp; &amp; \\text{Evaluate the integral.}\\hfill \\\\ &amp; =\\theta +C\\hfill &amp; &amp; &amp; \\text{Since}x=\\text{sinh}\\theta ,\\text{we know}\\theta ={\\text{sinh}}^{-1}x.\\hfill \\\\ &amp; ={\\text{sinh}}^{-1}x+C.\\hfill &amp; &amp; &amp; \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1165042063682\" data-type=\"commentary\">\r\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\r\n<p id=\"fs-id1165042063687\">This answer looks quite different from the answer obtained using the substitution [latex]x=\\tan\\theta [\/latex]. To see that the solutions are the same, set [latex]y={\\text{sinh}}^{-1}x[\/latex]. Thus, [latex]\\text{sinh}y=x[\/latex]. From this equation we obtain:<\/p>\r\n\r\n<div id=\"fs-id1165041843069\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{e}^{y}-{e}^{\\text{-}y}}{2}=x[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042277014\">After multiplying both sides by [latex]2{e}^{y}[\/latex] and rewriting, this equation becomes:<\/p>\r\n\r\n<div id=\"fs-id1165042202678\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{e}^{2y}-2x{e}^{y}-1=0[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042232174\">Use the quadratic equation to solve for [latex]{e}^{y}\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165040645138\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{e}^{y}=\\frac{2x\\pm \\sqrt{4{x}^{2}+4}}{2}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040644008\">Simplifying, we have:<\/p>\r\n\r\n<div id=\"fs-id1165040644011\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{e}^{y}=x\\pm \\sqrt{{x}^{2}+1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040768460\">Since [latex]x-\\sqrt{{x}^{2}+1}&lt;0[\/latex], it must be the case that [latex]{e}^{y}=x+\\sqrt{{x}^{2}+1}[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1165041921710\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=\\text{ln}\\left(x+\\sqrt{{x}^{2}+1}\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040743112\">Last, we obtain<\/p>\r\n\r\n<div id=\"fs-id1165042092492\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\text{sinh}}^{-1}x=\\text{ln}\\left(x+\\sqrt{{x}^{2}+1}\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042096551\">After we make the final observation that, since [latex]x+\\sqrt{{x}^{2}+1}&gt;0[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1165040713985\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{ln}\\left(x+\\sqrt{{x}^{2}+1}\\right)=\\text{ln}|\\sqrt{1+{x}^{2}}+x|[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040639545\">we see that the two different methods produced equivalent solutions.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165040639551\" data-type=\"example\">\r\n<div id=\"fs-id1165040639553\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042279310\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Finding an Arc Length<\/h3>\r\n<div id=\"fs-id1165042279310\" data-type=\"problem\">\r\n<p id=\"fs-id1165042279315\">Find the length of the curve [latex]y={x}^{2}[\/latex] over the interval [latex]\\left[0,\\frac{1}{2}\\right][\/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-id1165040668429\" data-type=\"solution\">\r\n<p id=\"fs-id1165040668432\">Because [latex]\\frac{dy}{dx}=2x[\/latex], the arc length is given by<\/p>\r\n\r\n<div id=\"fs-id1165040744954\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{0}^{1\\text{\/}2}\\sqrt{1+{\\left(2x\\right)}^{2}}dx={\\displaystyle\\int }_{0}^{1\\text{\/}2}\\sqrt{1+4{x}^{2}}dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040765593\">To evaluate this integral, use the substitution [latex]x=\\frac{1}{2}\\tan\\theta [\/latex] and [latex]dx=\\frac{1}{2}{\\sec}^{2}\\theta d\\theta [\/latex]. We also need to change the limits of integration. If [latex]x=0[\/latex], then [latex]\\theta =0[\/latex] and if [latex]x=\\frac{1}{2}[\/latex], then [latex]\\theta =\\frac{\\pi }{4}[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1165040665687\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }_{0}^{1\\text{\/}2}\\sqrt{1+4{x}^{2}}dx&amp; ={\\displaystyle\\int }_{0}^{\\pi \\text{\/}4}\\sqrt{1+{\\tan}^{2}\\theta }\\frac{1}{2}{\\sec}^{2}\\theta d\\theta \\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{After substitution,}\\hfill \\\\ \\sqrt{1+4{x}^{2}}=\\tan\\theta .\\text{Substitute}\\hfill \\\\ 1+{\\tan}^{2}\\theta ={\\sec}^{2}\\theta \\text{ and simplify.}\\hfill \\end{array}\\hfill \\\\ &amp; =\\frac{1}{2}{\\displaystyle\\int }_{0}^{\\pi \\text{\/}4}{\\sec}^{3}\\theta d\\theta \\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{We derived this integral in the}\\hfill \\\\ \\text{previous section.}\\hfill \\end{array}\\hfill \\\\ &amp; =\\frac{1}{2}\\left(\\frac{1}{2}\\sec\\theta \\tan\\theta +\\text{ln}|\\sec\\theta +\\tan\\theta |\\right)|{}_{\\begin{array}{c}\\\\ 0\\end{array}}^{\\begin{array}{c}\\pi \\text{\/}4\\\\ \\end{array}}\\hfill &amp; &amp; &amp; \\text{Evaluate and simplify.}\\hfill \\\\ &amp; =\\frac{1}{4}\\left(\\sqrt{2}+\\text{ln}\\left(\\sqrt{2}+1\\right)\\right).\\hfill &amp; &amp; &amp; \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042041805\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1165042041808\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042041810\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1165042041810\" data-type=\"problem\">\r\n<p id=\"fs-id1165042041812\">Rewrite [latex]{\\displaystyle\\int }^{\\text{ }}{x}^{3}\\sqrt{{x}^{2}+4}dx[\/latex] by using a substitution involving [latex]\\tan\\theta [\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558890\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n<div id=\"fs-id1165041921699\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165041921706\">Use [latex]x=2\\tan\\theta [\/latex] and [latex]dx=2{\\sec}^{2}\\theta d\\theta [\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558891\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n<div id=\"fs-id1165041952815\" data-type=\"solution\">\r\n<p id=\"fs-id1165041952817\">[latex]{\\displaystyle\\int }^{\\text{ }}32{\\tan}^{3}\\theta {\\sec}^{3}\\theta d\\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/bCIrhw0sjlU?controls=0&amp;start=1688&amp;end=1797&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.3TrigonometricSubstitution1688to1797_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.3 Trigonometric Substitution\" here (opens in new window)<\/a>.\r\n\r\n<\/section><section id=\"fs-id1165040682763\" data-depth=\"1\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5524[\/ohm_question]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Integrating Expressions Involving [latex]\\sqrt{{x}^{2}-{a}^{2}}[\/latex]<\/h2>\r\n<p id=\"fs-id1165042231872\">The domain of the expression [latex]\\sqrt{{x}^{2}-{a}^{2}}[\/latex] is [latex]\\left(\\text{-}\\infty ,\\text{-}a\\right]\\cup \\left[a,\\text{+}\\infty \\right)[\/latex]. Thus, either [latex]x\\le \\text{-}a[\/latex] or [latex]x\\ge a[\/latex]. Hence, [latex]\\frac{x}{a}\\le -1[\/latex] or [latex]\\frac{x}{a}\\ge 1[\/latex]. Since these intervals correspond to the range of [latex]\\sec\\theta [\/latex] on the set [latex]\\left[0,\\frac{\\pi }{2}\\right)\\cup \\left(\\frac{\\pi }{2},\\pi \\right][\/latex], it makes sense to use the substitution [latex]\\sec\\theta =\\frac{x}{a}[\/latex] or, equivalently, [latex]x=a\\sec\\theta [\/latex], where [latex]0\\le \\theta &lt;\\frac{\\pi }{2}[\/latex] or [latex]\\frac{\\pi }{2}&lt;\\theta \\le \\pi [\/latex]. The corresponding substitution for [latex]dx[\/latex] is [latex]dx=a\\sec\\theta \\tan\\theta d\\theta [\/latex]. The procedure for using this substitution is outlined in the following problem-solving strategy.<\/p>\r\n\r\n<div id=\"fs-id1165042235602\" class=\"problem-solving\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox examples\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Problem-Solving Strategy: Integrals Involving [latex]\\sqrt{{x}^{2}-{a}^{2}}[\/latex]<\/h3>\r\n\r\n<hr \/>\r\n\r\n<ol id=\"fs-id1165042108861\" type=\"1\">\r\n \t<li>Check to see whether the integral cannot be evaluated using another method. If so, we may wish to consider applying an alternative technique.<\/li>\r\n \t<li>Substitute [latex]x=a\\sec\\theta [\/latex] and [latex]dx=a\\sec\\theta \\tan\\theta d\\theta [\/latex]. This substitution yields<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1165041836430\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\sqrt{{x}^{2}-{a}^{2}}=\\sqrt{{\\left(a\\sec\\theta \\right)}^{2}-{a}^{2}}=\\sqrt{{a}^{2}\\left({\\sec}^{2}\\theta -1\\right)}=\\sqrt{{a}^{2}{\\tan}^{2}\\theta }=|a\\tan\\theta |[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nFor [latex]x\\ge a[\/latex], [latex]|a\\tan\\theta |=a\\tan\\theta [\/latex] and for [latex]x\\le -a[\/latex], [latex]|a\\tan\\theta |=\\text{-}a\\tan\\theta [\/latex].<\/li>\r\n \t<li>Simplify the expression.<\/li>\r\n \t<li>Evaluate the integral using techniques from the section on trigonometric integrals.<\/li>\r\n \t<li>Use the reference triangles from Figure 6 to rewrite the result in terms of [latex]x[\/latex]. You may also need to use some trigonometric identities and the relationship [latex]\\theta ={\\sec}^{-1}\\left(\\frac{x}{a}\\right)[\/latex]. (<em data-effect=\"italics\">Note<\/em>: We need both reference triangles, since the values of some of the trigonometric ratios are different depending on whether [latex]x\\ge a[\/latex] or [latex]x\\le \\text{-}a.[\/latex])<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<figure id=\"CNX_Calc_Figure_07_03_006\">[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233815\/CNX_Calc_Figure_07_03_006.jpg\" alt=\"This figure has two right triangles. The first triangle is in the first quadrant of the xy coordinate system and has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled x, the vertical leg is labeled the square root of (x^2-a^2), and the horizontal leg is labeled a. The horizontal leg is on the x-axis. To the left of the triangle is the equation sec(theta) = x\/a, x&gt;a. There are also the equations sin(theta)= the square root of (x^2-a^2)\/x, cos(theta) = a\/x, and tan(theta) = the square root of (x^2-a^2)\/a. The second triangle is in the second quadrant, with the hypotenuse labeled \u2013x. The horizontal leg is labeled \u2013a and is on the negative x-axis. The vertical leg is labeled the square root of (x^2-a^2). To the right of the triangle is the equation sec(theta) = x\/a, x&lt;-a. There are also the equations sin(theta)= the negative square root of (x^2-a^2)\/x, cos(theta) = a\/x, and tan(theta) = the negative square root of (x^2-a^2)\/a.\" width=\"975\" height=\"392\" data-media-type=\"image\/jpeg\" \/> Figure 6. Use the appropriate reference triangle to express the trigonometric functions evaluated at [latex]\\theta [\/latex] in terms of [latex]x[\/latex].[\/caption]<\/figure>\r\n<div id=\"fs-id1165041831016\" data-type=\"example\">\r\n<div id=\"fs-id1165041831018\" data-type=\"exercise\">\r\n<div id=\"fs-id1165041831020\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n\r\nFor the next example, it is worthwhile to review the technique of how to precisely evaluate a trignometric function whose input is an inverse trigonometric function.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Evaluating Trigonometric Functions composed with Inverse Trigonometric Functions<\/h3>\r\nSuppose we wanted to evaluate a trigonometric function composed with an inverse trigonometric function, for example:\u00a0[latex]\\sin\\left(\\cos^{\u22121}\\left(\\frac{4}{5}\\right)\\right)[\/latex].\r\n\r\nBeginning with the inside, we can say there is some angle such that [latex]\\theta=\\cos^{\u22121}(\\frac{4}{5})[\/latex], which means [latex]\\cos\\theta=\\frac{4}{5}[\/latex], and we are looking for [latex]\\sin\\theta[\/latex]. We can use the Pythagorean identity to do this.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} &amp;\\sin^{2}\\theta+\\cos^{2}\\theta=1 &amp;&amp; \\text{Use our known value for cosine.} \\\\ &amp;\\sin^{2}\\theta+\\left(\\frac{4}{5}\\right)^{2}=1 &amp;&amp; \\text{Solve for sine.} \\\\ &amp;\\sin^{2}\\theta=1\u2212\\frac{16}{25} \\\\ &amp;\\sin\\theta=\\pm\\sqrt{\\frac{9}{25}}=\\pm\\frac{3}{5} \\end{align}[\/latex]<\/p>\r\nSince [latex]\\theta=\\cos^{\u22121}(\\frac{4}{5})[\/latex] is in quadrant I, [latex]\\sin{\\theta}[\/latex] must be positive, so the solution is [latex]\\frac{3}{5}[\/latex]. See Figure A below.\r\n<p style=\"text-align: center;\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164021\/CNX_Precalc_Figure_06_03_010.jpg\" alt=\"An illustration of a right triangle with an angle theta. Oppostie the angle theta is a side with length 3. Adjacent the angle theta is a side with length 4. The hypoteneuse has angle of length 5.\" \/><\/p>\r\n<p style=\"text-align: center;\"><strong>Figure A.<\/strong> Right triangle illustrating that if [latex]\\cos\\theta=\\frac{4}{5}[\/latex], then [latex]\\sin\\theta=\\frac{3}{5}[\/latex]<\/p>\r\nWe know that the inverse cosine always gives an angle on the interval [0, \u03c0], so we know that the sine of that angle must be positive; therefore [latex]\\sin\\left(\\cos^{\u22121}\\left(\\frac{4}{5}\\right)\\right)=\\sin\\theta=\\frac{3}{5}[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Finding the Area of a Region<\/h3>\r\n<div id=\"fs-id1165041831020\" data-type=\"problem\">\r\n<p id=\"fs-id1165041831026\">Find the area of the region between the graph of [latex]f\\left(x\\right)=\\sqrt{{x}^{2}-9}[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over the interval [latex]\\left[3,5\\right][\/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-id1165041962754\" data-type=\"solution\">\r\n<p id=\"fs-id1165041962757\">First, sketch a rough graph of the region described in the problem, as shown in the following figure.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_07_03_007\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233818\/CNX_Calc_Figure_07_03_007.jpg\" alt=\"This figure is the graph of the function f(x) = the square root of (x^2-9). It is an increasing curve that starts on the x-axis at 3 and is in the first quadrant. Under the curve above the x-axis is a shaded region bounded to the right at x = 5.\" width=\"487\" height=\"239\" data-media-type=\"image\/jpeg\" \/> Figure 7. Calculating the area of the shaded region requires evaluating an integral with a trigonometric substitution.[\/caption]<\/figure>\r\n<p id=\"fs-id1165040730057\">We can see that the area is [latex]A={\\displaystyle\\int }_{3}^{5}\\sqrt{{x}^{2}-9}dx[\/latex]. To evaluate this definite integral, substitute [latex]x=3\\sec\\theta [\/latex] and [latex]dx=3\\sec\\theta \\tan\\theta d\\theta [\/latex]. We must also change the limits of integration. If [latex]x=3[\/latex], then [latex]3=3\\sec\\theta [\/latex] and hence [latex]\\theta =0[\/latex]. If [latex]x=5[\/latex], then [latex]\\theta ={\\sec}^{-1}\\left(\\frac{5}{3}\\right)[\/latex]. After making these substitutions and simplifying, we have<\/p>\r\n\r\n<div id=\"fs-id1165040687696\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill \\text{Area}&amp; ={\\displaystyle\\int }_{3}^{5}\\sqrt{{x}^{2}-9}dx\\hfill &amp; &amp; &amp; \\\\ &amp; ={\\displaystyle\\int }_{0}^{{\\sec}^{-1}\\left(5\\text{\/}3\\right)}9{\\tan}^{2}\\theta \\sec\\theta d\\theta \\hfill &amp; &amp; &amp; \\text{Use }{\\tan}^{2}\\theta =1-{\\sec}^{2}\\theta .\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{{\\sec}^{-1}\\left(5\\text{\/}3\\right)}9\\left({\\sec}^{2}\\theta -1\\right)\\sec\\theta d\\theta \\hfill &amp; &amp; &amp; \\text{Expand.}\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{{\\sec}^{-1}\\left(5\\text{\/}3\\right)}9\\left({\\sec}^{3}\\theta -\\sec\\theta \\right)d\\theta \\hfill &amp; &amp; &amp; \\text{Evaluate the integral.}\\hfill \\\\ &amp; =\\left(\\frac{9}{2}\\text{ln}|\\sec\\theta +\\tan\\theta |+\\frac{9}{2}\\sec\\theta \\tan\\theta \\right)-9\\text{ln}|\\sec\\theta +\\tan\\theta ||{}_{\\begin{array}{c}\\\\ 0\\end{array}}^{\\begin{array}{c}{\\sec}^{-1}\\left(5\\text{\/}3\\right)\\\\ \\end{array}}\\hfill &amp; &amp; &amp; \\text{Simplify.}\\hfill \\\\ &amp; =\\frac{9}{2}\\sec\\theta \\tan\\theta -\\frac{9}{2}\\text{ln}|\\sec\\theta +\\tan\\theta ||{}_{\\begin{array}{c}\\\\ 0\\end{array}}^{\\begin{array}{c}{\\sec}^{-1}\\left(5\\text{\/}3\\right)\\\\ \\end{array}}\\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Evaluate. Use }\\sec\\left({\\sec}^{-1}\\frac{5}{3}\\right)=\\frac{5}{3}\\hfill \\\\ \\text{and }\\tan\\left({\\sec}^{-1}\\frac{5}{3}\\right)=\\frac{4}{3}.\\hfill \\end{array}\\hfill \\\\ &amp; =\\frac{9}{2}\\cdot \\frac{5}{3}\\cdot \\frac{4}{3}-\\frac{9}{2}\\text{ln}|\\frac{5}{3}+\\frac{4}{3}|-\\left(\\frac{9}{2}\\cdot 1\\cdot 0-\\frac{9}{2}\\text{ln}|1+0|\\right)\\hfill &amp; &amp; &amp; \\\\ &amp; =10-\\frac{9}{2}\\text{ln}3.\\hfill &amp; &amp; &amp; \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165041757006\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1165041757010\" data-type=\"exercise\">\r\n<div id=\"fs-id1165041757013\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1165041757013\" data-type=\"problem\">\r\n<p id=\"fs-id1165041757015\">Evaluate [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{{x}^{2}-4}}[\/latex]. Assume that [latex]x&gt;2[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558869\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558869\"]\r\n<div id=\"fs-id1165040743205\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165040743213\">Substitute [latex]x=2\\sec\\theta [\/latex] and [latex]dx=2\\sec\\theta \\tan\\theta d\\theta [\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558879\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558879\"]\r\n<div id=\"fs-id1165042272930\" data-type=\"solution\">\r\n<p id=\"fs-id1165042272932\">[latex]\\text{ln}|\\frac{x}{2}+\\frac{\\sqrt{{x}^{2}-4}}{2}|+C[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/bCIrhw0sjlU?controls=0&amp;start=2177&amp;end=2312&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.3TrigonometricSubstitution2177to2312_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.3 Trigonometric Substitution\" here (opens in new window)<\/a>.\r\n\r\n<\/section>\r\n<div data-type=\"glossary\"><\/div>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve integration problems involving the square root of a sum or difference of two squares<\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1165041917030\" data-depth=\"1\">\n<h2 data-type=\"title\">Integrals Involving [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex]<\/h2>\n<p id=\"fs-id1165041987630\">Before developing a general strategy for integrals containing [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex], consider the integral [latex]\\displaystyle\\int \\sqrt{9-{x}^{2}}dx[\/latex]. This integral cannot be evaluated using any of the techniques we have discussed so far. However, if we make the substitution [latex]x=3\\sin\\theta[\/latex], we have [latex]dx=3\\cos\\theta d\\theta[\/latex]. After substituting into the integral, we have<\/p>\n<div id=\"fs-id1165042199492\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\sqrt{9-{x}^{2}}dx={\\displaystyle\\int }^{\\text{ }}\\sqrt{9-{\\left(3\\sin\\theta \\right)}^{2}}3\\cos\\theta d\\theta[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042090183\">After simplifying, we have<\/p>\n<div id=\"fs-id1165042101997\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}\\sqrt{9-{x}^{2}}dx={\\displaystyle\\int }^{\\text{ }}9\\sqrt{1-{\\sin}^{2}\\theta }\\cos\\theta d\\theta[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041909368\">Letting [latex]1-{\\sin}^{2}\\theta ={\\cos}^{2}\\theta[\/latex], we now have<\/p>\n<div id=\"fs-id1165040752105\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}\\sqrt{9-{x}^{2}}dx={\\displaystyle\\int }^{\\text{ }}9\\sqrt{{\\cos}^{2}\\theta }\\cos\\theta d\\theta[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041781797\">Assuming that [latex]\\cos\\theta \\ge 0[\/latex], we have<\/p>\n<div id=\"fs-id1165042007513\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }^{\\text{ }}\\sqrt{9-{x}^{2}}dx={\\displaystyle\\int }^{\\text{ }}9{\\cos}^{2}\\theta d\\theta[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042126138\">At this point, we can evaluate the integral using the techniques developed for integrating powers and products of trigonometric functions. Before completing this example, let\u2019s take a look at the general theory behind this idea.<\/p>\n<p id=\"fs-id1165042128862\">To evaluate integrals involving [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex], we make the substitution [latex]x=a\\sin\\theta[\/latex] and [latex]dx=a\\cos\\theta[\/latex]. To see that this actually makes sense, consider the following argument: The domain of [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex] is [latex]\\left[\\text{-}a,a\\right][\/latex]. Thus, [latex]\\text{-}a\\le x\\le a[\/latex]. Consequently, [latex]-1\\le \\frac{x}{a}\\le 1[\/latex]. Since the range of [latex]\\sin{x}[\/latex] over [latex]\\left[\\text{-}\\frac{\\pi}{2},\\frac{\\pi}{2}\\right][\/latex] is [latex]\\left[-1,1\\right][\/latex], there is a unique angle [latex]\\theta[\/latex] satisfying [latex]\\text{-}\\frac{\\pi}{2}\\le \\theta \\le \\frac{\\pi}{2}[\/latex] so that [latex]\\sin\\theta =x\\text{\/}a[\/latex], or equivalently, so that [latex]x=a\\sin\\theta[\/latex]. If we substitute [latex]x=a\\sin\\theta[\/latex] into [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex], we get<\/p>\n<div id=\"fs-id1165041762738\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill \\sqrt{{a}^{2}-{x}^{2}}& =\\sqrt{{a}^{2}-{\\left(a\\sin\\theta \\right)}^{2}}\\hfill & & & \\text{Let }x=a\\sin\\theta \\text{ where }-\\frac{\\pi }{2}\\le \\theta \\le \\frac{\\pi }{2}.\\text{Simplify.}\\hfill \\\\ & =\\sqrt{{a}^{2}-{a}^{2}{\\sin}^{2}\\theta }\\hfill & & & \\text{Factor out }{a}^{2}.\\hfill \\\\ & =\\sqrt{{a}^{2}\\left(1-{\\sin}^{2}\\theta \\right)}\\hfill & & & \\text{Substitute }1-{\\sin}^{2}x={\\cos}^{2}x.\\hfill \\\\ & =\\sqrt{{a}^{2}{\\cos}^{2}\\theta }\\hfill & & & \\text{Take the square root.}\\hfill \\\\ & =|a\\cos\\theta |\\hfill & & & \\\\ & =a\\cos\\theta .\\hfill & & & \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041814740\">Since [latex]\\cos\\theta \\ge 0[\/latex] on [latex]-\\frac{\\pi }{2}\\le \\theta \\le \\frac{\\pi }{2}[\/latex] and [latex]a>0[\/latex], [latex]|a\\cos\\theta |=a\\cos\\theta[\/latex]. We can see, from this discussion, that by making the substitution [latex]x=a\\sin\\theta[\/latex], we are able to convert an integral involving a radical into an integral involving trigonometric functions. After we evaluate the integral, we can convert the solution back to an expression involving [latex]x[\/latex]. To see how to do this, let\u2019s begin by assuming that [latex]0<x<a[\/latex]. In this case, [latex]0<\\theta <\\frac{\\pi }{2}[\/latex]. Since [latex]\\sin\\theta =\\frac{x}{a}[\/latex], we can draw the reference triangle in Figure 1 to assist in expressing the values of [latex]\\cos\\theta[\/latex], [latex]\\tan\\theta[\/latex], and the remaining trigonometric functions in terms of [latex]x[\/latex].\u00a0 Before drawing the reference triangle, we recall the definition of trigonometric functions in terms of ratios of side lengths below.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Right Triangle Trigonometry Relationships<\/h3>\n<p>Given a right triangle with an acute angle of [latex]\\theta[\/latex],<\/p>\n<p>[latex]\\begin{align}&\\sin \\left(\\theta \\right)=\\frac{\\text{opposite}}{\\text{hypotenuse}} \\\\ &\\cos \\left(\\theta \\right)=\\frac{\\text{adjacent}}{\\text{hypotenuse}} \\\\ &\\tan \\left(\\theta \\right)=\\frac{\\text{opposite}}{\\text{adjacent}} \\end{align}[\/latex]<\/p>\n<p>A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of &#8220;Sine is opposite over hypotenuse, Cosine is adjacent over hypotenuse, Tangent is opposite over adjacent.&#8221;<\/p>\n<\/div>\n<p>It can be shown that this triangle actually produces the correct values of the trigonometric functions evaluated at [latex]\\theta[\/latex] for all [latex]\\theta[\/latex] satisfying [latex]-\\frac{\\pi }{2}\\le \\theta \\le \\frac{\\pi }{2}[\/latex]. It is useful to observe that the expression [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex] actually appears as the length of one side of the triangle. Last, should [latex]\\theta[\/latex] appear by itself, we use [latex]\\theta ={\\sin}^{-1}\\left(\\frac{x}{a}\\right)[\/latex].<\/p>\n<figure id=\"CNX_Calc_Figure_07_03_001\"><figcaption><\/figcaption><div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233757\/CNX_Calc_Figure_07_03_001.jpg\" alt=\"This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled a, the vertical leg is labeled x, and the horizontal leg is labeled as the square root of (a^2 \u2013 x^2). To the left of the triangle is the equation sin(theta) = x\/a.\" width=\"487\" height=\"181\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. A reference triangle can help express the trigonometric functions evaluated at [latex]\\theta [\/latex] in terms of [latex]x[\/latex].<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165040764019\">The essential part of this discussion is summarized in the following problem-solving strategy.<\/p>\n<div id=\"fs-id1165042314803\" class=\"problem-solving\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox examples\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Problem-Solving Strategy: Integrating Expressions Involving [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex]<\/h3>\n<hr \/>\n<ol id=\"fs-id1165040638957\" type=\"1\">\n<li>It is a good idea to make sure the integral cannot be evaluated easily in another way. For example, although this method can be applied to integrals of the form [latex]\\displaystyle\\int \\frac{1}{\\sqrt{{a}^{2}-{x}^{2}}}dx[\/latex], [latex]\\displaystyle\\int \\frac{x}{\\sqrt{{a}^{2}-{x}^{2}}}dx[\/latex], and [latex]\\displaystyle\\int x\\sqrt{{a}^{2}-{x}^{2}}dx[\/latex], they can each be integrated directly either by formula or by a simple <em data-effect=\"italics\">u<\/em>-substitution.<\/li>\n<li>Make the substitution [latex]x=a\\sin\\theta[\/latex] and [latex]dx=a\\cos\\theta d\\theta[\/latex]. <em data-effect=\"italics\">Note<\/em>: This substitution yields [latex]\\sqrt{{a}^{2}-{x}^{2}}=a\\cos\\theta[\/latex].<\/li>\n<li>Simplify the expression.<\/li>\n<li>Evaluate the integral using techniques from the section on trigonometric integrals.<\/li>\n<li>Use the reference triangle from Figure 1 to rewrite the result in terms of [latex]x[\/latex]. You may also need to use some trigonometric identities and the relationship [latex]\\theta ={\\sin}^{-1}\\left(\\frac{x}{a}\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165041837235\" data-type=\"example\">\n<div id=\"fs-id1165042128156\" data-type=\"exercise\">\n<div id=\"fs-id1165042093676\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Integrating an Expression Involving [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex]<\/h3>\n<div id=\"fs-id1165041837235\" data-type=\"example\">\n<div id=\"fs-id1165042128156\" data-type=\"exercise\">\n<div id=\"fs-id1165042093676\" data-type=\"problem\">\n<p id=\"fs-id1165042297418\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}\\sqrt{9-{x}^{2}}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558899\">Show Solution<\/span><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041913990\" data-type=\"solution\">\n<p id=\"fs-id1165041893409\">Begin by making the substitutions [latex]x=3\\sin\\theta[\/latex] and [latex]dx=3\\cos\\theta d\\theta[\/latex]. Since [latex]\\sin\\theta =\\frac{x}{3}[\/latex], we can construct the reference triangle shown in the following figure.<\/p>\n<figure id=\"CNX_Calc_Figure_07_03_002\"><figcaption><\/figcaption><div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233800\/CNX_Calc_Figure_07_03_002.jpg\" alt=\"This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled 3, the vertical leg is labeled x, and the horizontal leg is labeled as the square root of (9 \u2013 x^2). To the left of the triangle is the equation sin(theta) = x\/3.\" width=\"487\" height=\"181\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. A reference triangle can be constructed for this example.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165040796218\">Thus,<\/p>\n<div id=\"fs-id1165041949238\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }^{\\text{ }}\\sqrt{9-{x}^{2}}dx& ={\\displaystyle\\int }^{\\text{ }}\\sqrt{9-{\\left(3\\sin\\theta \\right)}^{2}}3\\cos\\theta d\\theta \\hfill & & & \\text{Substitute }x=3\\sin\\theta \\text{ and }dx=3\\cos\\theta d\\theta .\\hfill \\\\ & ={\\displaystyle\\int }^{\\text{ }}\\sqrt{9\\left(1-{\\sin}^{2}\\theta \\right)}3\\cos\\theta d\\theta \\hfill & & & \\text{Simplify.}\\hfill \\\\ & ={\\displaystyle\\int }^{\\text{ }}\\sqrt{9{\\cos}^{2}\\theta }3\\cos\\theta d\\theta \\hfill & & & \\text{Substitute }{\\cos}^{2}\\theta =1-{\\sin}^{2}\\theta .\\hfill \\\\ & ={\\displaystyle\\int }^{\\text{ }}3|\\cos\\theta |3\\cos\\theta d\\theta \\hfill & & & \\text{Take the square root.}\\hfill \\\\ & ={\\displaystyle\\int }^{\\text{ }}9{\\cos}^{2}\\theta d\\theta \\hfill & & & \\begin{array}{c}\\text{Simplify. Since}-\\frac{\\pi }{2}\\le \\theta \\le \\frac{\\pi }{2},\\cos\\theta \\ge 0\\text{ and }\\hfill \\\\ |\\cos\\theta |=\\cos\\theta .\\hfill \\end{array}\\hfill \\\\ & ={\\displaystyle\\int }^{\\text{ }}9\\left(\\frac{1}{2}+\\frac{1}{2}\\cos\\left(2\\theta \\right)\\right)d\\theta \\hfill & & & \\begin{array}{c}\\text{Use the strategy for integrating an even power}\\hfill \\\\ \\text{of }\\cos\\theta .\\hfill \\end{array}\\hfill \\\\ & =\\frac{9}{2}\\theta +\\frac{9}{4}\\sin\\left(2\\theta \\right)+C\\hfill & & & \\text{Evaluate the integral.}\\hfill \\\\ & =\\frac{9}{2}\\theta +\\frac{9}{4}\\left(2\\sin\\theta \\cos\\theta \\right)+C\\hfill & & & \\text{Substitute }\\sin\\left(2\\theta \\right)=2\\sin\\theta \\cos\\theta .\\hfill \\\\ & =\\frac{9}{2}{\\sin}^{-1}\\left(\\frac{x}{3}\\right)+\\frac{9}{2}\\cdot \\frac{x}{3}\\cdot \\frac{\\sqrt{9-{x}^{2}}}{3}+C\\hfill & & & \\begin{array}{c}\\text{Substitute }{\\sin}^{-1}\\left(\\frac{x}{3}\\right)=\\theta \\text{ and }\\sin\\theta =\\frac{x}{3}.\\text{Use}\\hfill \\\\ \\text{the reference triangle to see that}\\hfill \\\\ \\cos\\theta =\\frac{\\sqrt{9-{x}^{2}}}{3}\\text{and make this substitution.}\\hfill \\end{array}\\hfill \\\\ & =\\frac{9}{2}{\\sin}^{-1}\\left(\\frac{x}{3}\\right)+\\frac{x\\sqrt{9-{x}^{2}}}{2}+C.\\hfill & & & \\text{Simplify.}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042229562\" data-type=\"example\">\n<div id=\"fs-id1165042073728\" data-type=\"exercise\">\n<div id=\"fs-id1165042073730\" data-type=\"problem\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Integrating an Expression Involving [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex]<\/h3>\n<div id=\"fs-id1165042073730\" data-type=\"problem\">\n<p id=\"fs-id1165042004532\">Evaluate [latex]\\displaystyle\\int \\frac{\\sqrt{4-{x}^{2}}}{x}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558898\">Show Solution<\/span><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040797018\" data-type=\"solution\">\n<p id=\"fs-id1165041759427\">First make the substitutions [latex]x=2\\sin\\theta[\/latex] and [latex]dx=2\\cos\\theta d\\theta[\/latex]. Since [latex]\\sin\\theta =\\frac{x}{2}[\/latex], we can construct the reference triangle shown in the following figure.<\/p>\n<figure id=\"CNX_Calc_Figure_07_03_003\"><figcaption><\/figcaption><div style=\"width: 390px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233803\/CNX_Calc_Figure_07_03_003.jpg\" alt=\"This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The vertical leg is labeled x, and the horizontal leg is labeled as the square root of (4 \u2013 x^2). To the left of the triangle is the equation sin(theta) = x\/2.\" width=\"380\" height=\"177\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. A reference triangle can be constructed for this example.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165040739936\">Thus,<\/p>\n<div id=\"fs-id1165040697663\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int \\frac{\\sqrt{4-{x}^{2}}}{x}dx}& ={\\displaystyle\\int \\frac{\\sqrt{4-{\\left(2\\sin\\theta \\right)}^{2}}}{2\\sin\\theta }2\\cos\\theta d\\theta} \\hfill & & & \\text{Substitute}x=2\\sin\\theta \\text{and}=2\\cos\\theta d\\theta .\\hfill \\\\ & ={\\displaystyle\\int \\frac{2{\\cos}^{2}\\theta }{\\sin\\theta }d\\theta} \\hfill & & & \\text{Substitute}{\\cos}^{2}\\theta =1-{\\sin}^{2}\\theta \\text{and simplify.}\\hfill \\\\ & ={\\displaystyle\\int \\frac{2\\left(1-{\\sin}^{2}\\theta \\right)}{\\sin\\theta }d\\theta} \\hfill & & & \\text{Substitute}{\\sin}^{2}\\theta =1-{\\cos}^{2}\\theta .\\hfill \\\\ & ={\\displaystyle\\int \\left(2\\csc\\theta -2\\sin\\theta \\right)d\\theta} \\hfill & & & \\begin{array}{c}\\text{Separate the numerator, simplify, and use}\\hfill \\\\ \\csc\\theta =\\frac{1}{\\sin\\theta }.\\hfill \\end{array}\\hfill \\\\ & =2\\text{ln}|\\csc\\theta -\\cot\\theta |+2\\cos\\theta +C\\hfill & & & \\text{Evaluate the integral.}\\hfill \\\\ & =2\\text{ln}|\\frac{2}{x}-\\frac{\\sqrt{4-{x}^{2}}}{x}|+\\sqrt{4-{x}^{2}}+C.\\hfill & & & \\begin{array}{c}\\text{Use the reference triangle to rewrite the}\\hfill \\\\ \\text{expression in terms of}x\\text{and simplify.}\\hfill \\end{array}\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165042092095\">In the next example, we see that we sometimes have a choice of methods.<\/p>\n<div id=\"fs-id1165042092098\" data-type=\"example\">\n<div id=\"fs-id1165040642691\" data-type=\"exercise\">\n<div id=\"fs-id1165040642694\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Integrating an Expression Involving [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex] Two Ways<\/h3>\n<div id=\"fs-id1165040642694\" data-type=\"problem\">\n<p id=\"fs-id1165040744148\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{x}^{3}\\sqrt{1-{x}^{2}}dx[\/latex] two ways: first by using the substitution [latex]u=1-{x}^{2}[\/latex] and then by using a trigonometric substitution.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558897\">Show Solution<\/span><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041958780\" data-type=\"solution\">\n<p id=\"fs-id1167794138271\"><strong data-effect=\"bold\">Method 1<\/strong><\/p>\n<p id=\"fs-id1165041958783\">Let [latex]u=1-{x}^{2}[\/latex] and hence [latex]{x}^{2}=1-u[\/latex]. Thus, [latex]du=-2xdx[\/latex]. In this case, the integral becomes<\/p>\n<div id=\"fs-id1165042301844\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int ^{\\text{ }}{x}^{3}\\sqrt{1-{x}^{2}}dx}& =-\\frac{1}{2}{\\displaystyle\\int ^{\\text{ }}{x}^{2}\\sqrt{1-{x}^{2}}\\left(-2xdx\\right)}\\hfill & & & \\text{Make the substitution.}\\hfill \\\\ & =-\\frac{1}{2}{\\displaystyle\\int ^{\\text{ }}\\left(1-u\\right)\\sqrt{u}du}\\hfill & & & \\text{Expand the expression.}\\hfill \\\\ & =-\\frac{1}{2}{\\displaystyle\\int \\left({u}^{1\\text{\/}2}-{u}^{3\\text{\/}2}\\right)du}\\hfill & & & \\text{Evaluate the integral.}\\hfill \\\\ & =-\\frac{1}{2}\\left(\\frac{2}{3}{u}^{3\\text{\/}2}-\\frac{2}{5}{u}^{5\\text{\/}2}\\right)+C\\hfill & & & \\text{Rewrite in terms of}x.\\hfill \\\\ & =-\\frac{1}{3}{\\left(1-{x}^{2}\\right)}^{3\\text{\/}2}+\\frac{1}{5}{\\left(1-{x}^{2}\\right)}^{5\\text{\/}2}+C.\\hfill & & & \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793929406\"><strong data-effect=\"bold\">Method 2<\/strong><\/p>\n<p id=\"fs-id1165041766752\">Let [latex]x=\\sin\\theta[\/latex]. In this case, [latex]dx=\\cos\\theta d\\theta[\/latex]. Using this substitution, we have<\/p>\n<div id=\"fs-id1165040793998\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }^{\\text{ }}{x}^{3}\\sqrt{1-{x}^{2}}dx& ={\\displaystyle\\int }^{\\text{ }}{\\sin}^{3}\\theta {\\cos}^{2}\\theta d\\theta \\hfill & & & \\\\ & ={\\displaystyle\\int }^{\\text{ }}\\left(1-{\\cos}^{2}\\theta \\right){\\cos}^{2}\\theta \\sin\\theta d\\theta \\hfill & & & \\text{Let }u=\\cos\\theta .\\text{Thus, }du=\\text{-}\\sin\\theta d\\theta .\\hfill \\\\ & ={\\displaystyle\\int }^{\\text{ }}\\left({u}^{4}-{u}^{2}\\right)du\\hfill & & & \\\\ & =\\frac{1}{5}{u}^{5}-\\frac{1}{3}{u}^{3}+C\\hfill & & & \\text{Substitute }\\cos\\theta =u.\\hfill \\\\ & =\\frac{1}{5}{\\cos}^{5}\\theta -\\frac{1}{3}{\\cos}^{3}\\theta +C\\hfill & & & \\begin{array}{c}\\text{Use a reference triangle to see that}\\hfill \\\\ \\cos\\theta =\\sqrt{1-{x}^{2}}.\\hfill \\end{array}\\hfill \\\\ & =\\frac{1}{5}{\\left(1-{x}^{2}\\right)}^{5\\text{\/}2}-\\frac{1}{3}{\\left(1-{x}^{2}\\right)}^{3\\text{\/}2}+C.\\hfill & & & \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042094001\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1165042094004\" data-type=\"exercise\">\n<div id=\"fs-id1165042033160\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1165042033160\" data-type=\"problem\">\n<p id=\"fs-id1165042033163\">Rewrite the integral [latex]\\displaystyle\\int \\frac{{x}^{3}}{\\sqrt{25-{x}^{2}}}dx[\/latex] using the appropriate trigonometric substitution (do not evaluate the integral).<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558895\">Hint<\/span><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041841595\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165041951969\">Substitute [latex]x=5\\sin\\theta[\/latex] and [latex]dx=5\\cos\\theta d\\theta[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558896\">Show Solution<\/span><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042237758\" data-type=\"solution\">\n<p id=\"fs-id1165040640415\">[latex]{\\displaystyle\\int }^{\\text{ }}125{\\sin}^{3}\\theta d\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/bCIrhw0sjlU?controls=0&amp;start=1071&amp;end=1182&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.3TrigonometricSubstitution1071to1182_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.3 Trigonometric Substitution&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<section id=\"fs-id1165041813590\" data-depth=\"1\">\n<h2 data-type=\"title\">Integrating Expressions Involving [latex]\\sqrt{{a}^{2}+{x}^{2}}[\/latex]<\/h2>\n<p id=\"fs-id1165042278317\">For integrals containing [latex]\\sqrt{{a}^{2}+{x}^{2},}[\/latex] let\u2019s first consider the domain of this expression. Since [latex]\\sqrt{{a}^{2}+{x}^{2}}[\/latex] is defined for all real values of [latex]x[\/latex], we restrict our choice to those trigonometric functions that have a range of all real numbers. Thus, our choice is restricted to selecting either [latex]x=a\\tan\\theta[\/latex] or [latex]x=a\\cot\\theta[\/latex]. Either of these substitutions would actually work, but the standard substitution is [latex]x=a\\tan\\theta[\/latex] or, equivalently, [latex]\\tan\\theta =\\frac{x}{a}[\/latex]. With this substitution, we make the assumption that [latex]\\text{-}\\frac{\\pi}{2}<\\theta <\\frac{\\pi}{2}[\/latex], so that we also have [latex]\\theta ={\\tan}^{-1}\\left(x\\text{\/}a\\right)[\/latex]. The procedure for using this substitution is outlined in the following problem-solving strategy.<\/p>\n<div id=\"fs-id1165042110570\" class=\"problem-solving\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox examples\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Problem-Solving Strategy: Integrating Expressions Involving [latex]\\sqrt{{a}^{2}+{x}^{2}}[\/latex]<\/h3>\n<hr \/>\n<ol id=\"fs-id1165041979050\" type=\"1\">\n<li>Check to see whether the integral can be evaluated easily by using another method. In some cases, it is more convenient to use an alternative method.<\/li>\n<li>Substitute [latex]x=a\\tan\\theta[\/latex] and [latex]dx=a{\\sec}^{2}\\theta d\\theta[\/latex]. This substitution yields<span data-type=\"newline\"><br \/>\n<\/span><br \/>\n[latex]\\sqrt{{a}^{2}+{x}^{2}}=\\sqrt{{a}^{2}+{\\left(a\\tan\\theta \\right)}^{2}}=\\sqrt{{a}^{2}\\left(1+{\\tan}^{2}\\theta \\right)}=\\sqrt{{a}^{2}{\\sec}^{2}\\theta }=|a\\sec\\theta |=a\\sec\\theta[\/latex]. (Since [latex]-\\frac{\\pi }{2}<\\theta <\\frac{\\pi }{2}[\/latex] and [latex]\\sec\\theta >0[\/latex] over this interval, [latex]|a\\sec\\theta |=a\\sec\\theta[\/latex] )<\/li>\n<li>Simplify the expression.<\/li>\n<li>Evaluate the integral using techniques from the section on trigonometric integrals.<\/li>\n<li>Use the reference triangle from Figure 4 to rewrite the result in terms of [latex]x[\/latex]. You may also need to use some trigonometric identities and the relationship [latex]\\theta ={\\tan}^{-1}\\left(\\frac{x}{a}\\right)[\/latex]. (<em data-effect=\"italics\">Note<\/em>: The reference triangle is based on the assumption that [latex]x>0[\/latex]; however, the trigonometric ratios produced from the reference triangle are the same as the ratios for which [latex]x \\le{0}[\/latex] )<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<figure id=\"CNX_Calc_Figure_07_03_004\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233806\/CNX_Calc_Figure_07_03_004.jpg\" alt=\"This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled the square root of (a^2+x^2), the vertical leg is labeled x, and the horizontal leg is labeled a. To the left of the triangle is the equation tan(theta) = x\/a.\" width=\"487\" height=\"173\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. A reference triangle can be constructed to express the trigonometric functions evaluated at [latex]\\theta [\/latex] in terms of [latex]x[\/latex].<\/p>\n<\/div>\n<\/figure>\n<div id=\"fs-id1165042089656\" data-type=\"example\">\n<div id=\"fs-id1165042089658\" data-type=\"exercise\">\n<div id=\"fs-id1165042089660\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Integrating an Expression Involving [latex]\\sqrt{{a}^{2}+{x}^{2}}[\/latex]<\/h3>\n<div id=\"fs-id1165042089660\" data-type=\"problem\">\n<p id=\"fs-id1165040773006\">Evaluate [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}[\/latex] and check the solution by differentiating.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558894\">Show Solution<\/span><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041887608\" data-type=\"solution\">\n<p id=\"fs-id1165041887610\">Begin with the substitution [latex]x=\\tan\\theta[\/latex] and [latex]dx={\\sec}^{2}\\theta d\\theta[\/latex]. Since [latex]\\tan\\theta =x[\/latex], draw the reference triangle in the following figure.<\/p>\n<figure id=\"CNX_Calc_Figure_07_03_005\"><figcaption><\/figcaption><div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233810\/CNX_Calc_Figure_07_03_005.jpg\" alt=\"This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled the square root of (1+x^2), the vertical leg is labeled x, and the horizontal leg is labeled 1. To the left of the triangle is the equation tan(theta) = x\/1.\" width=\"487\" height=\"173\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. The reference triangle for this example.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165040750681\">Thus,<\/p>\n<div id=\"fs-id1165040750684\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}}& ={\\displaystyle\\int \\frac{{\\sec}^{2}\\theta }{\\sec\\theta }d\\theta }\\hfill & & & \\begin{array}{c}\\text{Substitute}x=\\tan\\theta \\text{and}dx={\\sec}^{2}\\theta d\\theta .\\text{This}\\hfill \\\\ \\text{substitution makes}\\sqrt{1+{x}^{2}}=\\sec\\theta .\\text{Simplify.}\\hfill \\end{array}\\hfill \\\\ & ={\\displaystyle\\int ^{\\text{ }}\\sec\\theta d\\theta }\\hfill & & & \\text{Evaluate the integral.}\\hfill \\\\ & =\\text{ln}|\\sec\\theta +\\tan\\theta |+C\\hfill & & & \\begin{array}{c}\\text{Use the reference triangle to express the result}\\hfill \\\\ \\text{in terms of}x.\\hfill \\end{array}\\hfill \\\\ & =\\text{ln}|\\sqrt{1+{x}^{2}}+x|+C.\\hfill & & & \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040744515\">To check the solution, differentiate:<\/p>\n<div id=\"fs-id1165040744518\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{d}{dx}\\left(\\text{ln}|\\sqrt{1+{x}^{2}}+x|\\right)& =\\frac{1}{\\sqrt{1+{x}^{2}}+x}\\cdot \\left(\\frac{x}{\\sqrt{1+{x}^{2}}}+1\\right)\\hfill \\\\ & =\\frac{1}{\\sqrt{1+{x}^{2}}+x}\\cdot \\frac{x+\\sqrt{1+{x}^{2}}}{\\sqrt{1+{x}^{2}}}\\hfill \\\\ & =\\frac{1}{\\sqrt{1+{x}^{2}}}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041980958\">Since [latex]\\sqrt{1+{x}^{2}}+x>0[\/latex] for all values of [latex]x[\/latex], we could rewrite [latex]\\text{ln}|\\sqrt{1+{x}^{2}}+x|+C=\\text{ln}\\left(\\sqrt{1+{x}^{2}}+x\\right)+C[\/latex], if desired.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042301696\" data-type=\"example\">\n<div id=\"fs-id1165042301698\" data-type=\"exercise\">\n<div id=\"fs-id1165042301700\" data-type=\"problem\">\n<div data-type=\"title\">\n<p>In the example below, we explore how to use hyperbolic trigonometric functions as an alternative substitution method. First, we briefly review the definition of these functions and their derivatives.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Definition and Derivatives of Hyperbolic Trigonometric Functions<\/h3>\n<p>For any real number [latex]x[\/latex], the hyperbolic sine and hyperbolic cosine are defined as:<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\sinh x = \\frac{e^x - e^{-x}}{2} \\: \\text{and}\\:\\cosh x = \\frac{e^x + e^{-x}}{2}[\/latex]<\/p>\n<p>Their derivatives are given by:<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\frac{d}{dx} \\left( \\sinh x \\right) = \\cosh x \\:\\text{and}\\:\\frac{d}{dx} \\left( \\cosh x \\right) = \\sinh x[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Evaluating [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}[\/latex] Using a Different Substitution<\/h3>\n<div id=\"fs-id1165042301700\" data-type=\"problem\">\n<p id=\"fs-id1165042272858\">Use the substitution [latex]x=\\text{sinh}\\theta[\/latex] to evaluate [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558893\">Show Solution<\/span><\/p>\n<div id=\"q44558893\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041828644\" data-type=\"solution\">\n<p id=\"fs-id1165041828646\">Because [latex]\\text{sinh}\\theta[\/latex] has a range of all real numbers, and [latex]1+{\\text{sinh}}^{2}\\theta ={\\text{cosh}}^{2}\\theta[\/latex], we may also use the substitution [latex]x=\\text{sinh}\\theta[\/latex] to evaluate this integral. In this case, [latex]dx=\\text{cosh}\\theta d\\theta[\/latex]. Consequently,<\/p>\n<div id=\"fs-id1165042004330\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}}& ={\\displaystyle\\int \\frac{\\text{cosh}\\theta }{\\sqrt{1+{\\text{sinh}}^{2}\\theta }}d\\theta }\\hfill & & & \\begin{array}{c}\\text{Substitute}x=\\text{sinh}\\theta \\text{and}dx=\\text{cosh}\\theta d\\theta .\\hfill \\\\ \\text{Substitute}1+{\\text{sinh}}^{2}\\theta ={\\text{cosh}}^{2}\\theta .\\hfill \\end{array}\\hfill \\\\ & ={\\displaystyle\\int \\frac{\\text{cosh}\\theta }{\\sqrt{{\\text{cosh}}^{2}\\theta }}d\\theta }\\hfill & & & \\sqrt{{\\text{cosh}}^{2}\\theta }=|\\text{cosh}\\theta |\\hfill \\\\ & ={\\displaystyle\\int \\frac{\\text{cosh}\\theta }{|\\text{cosh}\\theta |}d\\theta }\\hfill & & & |\\text{cosh}\\theta |=\\text{cosh}\\theta \\text{since}\\text{cosh}\\theta >0\\text{for all}\\theta .\\hfill \\\\ & ={\\displaystyle\\int \\frac{\\text{cosh}\\theta }{\\text{cosh}\\theta }d\\theta }\\hfill & & & \\text{Simplify.}\\hfill \\\\ & ={\\displaystyle\\int ^{\\text{ }}1d\\theta }\\hfill & & & \\text{Evaluate the integral.}\\hfill \\\\ & =\\theta +C\\hfill & & & \\text{Since}x=\\text{sinh}\\theta ,\\text{we know}\\theta ={\\text{sinh}}^{-1}x.\\hfill \\\\ & ={\\text{sinh}}^{-1}x+C.\\hfill & & & \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1165042063682\" data-type=\"commentary\">\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\n<p id=\"fs-id1165042063687\">This answer looks quite different from the answer obtained using the substitution [latex]x=\\tan\\theta[\/latex]. To see that the solutions are the same, set [latex]y={\\text{sinh}}^{-1}x[\/latex]. Thus, [latex]\\text{sinh}y=x[\/latex]. From this equation we obtain:<\/p>\n<div id=\"fs-id1165041843069\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{e}^{y}-{e}^{\\text{-}y}}{2}=x[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042277014\">After multiplying both sides by [latex]2{e}^{y}[\/latex] and rewriting, this equation becomes:<\/p>\n<div id=\"fs-id1165042202678\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{e}^{2y}-2x{e}^{y}-1=0[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042232174\">Use the quadratic equation to solve for [latex]{e}^{y}\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1165040645138\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{e}^{y}=\\frac{2x\\pm \\sqrt{4{x}^{2}+4}}{2}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040644008\">Simplifying, we have:<\/p>\n<div id=\"fs-id1165040644011\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{e}^{y}=x\\pm \\sqrt{{x}^{2}+1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040768460\">Since [latex]x-\\sqrt{{x}^{2}+1}<0[\/latex], it must be the case that [latex]{e}^{y}=x+\\sqrt{{x}^{2}+1}[\/latex]. Thus,<\/p>\n<div id=\"fs-id1165041921710\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=\\text{ln}\\left(x+\\sqrt{{x}^{2}+1}\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040743112\">Last, we obtain<\/p>\n<div id=\"fs-id1165042092492\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\text{sinh}}^{-1}x=\\text{ln}\\left(x+\\sqrt{{x}^{2}+1}\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042096551\">After we make the final observation that, since [latex]x+\\sqrt{{x}^{2}+1}>0[\/latex],<\/p>\n<div id=\"fs-id1165040713985\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{ln}\\left(x+\\sqrt{{x}^{2}+1}\\right)=\\text{ln}|\\sqrt{1+{x}^{2}}+x|[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040639545\">we see that the two different methods produced equivalent solutions.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165040639551\" data-type=\"example\">\n<div id=\"fs-id1165040639553\" data-type=\"exercise\">\n<div id=\"fs-id1165042279310\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Finding an Arc Length<\/h3>\n<div id=\"fs-id1165042279310\" data-type=\"problem\">\n<p id=\"fs-id1165042279315\">Find the length of the curve [latex]y={x}^{2}[\/latex] over the interval [latex]\\left[0,\\frac{1}{2}\\right][\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558892\">Show Solution<\/span><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040668429\" data-type=\"solution\">\n<p id=\"fs-id1165040668432\">Because [latex]\\frac{dy}{dx}=2x[\/latex], the arc length is given by<\/p>\n<div id=\"fs-id1165040744954\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{0}^{1\\text{\/}2}\\sqrt{1+{\\left(2x\\right)}^{2}}dx={\\displaystyle\\int }_{0}^{1\\text{\/}2}\\sqrt{1+4{x}^{2}}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040765593\">To evaluate this integral, use the substitution [latex]x=\\frac{1}{2}\\tan\\theta[\/latex] and [latex]dx=\\frac{1}{2}{\\sec}^{2}\\theta d\\theta[\/latex]. We also need to change the limits of integration. If [latex]x=0[\/latex], then [latex]\\theta =0[\/latex] and if [latex]x=\\frac{1}{2}[\/latex], then [latex]\\theta =\\frac{\\pi }{4}[\/latex]. Thus,<\/p>\n<div id=\"fs-id1165040665687\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }_{0}^{1\\text{\/}2}\\sqrt{1+4{x}^{2}}dx& ={\\displaystyle\\int }_{0}^{\\pi \\text{\/}4}\\sqrt{1+{\\tan}^{2}\\theta }\\frac{1}{2}{\\sec}^{2}\\theta d\\theta \\hfill & & & \\begin{array}{c}\\text{After substitution,}\\hfill \\\\ \\sqrt{1+4{x}^{2}}=\\tan\\theta .\\text{Substitute}\\hfill \\\\ 1+{\\tan}^{2}\\theta ={\\sec}^{2}\\theta \\text{ and simplify.}\\hfill \\end{array}\\hfill \\\\ & =\\frac{1}{2}{\\displaystyle\\int }_{0}^{\\pi \\text{\/}4}{\\sec}^{3}\\theta d\\theta \\hfill & & & \\begin{array}{c}\\text{We derived this integral in the}\\hfill \\\\ \\text{previous section.}\\hfill \\end{array}\\hfill \\\\ & =\\frac{1}{2}\\left(\\frac{1}{2}\\sec\\theta \\tan\\theta +\\text{ln}|\\sec\\theta +\\tan\\theta |\\right)|{}_{\\begin{array}{c}\\\\ 0\\end{array}}^{\\begin{array}{c}\\pi \\text{\/}4\\\\ \\end{array}}\\hfill & & & \\text{Evaluate and simplify.}\\hfill \\\\ & =\\frac{1}{4}\\left(\\sqrt{2}+\\text{ln}\\left(\\sqrt{2}+1\\right)\\right).\\hfill & & & \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042041805\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1165042041808\" data-type=\"exercise\">\n<div id=\"fs-id1165042041810\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1165042041810\" data-type=\"problem\">\n<p id=\"fs-id1165042041812\">Rewrite [latex]{\\displaystyle\\int }^{\\text{ }}{x}^{3}\\sqrt{{x}^{2}+4}dx[\/latex] by using a substitution involving [latex]\\tan\\theta[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558890\">Hint<\/span><\/p>\n<div id=\"q44558890\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041921699\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165041921706\">Use [latex]x=2\\tan\\theta[\/latex] and [latex]dx=2{\\sec}^{2}\\theta d\\theta[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558891\">Show Solution<\/span><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041952815\" data-type=\"solution\">\n<p id=\"fs-id1165041952817\">[latex]{\\displaystyle\\int }^{\\text{ }}32{\\tan}^{3}\\theta {\\sec}^{3}\\theta d\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/bCIrhw0sjlU?controls=0&amp;start=1688&amp;end=1797&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.3TrigonometricSubstitution1688to1797_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.3 Trigonometric Substitution&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<section id=\"fs-id1165040682763\" data-depth=\"1\">\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5524\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5524&theme=oea&iframe_resize_id=ohm5524&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2 data-type=\"title\">Integrating Expressions Involving [latex]\\sqrt{{x}^{2}-{a}^{2}}[\/latex]<\/h2>\n<p id=\"fs-id1165042231872\">The domain of the expression [latex]\\sqrt{{x}^{2}-{a}^{2}}[\/latex] is [latex]\\left(\\text{-}\\infty ,\\text{-}a\\right]\\cup \\left[a,\\text{+}\\infty \\right)[\/latex]. Thus, either [latex]x\\le \\text{-}a[\/latex] or [latex]x\\ge a[\/latex]. Hence, [latex]\\frac{x}{a}\\le -1[\/latex] or [latex]\\frac{x}{a}\\ge 1[\/latex]. Since these intervals correspond to the range of [latex]\\sec\\theta[\/latex] on the set [latex]\\left[0,\\frac{\\pi }{2}\\right)\\cup \\left(\\frac{\\pi }{2},\\pi \\right][\/latex], it makes sense to use the substitution [latex]\\sec\\theta =\\frac{x}{a}[\/latex] or, equivalently, [latex]x=a\\sec\\theta[\/latex], where [latex]0\\le \\theta <\\frac{\\pi }{2}[\/latex] or [latex]\\frac{\\pi }{2}<\\theta \\le \\pi[\/latex]. The corresponding substitution for [latex]dx[\/latex] is [latex]dx=a\\sec\\theta \\tan\\theta d\\theta[\/latex]. The procedure for using this substitution is outlined in the following problem-solving strategy.<\/p>\n<div id=\"fs-id1165042235602\" class=\"problem-solving\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox examples\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Problem-Solving Strategy: Integrals Involving [latex]\\sqrt{{x}^{2}-{a}^{2}}[\/latex]<\/h3>\n<hr \/>\n<ol id=\"fs-id1165042108861\" type=\"1\">\n<li>Check to see whether the integral cannot be evaluated using another method. If so, we may wish to consider applying an alternative technique.<\/li>\n<li>Substitute [latex]x=a\\sec\\theta[\/latex] and [latex]dx=a\\sec\\theta \\tan\\theta d\\theta[\/latex]. This substitution yields<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1165041836430\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\sqrt{{x}^{2}-{a}^{2}}=\\sqrt{{\\left(a\\sec\\theta \\right)}^{2}-{a}^{2}}=\\sqrt{{a}^{2}\\left({\\sec}^{2}\\theta -1\\right)}=\\sqrt{{a}^{2}{\\tan}^{2}\\theta }=|a\\tan\\theta |[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nFor [latex]x\\ge a[\/latex], [latex]|a\\tan\\theta |=a\\tan\\theta[\/latex] and for [latex]x\\le -a[\/latex], [latex]|a\\tan\\theta |=\\text{-}a\\tan\\theta[\/latex].<\/li>\n<li>Simplify the expression.<\/li>\n<li>Evaluate the integral using techniques from the section on trigonometric integrals.<\/li>\n<li>Use the reference triangles from Figure 6 to rewrite the result in terms of [latex]x[\/latex]. You may also need to use some trigonometric identities and the relationship [latex]\\theta ={\\sec}^{-1}\\left(\\frac{x}{a}\\right)[\/latex]. (<em data-effect=\"italics\">Note<\/em>: We need both reference triangles, since the values of some of the trigonometric ratios are different depending on whether [latex]x\\ge a[\/latex] or [latex]x\\le \\text{-}a.[\/latex])<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<figure id=\"CNX_Calc_Figure_07_03_006\">\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233815\/CNX_Calc_Figure_07_03_006.jpg\" alt=\"This figure has two right triangles. The first triangle is in the first quadrant of the xy coordinate system and has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled x, the vertical leg is labeled the square root of (x^2-a^2), and the horizontal leg is labeled a. The horizontal leg is on the x-axis. To the left of the triangle is the equation sec(theta) = x\/a, x&gt;a. There are also the equations sin(theta)= the square root of (x^2-a^2)\/x, cos(theta) = a\/x, and tan(theta) = the square root of (x^2-a^2)\/a. The second triangle is in the second quadrant, with the hypotenuse labeled \u2013x. The horizontal leg is labeled \u2013a and is on the negative x-axis. The vertical leg is labeled the square root of (x^2-a^2). To the right of the triangle is the equation sec(theta) = x\/a, x&lt;-a. There are also the equations sin(theta)= the negative square root of (x^2-a^2)\/x, cos(theta) = a\/x, and tan(theta) = the negative square root of (x^2-a^2)\/a.\" width=\"975\" height=\"392\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. Use the appropriate reference triangle to express the trigonometric functions evaluated at [latex]\\theta [\/latex] in terms of [latex]x[\/latex].<\/p>\n<\/div>\n<\/figure>\n<div id=\"fs-id1165041831016\" data-type=\"example\">\n<div id=\"fs-id1165041831018\" data-type=\"exercise\">\n<div id=\"fs-id1165041831020\" data-type=\"problem\">\n<div data-type=\"title\">\n<p>For the next example, it is worthwhile to review the technique of how to precisely evaluate a trignometric function whose input is an inverse trigonometric function.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Evaluating Trigonometric Functions composed with Inverse Trigonometric Functions<\/h3>\n<p>Suppose we wanted to evaluate a trigonometric function composed with an inverse trigonometric function, for example:\u00a0[latex]\\sin\\left(\\cos^{\u22121}\\left(\\frac{4}{5}\\right)\\right)[\/latex].<\/p>\n<p>Beginning with the inside, we can say there is some angle such that [latex]\\theta=\\cos^{\u22121}(\\frac{4}{5})[\/latex], which means [latex]\\cos\\theta=\\frac{4}{5}[\/latex], and we are looking for [latex]\\sin\\theta[\/latex]. We can use the Pythagorean identity to do this.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} &\\sin^{2}\\theta+\\cos^{2}\\theta=1 && \\text{Use our known value for cosine.} \\\\ &\\sin^{2}\\theta+\\left(\\frac{4}{5}\\right)^{2}=1 && \\text{Solve for sine.} \\\\ &\\sin^{2}\\theta=1\u2212\\frac{16}{25} \\\\ &\\sin\\theta=\\pm\\sqrt{\\frac{9}{25}}=\\pm\\frac{3}{5} \\end{align}[\/latex]<\/p>\n<p>Since [latex]\\theta=\\cos^{\u22121}(\\frac{4}{5})[\/latex] is in quadrant I, [latex]\\sin{\\theta}[\/latex] must be positive, so the solution is [latex]\\frac{3}{5}[\/latex]. See Figure A below.<\/p>\n<p style=\"text-align: center;\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164021\/CNX_Precalc_Figure_06_03_010.jpg\" alt=\"An illustration of a right triangle with an angle theta. Oppostie the angle theta is a side with length 3. Adjacent the angle theta is a side with length 4. The hypoteneuse has angle of length 5.\" \/><\/p>\n<p style=\"text-align: center;\"><strong>Figure A.<\/strong> Right triangle illustrating that if [latex]\\cos\\theta=\\frac{4}{5}[\/latex], then [latex]\\sin\\theta=\\frac{3}{5}[\/latex]<\/p>\n<p>We know that the inverse cosine always gives an angle on the interval [0, \u03c0], so we know that the sine of that angle must be positive; therefore [latex]\\sin\\left(\\cos^{\u22121}\\left(\\frac{4}{5}\\right)\\right)=\\sin\\theta=\\frac{3}{5}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Finding the Area of a Region<\/h3>\n<div id=\"fs-id1165041831020\" data-type=\"problem\">\n<p id=\"fs-id1165041831026\">Find the area of the region between the graph of [latex]f\\left(x\\right)=\\sqrt{{x}^{2}-9}[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over the interval [latex]\\left[3,5\\right][\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558889\">Show Solution<\/span><\/p>\n<div id=\"q44558889\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041962754\" data-type=\"solution\">\n<p id=\"fs-id1165041962757\">First, sketch a rough graph of the region described in the problem, as shown in the following figure.<\/p>\n<figure id=\"CNX_Calc_Figure_07_03_007\"><figcaption><\/figcaption><div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233818\/CNX_Calc_Figure_07_03_007.jpg\" alt=\"This figure is the graph of the function f(x) = the square root of (x^2-9). It is an increasing curve that starts on the x-axis at 3 and is in the first quadrant. Under the curve above the x-axis is a shaded region bounded to the right at x = 5.\" width=\"487\" height=\"239\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7. Calculating the area of the shaded region requires evaluating an integral with a trigonometric substitution.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165040730057\">We can see that the area is [latex]A={\\displaystyle\\int }_{3}^{5}\\sqrt{{x}^{2}-9}dx[\/latex]. To evaluate this definite integral, substitute [latex]x=3\\sec\\theta[\/latex] and [latex]dx=3\\sec\\theta \\tan\\theta d\\theta[\/latex]. We must also change the limits of integration. If [latex]x=3[\/latex], then [latex]3=3\\sec\\theta[\/latex] and hence [latex]\\theta =0[\/latex]. If [latex]x=5[\/latex], then [latex]\\theta ={\\sec}^{-1}\\left(\\frac{5}{3}\\right)[\/latex]. After making these substitutions and simplifying, we have<\/p>\n<div id=\"fs-id1165040687696\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill \\text{Area}& ={\\displaystyle\\int }_{3}^{5}\\sqrt{{x}^{2}-9}dx\\hfill & & & \\\\ & ={\\displaystyle\\int }_{0}^{{\\sec}^{-1}\\left(5\\text{\/}3\\right)}9{\\tan}^{2}\\theta \\sec\\theta d\\theta \\hfill & & & \\text{Use }{\\tan}^{2}\\theta =1-{\\sec}^{2}\\theta .\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{{\\sec}^{-1}\\left(5\\text{\/}3\\right)}9\\left({\\sec}^{2}\\theta -1\\right)\\sec\\theta d\\theta \\hfill & & & \\text{Expand.}\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{{\\sec}^{-1}\\left(5\\text{\/}3\\right)}9\\left({\\sec}^{3}\\theta -\\sec\\theta \\right)d\\theta \\hfill & & & \\text{Evaluate the integral.}\\hfill \\\\ & =\\left(\\frac{9}{2}\\text{ln}|\\sec\\theta +\\tan\\theta |+\\frac{9}{2}\\sec\\theta \\tan\\theta \\right)-9\\text{ln}|\\sec\\theta +\\tan\\theta ||{}_{\\begin{array}{c}\\\\ 0\\end{array}}^{\\begin{array}{c}{\\sec}^{-1}\\left(5\\text{\/}3\\right)\\\\ \\end{array}}\\hfill & & & \\text{Simplify.}\\hfill \\\\ & =\\frac{9}{2}\\sec\\theta \\tan\\theta -\\frac{9}{2}\\text{ln}|\\sec\\theta +\\tan\\theta ||{}_{\\begin{array}{c}\\\\ 0\\end{array}}^{\\begin{array}{c}{\\sec}^{-1}\\left(5\\text{\/}3\\right)\\\\ \\end{array}}\\hfill & & & \\begin{array}{c}\\text{Evaluate. Use }\\sec\\left({\\sec}^{-1}\\frac{5}{3}\\right)=\\frac{5}{3}\\hfill \\\\ \\text{and }\\tan\\left({\\sec}^{-1}\\frac{5}{3}\\right)=\\frac{4}{3}.\\hfill \\end{array}\\hfill \\\\ & =\\frac{9}{2}\\cdot \\frac{5}{3}\\cdot \\frac{4}{3}-\\frac{9}{2}\\text{ln}|\\frac{5}{3}+\\frac{4}{3}|-\\left(\\frac{9}{2}\\cdot 1\\cdot 0-\\frac{9}{2}\\text{ln}|1+0|\\right)\\hfill & & & \\\\ & =10-\\frac{9}{2}\\text{ln}3.\\hfill & & & \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165041757006\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1165041757010\" data-type=\"exercise\">\n<div id=\"fs-id1165041757013\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1165041757013\" data-type=\"problem\">\n<p id=\"fs-id1165041757015\">Evaluate [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{{x}^{2}-4}}[\/latex]. Assume that [latex]x>2[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558869\">Hint<\/span><\/p>\n<div id=\"q44558869\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040743205\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165040743213\">Substitute [latex]x=2\\sec\\theta[\/latex] and [latex]dx=2\\sec\\theta \\tan\\theta d\\theta[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558879\">Show Solution<\/span><\/p>\n<div id=\"q44558879\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042272930\" data-type=\"solution\">\n<p id=\"fs-id1165042272932\">[latex]\\text{ln}|\\frac{x}{2}+\\frac{\\sqrt{{x}^{2}-4}}{2}|+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/bCIrhw0sjlU?controls=0&amp;start=2177&amp;end=2312&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.3TrigonometricSubstitution2177to2312_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.3 Trigonometric Substitution&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<div data-type=\"glossary\"><\/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-813\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>3.3 Trigonometric Substitution. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":10,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"3.3 Trigonometric Substitution\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-813","chapter","type-chapter","status-publish","hentry"],"part":158,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/813","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":28,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/813\/revisions"}],"predecessor-version":[{"id":2629,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/813\/revisions\/2629"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/158"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/813\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=813"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=813"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=813"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=813"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}