{"id":176,"date":"2019-08-27T17:22:11","date_gmt":"2019-08-27T17:22:11","guid":{"rendered":"http:\/\/104.42.120.246.xip.io\/calculus-tutorials\/?page_id=176"},"modified":"2019-12-02T21:00:39","modified_gmt":"2019-12-02T21:00:39","slug":"integration-by-parts","status":"publish","type":"page","link":"https:\/\/math.hmc.edu\/calculus\/hmc-mathematics-calculus-online-tutorials\/single-variable-calculus\/integration-by-parts\/","title":{"rendered":"Integration by Parts"},"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>Integration by Parts &#8211; HMC Calculus Tutorial<\/title>\n\n\n\n<p><\/p>\n\n\n\n<p>  We will use the <b><a href=\"https:\/\/math.hmc.edu\/calculus-tutorials\/hmc-mathematics-calculus-online-tutorials\/single-variable-calculus\/product-rule\/\">Product Rule<\/a><\/b> for derivatives to  derive a powerful integration formula: <\/p>\n\n\n\n<ul class=\"wp-block-list\"><li> Start with $(f(x)g(x))&#8217;=f(x)g'(x)+f'(x)g(x)$. <\/li><li> Integrate both sides to get $\\displaystyle  f(x)g(x)=\\int\\! f(x)g'(x)\\, dx +\\int\\! f'(x)g(x)\\, dx$.  We need not include a constant of integration on the left, since the integrals on the right will also have integration constants.<\/li><li> Solve for $\\displaystyle\\int\\! f(x)g'(x)\\, dx$, obtaining \\[\\int f(x)g'(x)\\, dx=f(x)g(x)-\\int f'(x)g(x)\\, dx.\\] <\/li><\/ul>\n\n\n\n<p>\nThis formula frequently allows us to compute a difficult integral by\ncomputing a much simpler integral.  We often express the Integration\nby Parts formula as follows:\n\n<\/p>\n\n\n\n<p>\nLet\n\\[\n\\begin{array}{ll}\n\tu  =  f(x)\\qquad\\qquad &amp;\n\tdv  =  g'(x)\\, dx\\\\\n\tdu  =  f'(x)\\, dx &amp;\n\tv  =  g(x)\n\\end{array}\n\\]\nThen the formula becomes\n\\[\\int u\\, dv=uv-\\int v\\, du.\\]\nTo integrate by parts, strategically choose $u$, $dv$ and then apply\nthe formula.\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\nLet&#8217;s evaluate $\\displaystyle\\int\\! xe^x\\, dx$.\n\n<\/p>\n\n\n\n<p>\nLet\n\\[\n\\begin{array}{ll}\n\tu  =  x \\qquad\\qquad &amp;\n\tdv  =  e^x\\, dx\\\\\n\tdu  =  dx &amp;\n\tv  =  e^x\n\\end{array}\n\\]\nThen by integration by parts, \n\\begin{eqnarray*}\n\t\\int xe^x&amp;=&amp;xe^x-\\int e^x\\, dx\\\\\n\t         &amp;=&amp;xe^x-e^x+C.\n\\end{eqnarray*}\n\n<\/p>\n\n\n<style>.kt-accordion-id_75fe62-33 .kt-accordion-inner-wrap{column-gap:var(--global-kb-gap-md, 2rem);row-gap:0px;}.kt-accordion-id_75fe62-33 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.kt-blocks-accordion-icon-trigger:before{background:#444444;}@media all and (max-width: 767px){.kt-accordion-id_75fe62-33 .kt-accordion-inner-wrap{display:block;}.kt-accordion-id_75fe62-33 .kt-accordion-inner-wrap .kt-accordion-pane:not(:first-child){margin-top:0px;}}<\/style>\n<div class=\"wp-block-kadence-accordion alignnone\"><div class=\"kt-accordion-wrap kt-accordion-wrap kt-accordion-id_75fe62-33 kt-accordion-has-2-panes kt-active-pane-0 kt-accordion-block kt-pane-header-alignment-left kt-accodion-icon-style-basic kt-accodion-icon-side-right\" style=\"max-width:none\"><div class=\"kt-accordion-inner-wrap\" data-allow-multiple-open=\"true\" data-start-open=\"none\">\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-1 kt-pane_e5d4d3-a8\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><div class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">A Faulty Choice<\/span><\/div><div class=\"kt-blocks-accordion-icon-trigger\"><\/div><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<script type=\"text\/x-mathjax-config\">\n  MathJax.Hub.Config({ tex2jax: { inlineMath: [['$','$'], [\"\\(\",\"\\)\"]] } });\n<\/script>\n<script type=\"text\/javascript\" src=\"http:\/\/cdn.mathjax.org\/mathjax\/latest\/MathJax.js?config=TeX-AMS_HTML\">\n<\/script>\n\n\n\n<p>\n<!------------------------>\n\nIn integrating $\\displaystyle\\int\\! xe^x\\, dx$, \nwhat if we had chosen $u$, $dv$ as follows?\n\\[\n\\begin{array}{ll}\n\tu  =  e^x\\qquad\\qquad &amp;\n\tdv  =  x\\, dx\\\\\n\tdu  =  e^x\\, dx &amp;\n\tv  =  \\frac{1}{2}x^2\n\\end{array}\n\\]\nThen by integration by parts,\n\\[\\int xe^x=\\frac{1}{2}x^2e^x-\\int \\frac{1}{2}x^2e^x\\, dx.\\]\nBut $\\displaystyle\\int\\! \\frac{1}{2}x^2e^x\\, dx$ is even more \ncomplicated than the original integral!\n\n<\/p>\n\n\n\n<p>\n<b>It takes intuition and practice to make a good choice of $u$\nand $dv$!<\/b>\n\n<!------------------------>\n\n\n<\/p>\n<\/div><\/div><\/div>\n\n\n\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-2 kt-pane_488750-af\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><div class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">A Reduction Formula<\/span><\/div><div class=\"kt-blocks-accordion-icon-trigger\"><\/div><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<script type=\"text\/x-mathjax-config\">\n  MathJax.Hub.Config({ tex2jax: { inlineMath: [['$','$'], [\"\\(\",\"\\)\"]] } });\n<\/script>\n<script type=\"text\/javascript\" src=\"http:\/\/cdn.mathjax.org\/mathjax\/latest\/MathJax.js?config=TeX-AMS_HTML\">\n<\/script>\n\n\n\n<p>\n<!------------------------>\n\nWe may use the method of integration by parts to obtain a <b>reduction\nformula<\/b> for $x^ne^x\\, dx$:\n\\[\\int x^ne^x\\, dx=x^ne^x-n\\int x^{n-1}e^x\\, dx.\\]\nWhat would be good choices for $u$ and $dv$ to obtain this formula?\n\n<\/p>\n\n\n\n<p>\nReduction formulas may also be found for integrals of trigonometric\nfunctions such as $\\displaystyle\\int\\! \\cos^n(x)\\, dx$.\n\n<!------------------------>\n\n\n<\/p>\n<\/div><\/div><\/div>\n<\/div><\/div><\/div>\n\n\n\n<p><\/p>\n\n\n\n<p>\n\nIntegration by parts &#8220;works&#8221; on definite integrals as well:\n\\[\\int^b_a u\\, dv=\\left.uv\\right|^b_a-\\int^b_a v\\, du.\\]\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\nWe will evaluate $\\displaystyle\\int^1_0\\! \\arctan (x)\\, dx$.\n\n<\/p>\n\n\n\n<p>\nLet\n\\[\n\\begin{array}{ll}\n\tu  =  \\arctan(x) \\qquad\\qquad &amp;\n\tdv  =  dx\\\\\n\tdu  =  \\displaystyle\\frac{1}{1+x^2}\\, dx &amp;\n\tv  =  x\n\\end{array}\n\\]\nThen by integration by parts, \n\\begin{eqnarray*}\n\t\\int^1_0 \\arctan(x)&amp;=&amp;\\left.x\\arctan(x)\\right|^1_0-\\int^1_0\n\t                       \\frac{x}{1+x^2}\\, dx\\\\\n\t                    &amp;=&amp;\\left.x\\arctan(x)\\right|^1_0-\\left.\\frac{1}{2}\\ln\n\t                       (1+x^2)\\right|^1_0\\\\\n\t\t\t    &amp;=&amp;\\left(\\frac{\\pi}{4}-0\\right)-\\left(\\frac{1}{2}\n\t                       \\ln (2)-0\\right)\\\\\n\t                    &amp;=&amp;\\frac{\\pi}{4}-\\ln (\\sqrt{2}).\n\\end{eqnarray*}\n\nSometimes it is necessary to integrate twice by parts in order to\ncompute an integral:\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n\nLet&#8217;s compute $\\displaystyle\\int\\! e^x\\cos x\\, dx$.\n\n<\/p>\n\n\n\n<p>\nLet\n\\[\n\\begin{array}{ll}\n\tu  =  e^x\\qquad\\qquad &amp;\n\tdv  =  \\cos x\\, dx\\\\\n\tdu  =  e^x\\, dx &amp;\n\tv  =  \\sin x\n\\end{array}\n\\]\nThen $\\displaystyle\\int\\! e^x\\cos x\\, dx=e^x\\sin x-\\int\\! e^x\\sin x\\, dx$.\n\n<\/p>\n\n\n\n<p>\nIt is not clear yet that we&#8217;ve accomplished anything, but now let&#8217;s\nintegrate the integral on the right-hand side by parts:\n\n<\/p>\n\n\n\n<p>\nNow let\n\\[\n\\begin{array}{ll}\n\tu  =  e^x\\qquad\\qquad &amp;\n\tdv  =  \\sin x\\, dx\\\\\n\tdu  =  e^x\\, dx &amp;\n\tv  =  -\\cos x\n\\end{array}\n\\]\nSo $\\displaystyle\\int\\! e^x\\sin x\\, dx=-e^x\\cos x+\\int e^x\\cos x\\, dx$.\n\n<\/p>\n\n\n\n<p>\nSubstituting this into $\\displaystyle\\int\\! e^x\\cos x\\, \ndx=e^x\\sin x-\\int\\! e^x\\sin x\\, dx$,\n\\begin{eqnarray*}\n\t\\int e^x\\cos x\\, dx&amp;=&amp;e^x\\sin x-\\left[-e^x\\cos x+\\int e^x\\cos x\\, dx\\right]\\\\\n\t                   &amp;=&amp;e^x\\sin x+ e^x\\cos x -\\int e^x\\cos x\\, dx.\n\\end{eqnarray*}\nThe integal $\\displaystyle\\int\\! e^x\\cos x\\, dx$ appears on both sides on the\nequation, so we can solve for it:\n\\[2\\int e^x\\cos x\\, dx=e^x\\sin x+e^x\\cos x.\\]\nFinally,\n\\[\\int e^x\\cos x\\, dx =\\frac{1}{2}e^x\\sin x+\\frac{1}{2}e^x\\cos x +C.\\]\n\n<\/p>\n\n\n<style>.kt-accordion-id_250ec8-f3 .kt-accordion-inner-wrap{column-gap:var(--global-kb-gap-md, 2rem);row-gap:0px;}.kt-accordion-id_250ec8-f3 .kt-accordion-panel-inner{border-top-width:0px;border-right-width:0px;border-bottom-width:0px;border-left-width:0px;padding-top:var(--global-kb-spacing-sm, 1.5rem);padding-right:var(--global-kb-spacing-sm, 1.5rem);padding-bottom:var(--global-kb-spacing-sm, 1.5rem);padding-left:var(--global-kb-spacing-sm, 1.5rem);}.kt-accordion-id_250ec8-f3 > .kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > 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.kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > .kt-blocks-accordion-header.kt-accordion-panel-active{color:#ffffff;background:#444444;border-top-color:#444444;border-right-color:#444444;border-bottom-color:#444444;border-left-color:#444444;}.kt-accordion-id_250ec8-f3:not( .kt-accodion-icon-style-basiccircle ):not( .kt-accodion-icon-style-xclosecircle ):not( .kt-accodion-icon-style-arrowcircle )  > .kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:after, .kt-accordion-id_250ec8-f3:not( .kt-accodion-icon-style-basiccircle ):not( .kt-accodion-icon-style-xclosecircle ):not( .kt-accodion-icon-style-arrowcircle )  > .kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:before{background:#ffffff;}.kt-accordion-id_250ec8-f3:not( .kt-accodion-icon-style-basic ):not( .kt-accodion-icon-style-xclose ):not( .kt-accodion-icon-style-arrow ) .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger{background:#ffffff;}.kt-accordion-id_250ec8-f3:not( .kt-accodion-icon-style-basic ):not( .kt-accodion-icon-style-xclose ):not( .kt-accodion-icon-style-arrow ) .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:after, .kt-accordion-id_250ec8-f3:not( .kt-accodion-icon-style-basic ):not( .kt-accodion-icon-style-xclose ):not( .kt-accodion-icon-style-arrow ) .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:before{background:#444444;}@media all and (max-width: 767px){.kt-accordion-id_250ec8-f3 .kt-accordion-inner-wrap{display:block;}.kt-accordion-id_250ec8-f3 .kt-accordion-inner-wrap .kt-accordion-pane:not(:first-child){margin-top:0px;}}<\/style>\n<div class=\"wp-block-kadence-accordion alignnone\"><div class=\"kt-accordion-wrap kt-accordion-wrap kt-accordion-id_250ec8-f3 kt-accordion-has-2-panes kt-active-pane-0 kt-accordion-block kt-pane-header-alignment-left kt-accodion-icon-style-basic kt-accodion-icon-side-right\" style=\"max-width:none\"><div class=\"kt-accordion-inner-wrap\" data-allow-multiple-open=\"true\" data-start-open=\"none\">\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-1 kt-pane_991246-5b\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><div class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">Check by Differentiating<\/span><\/div><div class=\"kt-blocks-accordion-icon-trigger\"><\/div><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<script type=\"text\/x-mathjax-config\">\n  MathJax.Hub.Config({ tex2jax: { inlineMath: [['$','$'], [\"\\(\",\"\\)\"]] } });\n<\/script>\n<script type=\"text\/javascript\" src=\"http:\/\/cdn.mathjax.org\/mathjax\/latest\/MathJax.js?config=TeX-AMS_HTML\">\n<\/script>\n\n\n\n<p>\n\nWe may verify $\\displaystyle\\int\\! e^x\\cos x\\, dx=\\frac{1}{2}e^x\\sin x+\\frac{1}{2}e^x\\cos x+C$ by differentiating the right-hand side: \\begin{eqnarray*} \\frac{d}{dx}\\left(\\frac{1}{2}e^x\\sin x+\\frac{1}{2}e^x\\cos x+C\\right)&amp;=&amp;\\frac{1}{2}e^x\\cos x+\\frac{1}{2}e^x\\sin x-\\frac{1}{2}e^x\\sin +\\frac{1}{2}e^x\\cos x\\\\ &amp;=&amp;e^x\\cos x. \\end{eqnarray*}  \n\n<\/p>\n<\/div><\/div><\/div>\n<\/div><\/div><\/div>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<center>\n<h4>Key Concept<\/h4>\n<\/center>\n\n\n\n<p>\n\n\\[\\int u\\, dv=uv-\\int v\\, du.\\]\n<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li> Choose $u$, $dv$ in such a way that:\n\t\t<ol>\n\t\t\t<li> $u$ is easy to <i>differentiate<\/i>.\n\t\t\t<\/li><li> $dv$ is easy to <i>integrate<\/i>.\n\t\t\t<\/li><li> $\\displaystyle\\int\\! v\\, du$ is easier to compute that\n\t\t\t\t$\\displaystyle\\int\\! u\\, dv$.\n\t\t<\/li><\/ol>\n\t<\/li><li> Sometimes it is necessary to integrate by parts more than once.\n<\/li><\/ul>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p>\n\n[<a href=\"https:\/\/physics.hmc.edu\/ct\/quiz\/QZ1210\/\">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>Integration by Parts &#8211; HMC Calculus Tutorial We will use the Product Rule for derivatives to derive a powerful integration formula: Start with $(f(x)g(x))&#8217;=f(x)g'(x)+f'(x)g(x)$. Integrate both sides to get $\\displaystyle f(x)g(x)=\\int\\! f(x)g'(x)\\, dx +\\int\\! f'(x)g(x)\\, dx$. We need not include a constant of integration on the left, since the integrals on the right will also&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-176","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/176","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=176"}],"version-history":[{"count":5,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/176\/revisions"}],"predecessor-version":[{"id":1086,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/176\/revisions\/1086"}],"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=176"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/tags?post=176"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}