{"id":202,"date":"2019-08-27T21:09:49","date_gmt":"2019-08-27T21:09:49","guid":{"rendered":"http:\/\/104.42.120.246.xip.io\/calculus-tutorials\/?page_id=202"},"modified":"2020-06-18T16:27:55","modified_gmt":"2020-06-18T16:27:55","slug":"trigonometric-substitution","status":"publish","type":"page","link":"https:\/\/math.hmc.edu\/calculus\/hmc-mathematics-calculus-online-tutorials\/single-variable-calculus\/trigonometric-substitution\/","title":{"rendered":"Trigonometric Substitution"},"content":{"rendered":"\n<script type=\"text\/x-mathjax-config\">\n  MathJax.Hub.Config({ tex2jax: { inlineMath: [['$','$'], [\"\\(\",\"\\)\"]] } });\n<\/script>\n\n\n\n<script type=\"text\/javascript\" src=\"http:\/\/cdn.mathjax.org\/mathjax\/latest\/MathJax.js?config=TeX-AMS_HTML\">\n<\/script>\n\n\n\n<meta http-equiv=\"X-UA-Compatible\" content=\"IE=EmulateIE7\">\n\n\n\n<title>Trigonometric Substitutions &#8211; HMC Calculus Tutorial<\/title>\n\n\n\n<p>\n<!------------------------>\n\nConsider the integral\n\\[\\int \\frac{dx}{\\sqrt{9-x^2}}.\\]\nAt first glance, we might try the substitution $u=9-x^2$, but this\nwill actually make the integral even more complicated!  \n\n<\/p>\n\n\n\n<p>\n<i>Let&#8217;s try a different approach:<\/i>\n\n<\/p>\n\n\n\n<p>\nThe radical $\\displaystyle\\sqrt{9-x^2}$ represents the length of the\nbase of a right triangle with height $x$ and hypotenuse of length $3$:\n\n<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"alignleft is-resized\"><img data-recalc-dims=\"1\" loading=\"lazy\" decoding=\"async\" src=\"https:\/\/i0.wp.com\/math.hmc.edu\/calculus\/wp-content\/uploads\/sites\/3\/2019\/08\/trig_figure1.gif?resize=154%2C111&#038;ssl=1\" alt=\"Follow the image link for a complete description of the image\" class=\"wp-image-416\" width=\"154\" height=\"111\"\/><\/figure><\/div>\n\n\n\n<p>\n\tFor this triangle, $\\displaystyle\\sin \\theta =\\frac{x}{3}$, suggesting\n\tthe substitution $x=3\\sin\\theta$.  Then $\\displaystyle\\theta\n\t=\\arcsin\\left(\\frac{x}{3}\\right)$, where we specify $-\\pi\/2\\leq\n\t\\theta\\leq \\pi\/2$.  Note that $dx=3\\cos \\theta\\, d\\theta$ and that\n\t$\\sqrt{9-x^2}=3\\cos\\theta$.  \n<\/p>\n\n\n\n<p>\nWith this change of variables,\n\\[\\int \\frac{dx}{\\sqrt{9-x^2}}= \\int \\frac{3\\cos \\theta \\, d\\theta}\n{3\\cos \\theta}=\\int d\\theta = \\theta+C=\\arcsin\\left(\\frac{x}{3}\\right)+C.\\]\n\n<\/p>\n\n\n\n<p>\n<b>Caution!<\/b>\n<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>The sketch of the triangle is very useful for determining what substitution should be made.  Note, though, that the sketch only has meaning for $x > 0$ and $\\theta > 0$. <\/li><li>It is important to be careful about how the angle $\\theta$ is defined.  With the restrictions on $\\theta$ mentioned in the examples here, we avoid sign difficulties even when $x &lt; 0$. <\/li><\/ul>\n\n\n\n<p>\nThere are two other trigonometric substitutions useful in integrals\nwith different forms:\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n\nLet&#8217;s evaluate \n\\[\\int \\frac{dx}{x^2\\sqrt{x^2-4}}.\\]\nThe radical $\\sqrt{x^2-4}$ suggests a triangle with hypotenuse of length\n$x$ and base of length $2$:\n\n<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"alignleft\"><img data-recalc-dims=\"1\" loading=\"lazy\" decoding=\"async\" width=\"205\" height=\"95\" src=\"https:\/\/i0.wp.com\/math.hmc.edu\/calculus\/wp-content\/uploads\/sites\/3\/2019\/08\/trig_figure2.gif?resize=205%2C95&#038;ssl=1\" alt=\"Follow the image link for a complete description of the image\" class=\"wp-image-417\"\/><\/figure><\/div>\n\n\n\n<p>\n\tFor this triangle, $\\displaystyle\\sec \\theta =\\frac{x}{2}$, we will\n\ttry the substitution $x=2\\sec \\theta$.  Then\n\t$\\displaystyle\\theta=\\sec^{-1} \\left(\\frac{x}{2}\\right)$, where we\n\tspecify $0\\leq\\theta&lt;\\pi\/2$ or $\\pi\\leq \\theta &lt;3\\pi\/2$.  Note that\n\t$dx=2\\sec\\theta \\tan\\theta \\, d\\theta$ and that\n\t$\\sqrt{x^2-4}=2\\tan\\theta$.\n<\/p>\n\n\n\n<p>\nThen \n\\[\\int \\frac{dx}{x^2\\sqrt{x^2-4}}=\\int\\frac{2\\sec\\theta\\tan\\theta}\n{(2\\sec\\theta)^2(2\\tan\\theta)}\\,d\\theta=\\int\\frac{1}{4}\\cos\\theta\\,\nd\\theta=\\frac{1}{4}\\sin\\theta+C.\\] \nBut we see from the sketch that $\\displaystyle\n\\sin\\theta=\\frac{\\sqrt{x^2-4}}{x}$, so \n\\[\\int \\frac{dx}{x^2\\sqrt{x^2-4}}=\\frac{\\sqrt{x^2-4}}{4x}+C.\\]\n\n<\/p>\n\n\n\n<p>\nWe may also use a trigonometric substitution to evaluate a\n<i>definite<\/i> integral, as long as care is taken in working with\nthe limits of integration:\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n\nWe will evaluate\n\\[\\int^1_{-1}\\frac{dx}{(1+x^2)^2}.\\]\nThe factor $(1+x^2)$ suggests a triangle with base of length $1$ and\nheight $x$:\n\n<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"alignleft\"><img data-recalc-dims=\"1\" loading=\"lazy\" decoding=\"async\" width=\"150\" height=\"95\" src=\"https:\/\/i0.wp.com\/math.hmc.edu\/calculus\/wp-content\/uploads\/sites\/3\/2019\/08\/trig_figure3.gif?resize=150%2C95&#038;ssl=1\" alt=\"Follow the image link for a complete description of the image\" class=\"wp-image-418\"\/><\/figure><\/div>\n\n\n\n<p>\n\t\tFor this triangle, $\\tan \\theta =x$, so we will try the substitution\n\t\t$x=\\tan \\theta$.  Then $\\theta =\\tan^{-1}(x)$, where we specify\n\t\t$-\\pi\/2&lt;\\theta &lt;\\pi\/2$.  Here, $dx=\\sec^2\\theta \\, d\\theta$.  Also,\n\t\t$\\sqrt{1+x^2}=\\sec\\theta$\n\t\tso $(1+x^2)^2 = \\sec^4 \\theta$.\n<\/p>\n\n\n\n<p>\nThen \n\\begin{eqnarray*}\n\t\\int^1_{-1}\\frac{dx}{(1+x^2)^2}&amp;=&amp;\\int^{\\pi\/4}_{-\\pi\/4}\n\t\\frac{\\sec^2\\theta}{\\sec^4\\theta}\\, d\\theta\\\\\n\t&amp;=&amp;\\int^{\\pi\/4}_{-\\pi\/4}\\cos^2\\theta \\, d\\theta\\\\\n\t&amp;=&amp;\\int^{\\pi\/4}_{-\\pi\/4}\\frac{1}{2}(1+\\cos 2\\theta)\\, d\\theta\\\\\n\t&amp;=&amp;\\left.\\frac{1}{2}\\left(\\theta +\\frac{1}{2}\\sin\n\t2\\theta\\right)\\right|^{\\pi\/4}_{-\\pi\/4}\\\\\n\t&amp;=&amp;=\\frac{1}{2}\\left(\\frac{\\pi}{4}+\n\t\\frac{1}{2}\\right)-\\frac{1}{2}\\left(-\\frac{\\pi}{4}-\\frac{1}{2}\\right)\\\\\n\t&amp;=&amp;\\frac{\\pi}{4}+\\frac{1}{2}.\n\\end{eqnarray*}\n\n<\/p>\n\n\n\n<p>\nThere is often more than one way to solve a particular integral.  A\ntrigonometric substitution will not always be necessary, even when the\ntypes of factors seen above appear.  With practice, you will gain\ninsight into what kind of substitution will work best for a particular\nintegral.\n\n<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<center>\n<p>\n<\/p><h4>Key Concepts<\/h4>\n<p>\n<\/p><\/center>\n\n\n\n<p>\n\nTrigonometric substitutions are often useful for integrals containing\nfactors of the form \n\\[(a^2-x^2)^n,\\qquad\\qquad (x^2+a^2)^n,\\qquad {\\small\\textrm{or}}\\qquad\n(x^2-a^2)^n.\\]\nThe exact substitution used depends on the form of the integral:\n\n<br><\/p>\n\n\n\n<table class=\"wp-block-table\"><tbody><tr><td> $(a^2-x^2)^n $ <\/td><td> $(x^2+a^2)^n $ <\/td><td> $(x^2-a^2)^n$ <\/td><\/tr><tr><td><img data-recalc-dims=\"1\" loading=\"lazy\" decoding=\"async\" width=\"135\" height=\"103\" class=\"wp-image-419\" style=\"width: 135px;\" src=\"https:\/\/i0.wp.com\/104.42.120.246.xip.io\/calculus-tutorials\/wp-content\/uploads\/sites\/3\/2019\/08\/trig_figure4.gif?resize=135%2C103\" alt=\"The (a^2\u2212x^2)^2 form can be applied to a leg of a triangle.\"><\/td><td><img data-recalc-dims=\"1\" loading=\"lazy\" decoding=\"async\" width=\"135\" height=\"103\" class=\"wp-image-420\" style=\"width: 135px;\" src=\"https:\/\/i0.wp.com\/104.42.120.246.xip.io\/calculus-tutorials\/wp-content\/uploads\/sites\/3\/2019\/08\/trig_figure5.gif?resize=135%2C103\" alt=\"The (x^2+a^2)^2 form can be applied to a hypotenuse of a triangle.\"><\/td><td><img data-recalc-dims=\"1\" loading=\"lazy\" decoding=\"async\" width=\"150\" height=\"80\" class=\"wp-image-415\" style=\"width: 150px;\" src=\"https:\/\/i0.wp.com\/104.42.120.246.xip.io\/calculus-tutorials\/wp-content\/uploads\/sites\/3\/2019\/08\/trig_figure6.gif?resize=150%2C80\" alt=\"The (x^2\u2212a^2)^2 form can be applied to a leg of a triangle.\"><\/td><\/tr><tr><td> $x=a\\sin\\theta $ <br> $dx=a\\cos\\theta $ <br> $\\sqrt{a^2-x^2}=a\\cos\\theta\\qquad $  <\/td><td> $x=a\\tan\\theta $  <br> $dx=a\\sec^2\\theta $ <br> $\\sqrt{x^2+a^2}=a\\sec\\theta\\qquad $ <\/td><td>$x=a\\sec\\theta$ <br>$dx=a\\sec\\theta\\tan\\theta$ <br>$\\sqrt{x^2-a^2}=a\\tan\\theta  $ <\/td><\/tr><\/tbody><\/table>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p>\n\n[<a href=\"https:\/\/physics.hmc.edu\/ct\/quiz\/QZ0810\/\">I&#8217;m ready to take the quiz.<\/a>]\n[<a href=\"#top\">I need to review more.<\/a>]<br>\n\n\n<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Trigonometric Substitutions &#8211; HMC Calculus Tutorial Consider the integral \\[\\int \\frac{dx}{\\sqrt{9-x^2}}.\\] At first glance, we might try the substitution $u=9-x^2$, but this will actually make the integral even more complicated! Let&#8217;s try a different approach: The radical $\\displaystyle\\sqrt{9-x^2}$ represents the length of the base of a right triangle with height $x$ and hypotenuse of length&hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"parent":57,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"tags":[],"class_list":["post-202","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/202","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/comments?post=202"}],"version-history":[{"count":7,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/202\/revisions"}],"predecessor-version":[{"id":1223,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/202\/revisions\/1223"}],"up":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/57"}],"wp:attachment":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/media?parent=202"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/tags?post=202"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}