{"id":412,"date":"2025-02-13T19:44:42","date_gmt":"2025-02-13T19:44:42","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integrals-resulting-in-inverse-trigonometric-functions-learn-it-1\/"},"modified":"2025-02-13T19:44:42","modified_gmt":"2025-02-13T19:44:42","slug":"integrals-resulting-in-inverse-trigonometric-functions-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integrals-resulting-in-inverse-trigonometric-functions-learn-it-1\/","title":{"raw":"Integrals Resulting in Inverse Trigonometric Functions: Learn It 1","rendered":"Integrals Resulting in Inverse Trigonometric Functions: Learn It 1"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li>Calculate integrals that lead to inverse trigonometric function solutions<\/li>\n<\/ul>\n<\/section>\n<p>In this section we focus on integrals that result in inverse trigonometric functions. We have worked with these functions before.<\/p>\n<section class=\"textbox recall\">\n<p>Recall that trigonometric functions are not one-to-one unless the domains are restricted.<\/p>\n<\/section>\n<p>When working with inverses of trigonometric functions, we always need to be careful to take these restrictions into account. Also, we developed formulas for derivatives of inverse trigonometric functions. The formulas developed there give rise directly to integration formulas involving inverse trigonometric functions.<\/p>\n<h2>Integrals Resulting in Inverse Trigonometric 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<section class=\"textbox keyTakeaway\">\n<h3>integration formulas resulting in inverse trigonometric Functions<\/h3>\n<p id=\"fs-id1170571528516\">The following integration formulas yield inverse trigonometric functions:<\/p>\n<ol id=\"fs-id1170572178178\">\n\t<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\t<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\t<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<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr>\n<p id=\"fs-id1170571652151\">Let [latex]y={ \\sin }^{-1}\\dfrac{x}{a}.[\/latex] Then [latex]a \\sin y=x.[\/latex]<\/p>\n<p>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)&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>\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]<\/p>\n<p>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]<\/p>\n<p>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 &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>\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<\/section>\n<section class=\"textbox example\">\n<p>Evaluate the definite integral [latex]{\\displaystyle\\int }_{0}^{\\frac{1}{2}}\\dfrac{dx}{\\sqrt{1-{x}^{2}}}.[\/latex]<\/p>\n<p>[reveal-answer q=\"7116223\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"7116223\"]<\/p>\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.<\/p>\n<p>We have,<\/p>\n<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]<\/center>[\/hidden-answer]<\/section>\n<section class=\"textbox example\">\n<p>Evaluate the integral [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{4-9{x}^{2}}}.[\/latex]<\/p>\n\n[reveal-answer q=\"fs-id1170572449550\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572449550\"]Substitute [latex]u=3x.[\/latex] Then [latex]du=3dx[\/latex] and we have,\n\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 &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>\n<p id=\"fs-id1170572449549\"><span style=\"font-size: 0.9em;\">[\/hidden-answer]<\/span><\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Evaluate the definite integral [latex]{\\displaystyle\\int }_{0}^{\\sqrt{3}\\text{\/}2}\\dfrac{du}{\\sqrt{1-{u}^{2}}}.[\/latex]<\/p>\n<p>[reveal-answer q=\"fs-id1170572150550\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572150550\"]<\/p>\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 &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>\n<p style=\"text-align: left;\">[\/hidden-answer]<\/p>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Calculate integrals that lead to inverse trigonometric function solutions<\/li>\n<\/ul>\n<\/section>\n<p>In this section we focus on integrals that result in inverse trigonometric functions. We have worked with these functions before.<\/p>\n<section class=\"textbox recall\">\n<p>Recall that trigonometric functions are not one-to-one unless the domains are restricted.<\/p>\n<\/section>\n<p>When working with inverses of trigonometric functions, we always need to be careful to take these restrictions into account. Also, we developed formulas for derivatives of inverse trigonometric functions. The formulas developed there give rise directly to integration formulas involving inverse trigonometric functions.<\/p>\n<h2>Integrals Resulting in Inverse Trigonometric 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<section class=\"textbox keyTakeaway\">\n<h3>integration formulas resulting in inverse trigonometric Functions<\/h3>\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<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\n<p id=\"fs-id1170571652151\">Let [latex]y={ \\sin }^{-1}\\dfrac{x}{a}.[\/latex] Then [latex]a \\sin y=x.[\/latex]<\/p>\n<p>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]<\/p>\n<p>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]<\/p>\n<p>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 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<\/section>\n<section class=\"textbox example\">\n<p>Evaluate the definite integral [latex]{\\displaystyle\\int }_{0}^{\\frac{1}{2}}\\dfrac{dx}{\\sqrt{1-{x}^{2}}}.[\/latex]<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q7116223\">Show Solution<\/button><\/p>\n<div id=\"q7116223\" class=\"hidden-answer\" style=\"display: none\">\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.<\/p>\n<p>We have,<\/p>\n<div 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>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Evaluate the integral [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{4-9{x}^{2}}}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572449550\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox example\">\n<p>Evaluate the definite integral [latex]{\\displaystyle\\int }_{0}^{\\sqrt{3}\\text{\/}2}\\dfrac{du}{\\sqrt{1-{u}^{2}}}.[\/latex]<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572150550\">Show Solution<\/button><\/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<\/section>\n","protected":false},"author":6,"menu_order":13,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/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\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":399,"module-header":"","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/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":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/412"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/412\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/399"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/412\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=412"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=412"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=412"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=412"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}