{"id":1143,"date":"2021-06-30T17:01:59","date_gmt":"2021-06-30T17:01:59","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/integrals-that-result-in-inverse-trig-functions\/"},"modified":"2022-03-19T03:24:02","modified_gmt":"2022-03-19T03:24:02","slug":"integrals-that-result-in-inverse-trig-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/integrals-that-result-in-inverse-trig-functions\/","title":{"raw":"Integrals that Result in Inverse Trig Functions","rendered":"Integrals that Result in Inverse Trig Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Integrate functions resulting in inverse trigonometric functions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Integrals that Result in Inverse Sine Functions<\/h2>\r\n<p id=\"fs-id1170571596362\">Let us begin with the three formulas. Along with these formulas, we use substitution to evaluate the integrals. We prove the formula for the inverse sine integral.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Integration Formulas Resulting in Inverse Trigonometric Functions<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170571528516\">The following integration formulas yield inverse trigonometric functions:<\/p>\r\n\r\n<ol id=\"fs-id1170572178178\">\r\n \t<li>\r\n<div id=\"fs-id1170572554001\" class=\"equation\">[latex]\\displaystyle\\int \\frac{du}{\\sqrt{{a}^{2}-{u}^{2}}}={ \\sin }^{-1}\\frac{u}{|a|}+C[\/latex]<\/div><\/li>\r\n \t<li>\r\n<div id=\"fs-id1170572607870\" class=\"equation\">[latex]\\displaystyle\\int \\frac{du}{{a}^{2}+{u}^{2}}=\\frac{1}{a}\\phantom{\\rule{0.05em}{0ex}}{ \\tan }^{-1}\\frac{u}{a}+C[\/latex]<\/div><\/li>\r\n \t<li>\r\n<div class=\"equation\">[latex]\\displaystyle\\int \\frac{du}{u\\sqrt{{u}^{2}-{a}^{2}}}=\\frac{1}{|a|}\\phantom{\\rule{0.05em}{0ex}}{ \\sec }^{-1}\\frac{|u|}{a}+C[\/latex]<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1170572216525\" class=\"bc-section section\">\r\n<h3>Proof<\/h3>\r\n<p id=\"fs-id1170571652151\">Let [latex]y={ \\sin }^{-1}\\dfrac{x}{a}.[\/latex] Then [latex]a \\sin y=x.[\/latex] Now let\u2019s use implicit differentiation. We obtain<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\frac{d}{dx}(a \\sin y)&amp; =\\hfill &amp; \\frac{d}{dx}(x)\\hfill \\\\ \\\\ \\hfill a \\cos y\\frac{dy}{dx}&amp; =\\hfill &amp; 1\\hfill \\\\ \\hfill \\frac{dy}{dx}&amp; =\\hfill &amp; \\frac{1}{a \\cos y}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572163738\">For [latex]-\\frac{\\pi }{2}\\le y\\le \\frac{\\pi }{2}, \\cos y\\ge 0.[\/latex] Thus, applying the Pythagorean identity [latex]{ \\sin }^{2}y+{ \\cos }^{2}y=1,[\/latex] we have [latex] \\cos y=\\sqrt{1={ \\sin }^{2}y}.[\/latex] This gives<\/p>\r\n\r\n<div id=\"fs-id1170571660097\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\frac{1}{a \\cos y}\\hfill &amp; =\\frac{1}{a\\sqrt{1-{ \\sin }^{2}y}}\\hfill \\\\ \\\\ &amp; =\\frac{1}{\\sqrt{{a}^{2}-{a}^{2}{ \\sin }^{2}y}}\\hfill \\\\ &amp; =\\frac{1}{\\sqrt{{a}^{2}-{x}^{2}}}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572101851\">Then for [latex]\\text{\u2212}a\\le x\\le a,[\/latex] we have<\/p>\r\n\r\n<div id=\"fs-id1170572175143\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{1}{\\sqrt{{a}^{2}-{u}^{2}}}du={ \\sin }^{-1}\\left(\\frac{u}{a}\\right)+C.[\/latex]<\/div>\r\n<p id=\"fs-id1170572346856\">[latex]_\\blacksquare[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Evaluating a Definite Integral Using Inverse Trigonometric Functions<\/h3>\r\nEvaluate the definite integral [latex]{\\displaystyle\\int }_{0}^{\\frac{1}{2}}\\dfrac{dx}{\\sqrt{1-{x}^{2}}}.[\/latex]\r\n\r\n[reveal-answer q=\"7116223\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"7116223\"]\r\n<div id=\"fs-id1170571596198\" class=\"exercise\">\r\n<div class=\"solution\">\r\n\r\nWe can go directly to the formula for the antiderivative in the rule on integration formulas resulting in inverse trigonometric functions, and then evaluate the definite integral. We have\r\n<div id=\"fs-id1170572203987\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int }_{0}^{\\frac{1}{2}}\\dfrac{dx}{\\sqrt{1-{x}^{2}}}\\hfill &amp; ={ \\sin }^{-1}x{|}_{0}^{\\frac{1}{2}}\\hfill \\\\ &amp; ={ \\sin }^{-1}{\\frac{1}{2}}-{ \\sin }^{-1}0\\hfill \\\\ &amp; =\\frac{\\pi }{6}-0\\hfill \\\\ &amp; =\\frac{\\pi }{6}.\\hfill \\end{array}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572216525\" class=\"bc-section section\">\r\n<div class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the antiderivative of [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1-16{x}^{2}}}.[\/latex]\r\n<div class=\"exercise\">[reveal-answer q=\"4782055\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"4782055\"]\r\n<p id=\"fs-id1170572089952\">Substitute [latex]u=4x[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572480280\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572480280\"]\r\n\r\n[latex]\\frac{1}{4}\\phantom{\\rule{0.05em}{0ex}}{ \\sin }^{-1}(4x)+C[\/latex]\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/yVwFRhfDP_Y?controls=0&amp;start=173&amp;end=278&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.7IntegralsResultingInInverseTrigonometricFunctions173to278_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"1.7 Integrals Resulting in Inverse Trigonometric Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170572206423\" class=\"textbook exercises\">\r\n<h3>Example: Finding an Antiderivative Involving an Inverse Trigonometric Function<\/h3>\r\nEvaluate the integral [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{4-9{x}^{2}}}.[\/latex]\r\n<div class=\"exercise\">[reveal-answer q=\"fs-id1170572449550\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572449550\"]Substitute [latex]u=3x.[\/latex] Then [latex]du=3dx[\/latex] and we have\r\n<div id=\"fs-id1170571618996\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{dx}{\\sqrt{4-9{x}^{2}}}=\\frac{1}{3}\\displaystyle\\int \\frac{du}{\\sqrt{4-{u}^{2}}}.[\/latex]<\/div>\r\n<p id=\"fs-id1170572141145\">Applying the formula with [latex]a=2,[\/latex] we obtain<\/p>\r\n\r\n<div id=\"fs-id1170572130022\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int} \\dfrac{dx}{\\sqrt{4-9{x}^{2}}}\\hfill &amp; =\\frac{1}{3}{\\displaystyle\\int} \\dfrac{du}{\\sqrt{4-{u}^{2}}}\\hfill \\\\ \\\\ &amp; =\\frac{1}{3}{ \\sin }^{-1}\\left(\\frac{u}{2}\\right)+C\\hfill \\\\ &amp; =\\frac{1}{3}{ \\sin }^{-1}\\left(\\frac{3x}{2}\\right)+C.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170572449549\"><span style=\"font-size: 0.9em;\">[\/hidden-answer]<\/span><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572141583\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the indefinite integral using an inverse trigonometric function and substitution for [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{9-{x}^{2}}}.[\/latex]\r\n<div id=\"fs-id1170571698177\" class=\"exercise\">[reveal-answer q=\"833621\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"833621\"]\r\n<p id=\"fs-id1170572274575\">Use the formula in the rule on integration formulas resulting in inverse trigonometric functions.<\/p>\r\n[\/hidden-answer]\r\n<p id=\"fs-id1170572557808\">[reveal-answer q=\"fs-id1170572557808\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572557808\"][latex]{ \\sin }^{-1}\\left(\\frac{x}{3}\\right)+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/yVwFRhfDP_Y?controls=0&amp;start=365&amp;end=404&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.7IntegralsResultingInInverseTrigonometricFunctions365to404_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"1.7 Integrals Resulting in Inverse Trigonometric Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbox exercises\">\r\n<h3>Evaluating a Definite Integral<\/h3>\r\nEvaluate the definite integral [latex]{\\displaystyle\\int }_{0}^{\\sqrt{3}\\text{\/}2}\\dfrac{du}{\\sqrt{1-{u}^{2}}}.[\/latex]\r\n\r\n[reveal-answer q=\"fs-id1170572150550\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572150550\"]\r\n\r\nThe format of the problem matches the inverse sine formula. Thus,\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int }_{0}^{\\sqrt{3}\\text{\/}2}\\dfrac{du}{\\sqrt{1-{u}^{2}}}\\hfill &amp; ={ \\sin }^{-1}u{|}_{0}^{\\sqrt{3}\\text{\/}2}\\hfill \\\\ &amp; =\\left[{ \\sin }^{-1}(\\frac{\\sqrt{3}}{2})\\right]-\\left[{ \\sin }^{-1}(0)\\right]\\hfill \\\\ &amp; =\\frac{\\pi }{3}.\\hfill \\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<div>\r\n<h2>Integrals Resulting in Other Inverse Trigonometric Functions<\/h2>\r\n<p id=\"fs-id1170572106889\">There are six inverse trigonometric functions. However, only three integration formulas are noted in the rule on integration formulas resulting in inverse trigonometric functions because the remaining three are negative versions of the ones we use. The only difference is whether the integrand is positive or negative. Rather than memorizing three more formulas, if the integrand is negative, simply factor out \u22121 and evaluate the integral using one of the formulas already provided. To close this section, we examine one more formula: the integral resulting in the inverse tangent function.<\/p>\r\n\r\n<div id=\"fs-id1170572206423\" class=\"textbook exercises\">\r\n<h3>Example: Finding an Antiderivative Involving the Inverse Tangent Function<\/h3>\r\nFind an antiderivative of [latex]\\displaystyle\\int \\frac{1}{1+4{x}^{2}}dx.[\/latex]\r\n<div class=\"exercise\">[reveal-answer q=\"fs-id1170572449549\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572449549\"]\r\n<p id=\"fs-id1170572449549\">Comparing this problem with the formulas stated in the rule on integration formulas resulting in inverse trigonometric functions, the integrand looks similar to the formula for [latex]{ \\tan }^{-1}u+C.[\/latex] So we use substitution, letting [latex]u=2x,[\/latex] then [latex]du=2dx[\/latex] and [latex]\\frac{1}{2}du=dx.[\/latex] Then, we have<\/p>\r\n\r\n<div id=\"fs-id1170572548844\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{2}\\displaystyle\\int \\frac{1}{1+{u}^{2}}du=\\frac{1}{2}\\phantom{\\rule{0.05em}{0ex}}{ \\tan }^{-1}u+C=\\frac{1}{2}\\phantom{\\rule{0.05em}{0ex}}{ \\tan }^{-1}(2x)+C.[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572608152\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse substitution to find the antiderivative of [latex]\\displaystyle\\int \\frac{dx}{25+4{x}^{2}}.[\/latex]\r\n\r\n[reveal-answer q=\"745067\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"745067\"]Use the solving strategy from finding an antiderivative involving an inverse trigonometric function\u00a0and the rule on integration formulas resulting in inverse trigonometric functions.[\/hidden-answer]\r\n<div id=\"fs-id1170572308096\" class=\"exercise\">[reveal-answer q=\"fs-id1170571562665\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571562665\"]\r\n<p id=\"fs-id1170571562665\">[latex]\\frac{1}{10}\\phantom{\\rule{0.05em}{0ex}}{ \\tan }^{-1}\\left(\\frac{2x}{5}\\right)+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572350769\" class=\"textbook exercises\">\r\n<h3>Example: Applying the Integration Formulas<\/h3>\r\nFind the antiderivative of [latex]\\displaystyle\\int \\frac{1}{9+{x}^{2}}dx.[\/latex]\r\n<div id=\"fs-id1170572230273\" class=\"exercise\">\r\n<div class=\"solution\">\r\n<p id=\"fs-id1170572237513\">[reveal-answer q=\"7084662\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"7084662\"]<\/p>\r\nApply the formula with [latex]a=3.[\/latex] Then,\r\n<div id=\"fs-id1170572221498\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{dx}{9+{x}^{2}}=\\frac{1}{3}\\phantom{\\rule{0.05em}{0ex}}{ \\tan }^{-1}\\left(\\frac{x}{3}\\right)+C[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571681250\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the antiderivative of [latex]\\displaystyle\\int \\frac{dx}{16+{x}^{2}}.[\/latex]\r\n<div>[reveal-answer q=\"555236\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"555236\"]Follow the steps in the previous example.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1170572243063\" class=\"exercise\">[reveal-answer q=\"fs-id1170571609481\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571609481\"]\r\n<p id=\"fs-id1170571609481\">[latex]\\frac{1}{4}\\phantom{\\rule{0.05em}{0ex}}{ \\tan }^{-1}(\\frac{x}{4})+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571714255\" class=\"textbox exercises\">\r\n<h3>Evaluating a Definite Integral<\/h3>\r\nEvaluate the definite integral [latex]{\\displaystyle\\int }_{\\sqrt{3}\\text{\/}3}^{\\sqrt{3}}\\dfrac{dx}{1+{x}^{2}}.[\/latex]\r\n<div id=\"fs-id1170572220235\" class=\"exercise\">[reveal-answer q=\"fs-id1170572099768\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572099768\"]\r\n<p id=\"fs-id1170572099768\">Use the formula for the inverse tangent. We have<\/p>\r\n\r\n<div id=\"fs-id1170572099771\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int }_{\\sqrt{3}\\text{\/}3}^{\\sqrt{3}}\\dfrac{dx}{1+{x}^{2}}\\hfill &amp; ={ \\tan }^{-1}x{|}_{\\sqrt{3}\\text{\/}3}^{\\sqrt{3}}\\hfill \\\\ &amp; =\\left[{ \\tan }^{-1}(\\sqrt{3})\\right]-\\left[{ \\tan }^{-1}\\left(\\frac{\\sqrt{3}}{3}\\right)\\right]\\hfill \\\\ &amp; =\\frac{\\pi }{6}.\\hfill \\end{array}[\/latex]<\/div>\r\n<div><\/div>\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572369374\">Evaluate the definite integral [latex]{\\displaystyle\\int }_{0}^{2}\\dfrac{dx}{4+{x}^{2}}.[\/latex]<\/p>\r\n[reveal-answer q=\"3852\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"3852\"]Follow the procedures from the previous example to solve the problem.[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572176762\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572176762\"]\r\n<p id=\"fs-id1170572176762\">[latex]\\frac{\\pi }{8}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/yVwFRhfDP_Y?controls=0&amp;start=973&amp;end=1054&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266836\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266836\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.7IntegralsResultingInInverseTrigonometricFunctions973to1054_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"1.7 Integrals Resulting in Inverse Trigonometric Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]20248[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Integrate functions resulting in inverse trigonometric functions<\/li>\n<\/ul>\n<\/div>\n<h2>Integrals that Result in Inverse Sine Functions<\/h2>\n<p id=\"fs-id1170571596362\">Let us begin with the three formulas. Along with these formulas, we use substitution to evaluate the integrals. We prove the formula for the inverse sine integral.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Integration Formulas Resulting in Inverse Trigonometric Functions<\/h3>\n<hr \/>\n<p id=\"fs-id1170571528516\">The following integration formulas yield inverse trigonometric functions:<\/p>\n<ol id=\"fs-id1170572178178\">\n<li>\n<div id=\"fs-id1170572554001\" class=\"equation\">[latex]\\displaystyle\\int \\frac{du}{\\sqrt{{a}^{2}-{u}^{2}}}={ \\sin }^{-1}\\frac{u}{|a|}+C[\/latex]<\/div>\n<\/li>\n<li>\n<div id=\"fs-id1170572607870\" class=\"equation\">[latex]\\displaystyle\\int \\frac{du}{{a}^{2}+{u}^{2}}=\\frac{1}{a}\\phantom{\\rule{0.05em}{0ex}}{ \\tan }^{-1}\\frac{u}{a}+C[\/latex]<\/div>\n<\/li>\n<li>\n<div class=\"equation\">[latex]\\displaystyle\\int \\frac{du}{u\\sqrt{{u}^{2}-{a}^{2}}}=\\frac{1}{|a|}\\phantom{\\rule{0.05em}{0ex}}{ \\sec }^{-1}\\frac{|u|}{a}+C[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1170572216525\" class=\"bc-section section\">\n<h3>Proof<\/h3>\n<p id=\"fs-id1170571652151\">Let [latex]y={ \\sin }^{-1}\\dfrac{x}{a}.[\/latex] Then [latex]a \\sin y=x.[\/latex] Now let\u2019s use implicit differentiation. We obtain<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\frac{d}{dx}(a \\sin y)& =\\hfill & \\frac{d}{dx}(x)\\hfill \\\\ \\\\ \\hfill a \\cos y\\frac{dy}{dx}& =\\hfill & 1\\hfill \\\\ \\hfill \\frac{dy}{dx}& =\\hfill & \\frac{1}{a \\cos y}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572163738\">For [latex]-\\frac{\\pi }{2}\\le y\\le \\frac{\\pi }{2}, \\cos y\\ge 0.[\/latex] Thus, applying the Pythagorean identity [latex]{ \\sin }^{2}y+{ \\cos }^{2}y=1,[\/latex] we have [latex]\\cos y=\\sqrt{1={ \\sin }^{2}y}.[\/latex] This gives<\/p>\n<div id=\"fs-id1170571660097\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\frac{1}{a \\cos y}\\hfill & =\\frac{1}{a\\sqrt{1-{ \\sin }^{2}y}}\\hfill \\\\ \\\\ & =\\frac{1}{\\sqrt{{a}^{2}-{a}^{2}{ \\sin }^{2}y}}\\hfill \\\\ & =\\frac{1}{\\sqrt{{a}^{2}-{x}^{2}}}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572101851\">Then for [latex]\\text{\u2212}a\\le x\\le a,[\/latex] we have<\/p>\n<div id=\"fs-id1170572175143\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{1}{\\sqrt{{a}^{2}-{u}^{2}}}du={ \\sin }^{-1}\\left(\\frac{u}{a}\\right)+C.[\/latex]<\/div>\n<p id=\"fs-id1170572346856\">[latex]_\\blacksquare[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Evaluating a Definite Integral Using Inverse Trigonometric Functions<\/h3>\n<p>Evaluate the definite integral [latex]{\\displaystyle\\int }_{0}^{\\frac{1}{2}}\\dfrac{dx}{\\sqrt{1-{x}^{2}}}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q7116223\">Show Solution<\/span><\/p>\n<div id=\"q7116223\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571596198\" class=\"exercise\">\n<div class=\"solution\">\n<p>We can go directly to the formula for the antiderivative in the rule on integration formulas resulting in inverse trigonometric functions, and then evaluate the definite integral. We have<\/p>\n<div id=\"fs-id1170572203987\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int }_{0}^{\\frac{1}{2}}\\dfrac{dx}{\\sqrt{1-{x}^{2}}}\\hfill & ={ \\sin }^{-1}x{|}_{0}^{\\frac{1}{2}}\\hfill \\\\ & ={ \\sin }^{-1}{\\frac{1}{2}}-{ \\sin }^{-1}0\\hfill \\\\ & =\\frac{\\pi }{6}-0\\hfill \\\\ & =\\frac{\\pi }{6}.\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572216525\" class=\"bc-section section\">\n<div class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the antiderivative of [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1-16{x}^{2}}}.[\/latex]<\/p>\n<div class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q4782055\">Hint<\/span><\/p>\n<div id=\"q4782055\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572089952\">Substitute [latex]u=4x[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572480280\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572480280\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{1}{4}\\phantom{\\rule{0.05em}{0ex}}{ \\sin }^{-1}(4x)+C[\/latex]<\/p>\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\/yVwFRhfDP_Y?controls=0&amp;start=173&amp;end=278&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.7IntegralsResultingInInverseTrigonometricFunctions173to278_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;1.7 Integrals Resulting in Inverse Trigonometric Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170572206423\" class=\"textbook exercises\">\n<h3>Example: Finding an Antiderivative Involving an Inverse Trigonometric Function<\/h3>\n<p>Evaluate the integral [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{4-9{x}^{2}}}.[\/latex]<\/p>\n<div class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572449550\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572449550\" class=\"hidden-answer\" style=\"display: none\">Substitute [latex]u=3x.[\/latex] Then [latex]du=3dx[\/latex] and we have<\/p>\n<div id=\"fs-id1170571618996\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{dx}{\\sqrt{4-9{x}^{2}}}=\\frac{1}{3}\\displaystyle\\int \\frac{du}{\\sqrt{4-{u}^{2}}}.[\/latex]<\/div>\n<p id=\"fs-id1170572141145\">Applying the formula with [latex]a=2,[\/latex] we obtain<\/p>\n<div id=\"fs-id1170572130022\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int} \\dfrac{dx}{\\sqrt{4-9{x}^{2}}}\\hfill & =\\frac{1}{3}{\\displaystyle\\int} \\dfrac{du}{\\sqrt{4-{u}^{2}}}\\hfill \\\\ \\\\ & =\\frac{1}{3}{ \\sin }^{-1}\\left(\\frac{u}{2}\\right)+C\\hfill \\\\ & =\\frac{1}{3}{ \\sin }^{-1}\\left(\\frac{3x}{2}\\right)+C.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572449549\"><span style=\"font-size: 0.9em;\"><\/div>\n<\/div>\n<p><\/span><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572141583\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the indefinite integral using an inverse trigonometric function and substitution for [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{9-{x}^{2}}}.[\/latex]<\/p>\n<div id=\"fs-id1170571698177\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q833621\">Hint<\/span><\/p>\n<div id=\"q833621\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572274575\">Use the formula in the rule on integration formulas resulting in inverse trigonometric functions.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1170572557808\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572557808\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572557808\" class=\"hidden-answer\" style=\"display: none\">[latex]{ \\sin }^{-1}\\left(\\frac{x}{3}\\right)+C[\/latex]<\/p>\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\/yVwFRhfDP_Y?controls=0&amp;start=365&amp;end=404&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.7IntegralsResultingInInverseTrigonometricFunctions365to404_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;1.7 Integrals Resulting in Inverse Trigonometric Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Evaluating a Definite Integral<\/h3>\n<p>Evaluate the definite integral [latex]{\\displaystyle\\int }_{0}^{\\sqrt{3}\\text{\/}2}\\dfrac{du}{\\sqrt{1-{u}^{2}}}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572150550\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572150550\" class=\"hidden-answer\" style=\"display: none\">\n<p>The format of the problem matches the inverse sine formula. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int }_{0}^{\\sqrt{3}\\text{\/}2}\\dfrac{du}{\\sqrt{1-{u}^{2}}}\\hfill & ={ \\sin }^{-1}u{|}_{0}^{\\sqrt{3}\\text{\/}2}\\hfill \\\\ & =\\left[{ \\sin }^{-1}(\\frac{\\sqrt{3}}{2})\\right]-\\left[{ \\sin }^{-1}(0)\\right]\\hfill \\\\ & =\\frac{\\pi }{3}.\\hfill \\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\"><\/div>\n<\/div>\n<\/div>\n<div>\n<h2>Integrals Resulting in Other Inverse Trigonometric Functions<\/h2>\n<p id=\"fs-id1170572106889\">There are six inverse trigonometric functions. However, only three integration formulas are noted in the rule on integration formulas resulting in inverse trigonometric functions because the remaining three are negative versions of the ones we use. The only difference is whether the integrand is positive or negative. Rather than memorizing three more formulas, if the integrand is negative, simply factor out \u22121 and evaluate the integral using one of the formulas already provided. To close this section, we examine one more formula: the integral resulting in the inverse tangent function.<\/p>\n<div id=\"fs-id1170572206423\" class=\"textbook exercises\">\n<h3>Example: Finding an Antiderivative Involving the Inverse Tangent Function<\/h3>\n<p>Find an antiderivative of [latex]\\displaystyle\\int \\frac{1}{1+4{x}^{2}}dx.[\/latex]<\/p>\n<div class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572449549\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572449549\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572449549\">Comparing this problem with the formulas stated in the rule on integration formulas resulting in inverse trigonometric functions, the integrand looks similar to the formula for [latex]{ \\tan }^{-1}u+C.[\/latex] So we use substitution, letting [latex]u=2x,[\/latex] then [latex]du=2dx[\/latex] and [latex]\\frac{1}{2}du=dx.[\/latex] Then, we have<\/p>\n<div id=\"fs-id1170572548844\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{2}\\displaystyle\\int \\frac{1}{1+{u}^{2}}du=\\frac{1}{2}\\phantom{\\rule{0.05em}{0ex}}{ \\tan }^{-1}u+C=\\frac{1}{2}\\phantom{\\rule{0.05em}{0ex}}{ \\tan }^{-1}(2x)+C.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572608152\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use substitution to find the antiderivative of [latex]\\displaystyle\\int \\frac{dx}{25+4{x}^{2}}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q745067\">Hint<\/span><\/p>\n<div id=\"q745067\" class=\"hidden-answer\" style=\"display: none\">Use the solving strategy from finding an antiderivative involving an inverse trigonometric function\u00a0and the rule on integration formulas resulting in inverse trigonometric functions.<\/div>\n<\/div>\n<div id=\"fs-id1170572308096\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571562665\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571562665\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571562665\">[latex]\\frac{1}{10}\\phantom{\\rule{0.05em}{0ex}}{ \\tan }^{-1}\\left(\\frac{2x}{5}\\right)+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572350769\" class=\"textbook exercises\">\n<h3>Example: Applying the Integration Formulas<\/h3>\n<p>Find the antiderivative of [latex]\\displaystyle\\int \\frac{1}{9+{x}^{2}}dx.[\/latex]<\/p>\n<div id=\"fs-id1170572230273\" class=\"exercise\">\n<div class=\"solution\">\n<p id=\"fs-id1170572237513\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q7084662\">Show Solution<\/span><\/p>\n<div id=\"q7084662\" class=\"hidden-answer\" style=\"display: none\">\n<p>Apply the formula with [latex]a=3.[\/latex] Then,<\/p>\n<div id=\"fs-id1170572221498\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{dx}{9+{x}^{2}}=\\frac{1}{3}\\phantom{\\rule{0.05em}{0ex}}{ \\tan }^{-1}\\left(\\frac{x}{3}\\right)+C[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571681250\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the antiderivative of [latex]\\displaystyle\\int \\frac{dx}{16+{x}^{2}}.[\/latex]<\/p>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q555236\">Hint<\/span><\/p>\n<div id=\"q555236\" class=\"hidden-answer\" style=\"display: none\">Follow the steps in the previous example.<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1170572243063\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571609481\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571609481\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571609481\">[latex]\\frac{1}{4}\\phantom{\\rule{0.05em}{0ex}}{ \\tan }^{-1}(\\frac{x}{4})+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571714255\" class=\"textbox exercises\">\n<h3>Evaluating a Definite Integral<\/h3>\n<p>Evaluate the definite integral [latex]{\\displaystyle\\int }_{\\sqrt{3}\\text{\/}3}^{\\sqrt{3}}\\dfrac{dx}{1+{x}^{2}}.[\/latex]<\/p>\n<div id=\"fs-id1170572220235\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572099768\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572099768\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572099768\">Use the formula for the inverse tangent. We have<\/p>\n<div id=\"fs-id1170572099771\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int }_{\\sqrt{3}\\text{\/}3}^{\\sqrt{3}}\\dfrac{dx}{1+{x}^{2}}\\hfill & ={ \\tan }^{-1}x{|}_{\\sqrt{3}\\text{\/}3}^{\\sqrt{3}}\\hfill \\\\ & =\\left[{ \\tan }^{-1}(\\sqrt{3})\\right]-\\left[{ \\tan }^{-1}\\left(\\frac{\\sqrt{3}}{3}\\right)\\right]\\hfill \\\\ & =\\frac{\\pi }{6}.\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<div class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572369374\">Evaluate the definite integral [latex]{\\displaystyle\\int }_{0}^{2}\\dfrac{dx}{4+{x}^{2}}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3852\">Hint<\/span><\/p>\n<div id=\"q3852\" class=\"hidden-answer\" style=\"display: none\">Follow the procedures from the previous example to solve the problem.<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572176762\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572176762\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572176762\">[latex]\\frac{\\pi }{8}[\/latex]<\/p>\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\/yVwFRhfDP_Y?controls=0&amp;start=973&amp;end=1054&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266836\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266836\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.7IntegralsResultingInInverseTrigonometricFunctions973to1054_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;1.7 Integrals Resulting in Inverse Trigonometric Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm20248\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20248&theme=oea&iframe_resize_id=ohm20248&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1143\">\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>1.7 Integrals Resulting in Inverse Trigonometric Functions. <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":30,"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\":\"1.7 Integrals Resulting in Inverse Trigonometric Functions\",\"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-1143","chapter","type-chapter","status-publish","hentry"],"part":1113,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1143","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":2,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1143\/revisions"}],"predecessor-version":[{"id":1372,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1143\/revisions\/1372"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/1113"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1143\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1143"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1143"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1143"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1143"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}