{"id":404,"date":"2025-02-13T19:44:38","date_gmt":"2025-02-13T19:44:38","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integration-using-substitution-learn-it-1\/"},"modified":"2025-02-13T19:44:38","modified_gmt":"2025-02-13T19:44:38","slug":"integration-using-substitution-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integration-using-substitution-learn-it-1\/","title":{"raw":"Integration using Substitution: Learn It 1","rendered":"Integration using Substitution: Learn It 1"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li>Identify when to use substitution to simplify and solve integrals<\/li>\n\t<li>Apply substitution methods to find indefinite integrals<\/li>\n\t<li>Apply substitution methods to find definite integrals<\/li>\n<\/ul>\n<\/section>\n<p>The Fundamental Theorem of Calculus gave us a method to evaluate integrals without using Riemann sums. The drawback of this method, though, is that we must be able to find an antiderivative, and this is not always easy. In this section we examine a technique, called<strong>&nbsp;integration by substitution<\/strong>, to help us find antiderivatives. Specifically, this method helps us find antiderivatives when the integrand is the result of a chain-rule derivative.<\/p>\n<h2>Substitution for Indefinite Integrals<\/h2>\n<p>At first, the approach to the substitution procedure may not appear very obvious. However, it is primarily a visual task\u2014that is, the integrand shows you what to do; it is a matter of recognizing the form of the function. <br>\n<br>\nSo, what are we supposed to see? We are looking for an integrand of the form [latex]f\\left[g(x)\\right]{g}^{\\prime }(x)dx.[\/latex]<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1170573502792\">For example, in the integral [latex]\\displaystyle\\int {({x}^{2}-3)}^{3}2xdx,[\/latex] we have [latex]f(x)={x}^{3},g(x)={x}^{2}-3,[\/latex] and [latex]g\\text{\u2018}(x)=2x.[\/latex] Then,<\/p>\n<div id=\"fs-id1170573442602\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left[g(x)\\right]{g}^{\\prime }(x)={({x}^{2}-3)}^{3}(2x),[\/latex]<\/div>\n<p id=\"fs-id1170573209422\">and we see that our integrand is in the correct form.<\/p>\n<\/section>\n<p id=\"fs-id1170573431694\">The method is called <strong>substitution <\/strong>because we substitute part of the integrand with the variable [latex]u[\/latex] and part of the integrand with <em>du<\/em>. It is also referred to as <em>change of variables<\/em> because we are changing variables to obtain an expression that is easier to work with for applying the integration rules.<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>substitution with indefinite integrals<\/h3>\n<p id=\"fs-id1170571193831\">Let [latex]u=g(x),,[\/latex] where [latex]{g}^{\\prime }(x)[\/latex] is continuous over an interval, let [latex]f(x)[\/latex] be continuous over the corresponding range of [latex]g[\/latex], and let [latex]F(x)[\/latex] be an antiderivative of [latex]f(x).[\/latex] Then,<\/p>\n<div id=\"fs-id1170573306241\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{cc} {\\displaystyle\\int f\\left[g(x)\\right]{g}^{\\prime }(x)dx}\\hfill &amp; = {\\displaystyle\\int f(u)du}\\hfill \\\\ &amp; =F(u)+C\\hfill \\\\ &amp; =F(g(x))+C.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr>\n<p id=\"fs-id1170573351199\">Let [latex]f[\/latex], [latex]g[\/latex], [latex]u[\/latex], and <em>F<\/em> be as specified in the theorem. Then<\/p>\n<div id=\"fs-id1170573534008\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\frac{d}{dx}F(g(x))\\hfill &amp; ={F}^{\\prime }(g(x)){g}^{\\prime }(x)\\hfill \\\\ &amp; =f\\left[g(x)\\right]{g}^{\\prime }(x).\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573366538\">Integrating both sides with respect to [latex]x[\/latex], we see that<\/p>\n<div id=\"fs-id1170573355959\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int f\\left[g(x)\\right]{g}^{\\prime }(x)dx=F(g(x))+C.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573362143\">If we now substitute [latex]u=g(x),[\/latex] and [latex]du=g\\text{\u2018}(x)dx,[\/latex] we get<\/p>\n<div id=\"fs-id1170573437846\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc} {\\displaystyle\\int f\\left[g(x)\\right]{g}^{\\prime }(x)dx}\\hfill &amp; = {\\displaystyle\\int f(u)du}\\hfill \\\\ &amp; =F(u)+C\\hfill \\\\ &amp; =F(g(x))+C.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170573331564\">[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n<p id=\"fs-id1170573368234\">Returning to the problem we looked at originally, we let [latex]u={x}^{2}-3[\/latex] and then [latex]du=2xdx.[\/latex] Rewrite the integral in terms of [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170573366731\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int \\underset{u}{\\underbrace{({x}^{2}-3)}}}^{3}\\underset{du}{\\underbrace{(2xdx)}}=\\displaystyle\\int {u}^{3}du.[\/latex]<\/div>\n<p>Using the power rule for integrals, we have<\/p>\n<div id=\"fs-id1170571257696\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {u}^{3}du=\\frac{{u}^{4}}{4}+C[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571099748\">Substitute the original expression for [latex]x[\/latex] back into the solution:<\/p>\n<div id=\"fs-id1170573410936\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{{u}^{4}}{4}+C=\\dfrac{{({x}^{2}-3)}^{4}}{4}+C[\/latex]<\/div>\n<section class=\"textbox questionHelp\">\n<p><strong>How To: Integrate by Substitution<\/strong><\/p>\n<ol id=\"fs-id1170570997711\">\n\t<li>Look carefully at the integrand and select an expression [latex]g(x)[\/latex] within the integrand to set equal to [latex]u[\/latex]. Let\u2019s select [latex]g(x).[\/latex] such that [latex]{g}^{\\prime }(x)[\/latex] is also part of the integrand.<\/li>\n\t<li>Substitute [latex]u=g(x)[\/latex] and [latex]du={g}^{\\prime }(x)dx[\/latex] into the integral.<\/li>\n\t<li>We should now be able to evaluate the integral with respect to [latex]u[\/latex]. If the integral can\u2019t be evaluated we need to go back and select a different expression to use as [latex]u[\/latex].<\/li>\n\t<li>Evaluate the integral in terms of [latex]u[\/latex].<\/li>\n\t<li>Write the result in terms of [latex]x[\/latex] and the expression [latex]g(x).[\/latex]<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p>Use substitution to find the antiderivative of [latex]\\displaystyle\\int 6x{(3{x}^{2}+4)}^{4}dx.[\/latex]<br>\n[reveal-answer q=\"fs-id1170572587719\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572587719\"]<\/p>\n<p id=\"fs-id1170573569711\">The first step is to choose an expression for [latex]u[\/latex]. We choose [latex]u=3{x}^{2}+4.[\/latex] because then [latex]du=6xdx.,[\/latex] and we already have <em>du<\/em> in the integrand. Write the integral in terms of [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170570995689\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int 6x{(3{x}^{2}+4)}^{4}dx=\\displaystyle\\int {u}^{4}du.[\/latex]<\/div>\n<p id=\"fs-id1170573398341\">Remember that <em>du<\/em> is the derivative of the expression chosen for [latex]u[\/latex], regardless of what is inside the integrand. Now we can evaluate the integral with respect to [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170571000121\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} {\\displaystyle\\int {u}^{4}du}\\hfill &amp; =\\frac{{u}^{5}}{5}+C\\hfill \\\\ \\\\ \\\\ &amp; =\\frac{{(3{x}^{2}+4)}^{5}}{5}+C.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170573208347\"><strong>Analysis<\/strong><\/p>\n<p id=\"fs-id1170573569153\">We can check our answer by taking the derivative of the result of integration. We should obtain the integrand. Picking a value for <em>C<\/em> of 1, we let [latex]y=\\frac{1}{5}{(3{x}^{2}+4)}^{5}+1.[\/latex] We have<\/p>\n<div id=\"fs-id1170573336352\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\frac{1}{5}{(3{x}^{2}+4)}^{5}+1,[\/latex]<\/div>\n<p id=\"fs-id1170573411738\">so<\/p>\n<div id=\"fs-id1170573408431\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\hfill {y}^{\\prime }&amp; =(\\frac{1}{5})5{(3{x}^{2}+4)}^{4}6x\\hfill \\\\ &amp; =6x{(3{x}^{2}+4)}^{4}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170573371003\">This is exactly the expression we started with inside the integrand.<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Ak_y3lsBNfE?controls=0&amp;start=52&amp;end=163&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\n<p>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\/5.5Substitution52to163_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.5 Substitution\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\n<\/section>\n<p id=\"fs-id1170571121618\">Sometimes we need to adjust the constants in our integral if they don\u2019t match up exactly with the expressions we are substituting.<\/p>\n<section class=\"textbox proTip\">\n<p>As long as you select a [latex]g(x)[\/latex] for [latex]u[\/latex] such that&nbsp;<em>a multiple<\/em> of&nbsp;[latex]g'(x)[\/latex] exists in the integrand, it will work! In other words, make sure the exponents work - don't worry about the constants.<\/p>\n<p>For instance, in the example below, if we select [latex]{u={z}^{2}-5}[\/latex], [latex]g'(x)={2z}[\/latex]. Although [latex]g'(x)={2z}[\/latex] doesn't appear in the integrand, [latex]z[\/latex] does. Substitution can work here!&nbsp;<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Use substitution to find the antiderivative of [latex]\\displaystyle\\int z\\sqrt{{z}^{2}-5}dz.[\/latex]<\/p>\n<p>[reveal-answer q=\"fs-id1170572587720\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572587720\"]<\/p>\n<p id=\"fs-id1170573391210\">Rewrite the integral as [latex]\\displaystyle\\int z{({z}^{2}-5)}^{1\\text{\/}2}dz.[\/latex] Let [latex]u={z}^{2}-5[\/latex] and [latex]du=2zdz.[\/latex]<\/p>\n<p>Now we have a problem because [latex]du=2zdz[\/latex] and the original expression has only [latex]zdz.[\/latex] We have to alter our expression for <em>du<\/em> or the integral in [latex]u[\/latex] will be twice as large as it should be. If we multiply both sides of the <em>du<\/em> equation by [latex]\\frac{1}{2}.[\/latex] we can solve this problem. Thus,<\/p>\n<div id=\"fs-id1170573430996\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\hfill u&amp; ={z}^{2}-5\\hfill \\\\ \\hfill du&amp; =2zdz\\hfill \\\\ \\hfill \\frac{1}{2}du&amp; =\\frac{1}{2}(2z)dz=zdz.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170570976225\">Write the integral in terms of [latex]u[\/latex], but pull the [latex]\\frac{1}{2}[\/latex] outside the integration symbol:<\/p>\n<div id=\"fs-id1170573533836\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int z{({z}^{2}-5)}^{1\\text{\/}2}dz=\\frac{1}{2}\\displaystyle\\int {u}^{1\\text{\/}2}du.[\/latex]<\/div>\n<p id=\"fs-id1170573586341\">Integrate the expression in [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170570994415\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\frac{1}{2} {\\displaystyle\\int {u}^{1\\text{\/}2}du}\\hfill &amp; =(\\frac{1}{2})\\frac{{u}^{3\\text{\/}2}}{\\frac{3}{2}}+C\\hfill \\\\ \\\\ &amp; =(\\frac{1}{2})(\\frac{2}{3}){u}^{3\\text{\/}2}+C\\hfill \\\\ &amp; =\\frac{1}{3}{u}^{3\\text{\/}2}+C\\hfill \\\\ &amp; =\\frac{1}{3}{({z}^{2}-5)}^{3\\text{\/}2}+C.\\hfill \\end{array}[\/latex]&nbsp; &nbsp;<\/div>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Ak_y3lsBNfE?controls=0&amp;start=165&amp;end=266&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\n<p>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\/5.5Substitution165to266_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.5 Substitution\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Use substitution to evaluate the integral [latex]\\displaystyle\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt.[\/latex]<\/p>\n<p>[reveal-answer q=\"fs-id1170572587721\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572587721\"]<\/p>\n<div id=\"qfs-id1170573437523\" class=\"hidden-answer\">\n<p id=\"fs-id1170573437523\">We know the derivative of [latex] \\cos t[\/latex] is [latex]\\text{\u2212} \\sin t,[\/latex] so we set [latex]u= \\cos t.[\/latex] Then [latex]du=\\text{\u2212} \\sin tdt.[\/latex] Substituting into the integral, we have<\/p>\n<div id=\"fs-id1170573337947\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt=\\text{\u2212}\\displaystyle\\int \\frac{du}{{u}^{3}}.[\/latex]<\/div>\n<p id=\"fs-id1170573364147\">Evaluating the integral, we get<\/p>\n<div id=\"fs-id1170571136365\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ \\text{\u2212}{\\displaystyle\\int \\frac{du}{{u}^{3}}}\\hfill &amp; =\\text{\u2212} {\\displaystyle\\int {u}^{-3}du}\\hfill \\\\ &amp; =\\text{\u2212}(-\\frac{1}{2}){u}^{-2}+C.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170573409376\">Putting the answer back in terms of [latex]t[\/latex], we get<\/p>\n<div id=\"fs-id1170573587515\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc} {\\displaystyle\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt}\\hfill &amp; =\\frac{1}{2{u}^{2}}+C\\hfill \\\\ \\\\ &amp; =\\dfrac{1}{2{ \\cos }^{2}t}+C.\\hfill \\end{array}[\/latex] [\/hidden-answer]<\/div>\n<\/div>\n<\/section>\n<p id=\"fs-id1170573413569\">Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a constant. We need to eliminate all the expressions within the integrand that are in terms of the original variable. When we are done, [latex]u[\/latex] should be the only variable in the integrand.<\/p>\n<p>In some cases, this means solving for the original variable in terms of [latex]u[\/latex]. This technique should become clear in the next example.<\/p>\n<section class=\"textbox example\">\n<p>Use substitution to find the antiderivative of [latex]\\displaystyle\\int \\frac{x}{\\sqrt{x-1}}dx.[\/latex]<\/p>\n<p>[reveal-answer q=\"fs-id1170572587722\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572587722\"]<\/p>\n<p id=\"fs-id1170573435928\">If we let [latex]u=x-1,[\/latex] then [latex]du=dx.[\/latex] But this does not account for the [latex]x[\/latex] in the numerator of the integrand. We need to express [latex]x[\/latex] in terms of [latex]u[\/latex]. If [latex]u=x-1,[\/latex] then [latex]x=u+1.[\/latex] Now we can rewrite the integral in terms of [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170573333822\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} {\\displaystyle\\int \\frac{x}{\\sqrt{x-1}}dx}\\hfill &amp; = {\\displaystyle\\int \\frac{u+1}{\\sqrt{u}}du}\\hfill \\\\ \\\\ &amp; = {\\displaystyle\\int \\sqrt{u}+\\frac{1}{\\sqrt{u}}du}\\hfill \\\\ &amp; = {\\displaystyle\\int ({u}^{1\\text{\/}2}+{u}^{-1\\text{\/}2})du}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170573544033\">Then we integrate in the usual way, replace [latex]u[\/latex] with the original expression, and factor and simplify the result. Thus,<\/p>\n<div id=\"fs-id1170570997027\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc} {\\displaystyle\\int ({u}^{1\\text{\/}2}+{u}^{-1\\text{\/}2})du}\\hfill &amp; =\\frac{2}{3}{u}^{3\\text{\/}2}+2{u}^{1\\text{\/}2}+C\\hfill \\\\ \\\\ &amp; =\\frac{2}{3}{(x-1)}^{3\\text{\/}2}+2{(x-1)}^{1\\text{\/}2}+C\\hfill \\\\ &amp; ={(x-1)}^{1\\text{\/}2}\\left[\\frac{2}{3}(x-1)+2\\right]+C\\hfill \\\\ &amp; ={(x-1)}^{1\\text{\/}2}(\\frac{2}{3}x-\\frac{2}{3}+\\frac{6}{3})\\hfill \\\\ &amp; ={(x-1)}^{1\\text{\/}2}(\\frac{2}{3}x+\\frac{4}{3})\\hfill \\\\ &amp; =\\frac{2}{3}{(x-1)}^{1\\text{\/}2}(x+2)+C.\\hfill \\end{array}[\/latex] [\/hidden-answer]<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<p>[ohm_question hide_question_numbers=1]288435[\/ohm_question]<\/p>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Identify when to use substitution to simplify and solve integrals<\/li>\n<li>Apply substitution methods to find indefinite integrals<\/li>\n<li>Apply substitution methods to find definite integrals<\/li>\n<\/ul>\n<\/section>\n<p>The Fundamental Theorem of Calculus gave us a method to evaluate integrals without using Riemann sums. The drawback of this method, though, is that we must be able to find an antiderivative, and this is not always easy. In this section we examine a technique, called<strong>&nbsp;integration by substitution<\/strong>, to help us find antiderivatives. Specifically, this method helps us find antiderivatives when the integrand is the result of a chain-rule derivative.<\/p>\n<h2>Substitution for Indefinite Integrals<\/h2>\n<p>At first, the approach to the substitution procedure may not appear very obvious. However, it is primarily a visual task\u2014that is, the integrand shows you what to do; it is a matter of recognizing the form of the function. <\/p>\n<p>So, what are we supposed to see? We are looking for an integrand of the form [latex]f\\left[g(x)\\right]{g}^{\\prime }(x)dx.[\/latex]<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1170573502792\">For example, in the integral [latex]\\displaystyle\\int {({x}^{2}-3)}^{3}2xdx,[\/latex] we have [latex]f(x)={x}^{3},g(x)={x}^{2}-3,[\/latex] and [latex]g\\text{\u2018}(x)=2x.[\/latex] Then,<\/p>\n<div id=\"fs-id1170573442602\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left[g(x)\\right]{g}^{\\prime }(x)={({x}^{2}-3)}^{3}(2x),[\/latex]<\/div>\n<p id=\"fs-id1170573209422\">and we see that our integrand is in the correct form.<\/p>\n<\/section>\n<p id=\"fs-id1170573431694\">The method is called <strong>substitution <\/strong>because we substitute part of the integrand with the variable [latex]u[\/latex] and part of the integrand with <em>du<\/em>. It is also referred to as <em>change of variables<\/em> because we are changing variables to obtain an expression that is easier to work with for applying the integration rules.<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>substitution with indefinite integrals<\/h3>\n<p id=\"fs-id1170571193831\">Let [latex]u=g(x),,[\/latex] where [latex]{g}^{\\prime }(x)[\/latex] is continuous over an interval, let [latex]f(x)[\/latex] be continuous over the corresponding range of [latex]g[\/latex], and let [latex]F(x)[\/latex] be an antiderivative of [latex]f(x).[\/latex] Then,<\/p>\n<div id=\"fs-id1170573306241\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{cc} {\\displaystyle\\int f\\left[g(x)\\right]{g}^{\\prime }(x)dx}\\hfill & = {\\displaystyle\\int f(u)du}\\hfill \\\\ & =F(u)+C\\hfill \\\\ & =F(g(x))+C.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\n<p id=\"fs-id1170573351199\">Let [latex]f[\/latex], [latex]g[\/latex], [latex]u[\/latex], and <em>F<\/em> be as specified in the theorem. Then<\/p>\n<div id=\"fs-id1170573534008\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\frac{d}{dx}F(g(x))\\hfill & ={F}^{\\prime }(g(x)){g}^{\\prime }(x)\\hfill \\\\ & =f\\left[g(x)\\right]{g}^{\\prime }(x).\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573366538\">Integrating both sides with respect to [latex]x[\/latex], we see that<\/p>\n<div id=\"fs-id1170573355959\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int f\\left[g(x)\\right]{g}^{\\prime }(x)dx=F(g(x))+C.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573362143\">If we now substitute [latex]u=g(x),[\/latex] and [latex]du=g\\text{\u2018}(x)dx,[\/latex] we get<\/p>\n<div id=\"fs-id1170573437846\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc} {\\displaystyle\\int f\\left[g(x)\\right]{g}^{\\prime }(x)dx}\\hfill & = {\\displaystyle\\int f(u)du}\\hfill \\\\ & =F(u)+C\\hfill \\\\ & =F(g(x))+C.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170573331564\">[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n<p id=\"fs-id1170573368234\">Returning to the problem we looked at originally, we let [latex]u={x}^{2}-3[\/latex] and then [latex]du=2xdx.[\/latex] Rewrite the integral in terms of [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170573366731\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int \\underset{u}{\\underbrace{({x}^{2}-3)}}}^{3}\\underset{du}{\\underbrace{(2xdx)}}=\\displaystyle\\int {u}^{3}du.[\/latex]<\/div>\n<p>Using the power rule for integrals, we have<\/p>\n<div id=\"fs-id1170571257696\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {u}^{3}du=\\frac{{u}^{4}}{4}+C[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571099748\">Substitute the original expression for [latex]x[\/latex] back into the solution:<\/p>\n<div id=\"fs-id1170573410936\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{{u}^{4}}{4}+C=\\dfrac{{({x}^{2}-3)}^{4}}{4}+C[\/latex]<\/div>\n<section class=\"textbox questionHelp\">\n<p><strong>How To: Integrate by Substitution<\/strong><\/p>\n<ol id=\"fs-id1170570997711\">\n<li>Look carefully at the integrand and select an expression [latex]g(x)[\/latex] within the integrand to set equal to [latex]u[\/latex]. Let\u2019s select [latex]g(x).[\/latex] such that [latex]{g}^{\\prime }(x)[\/latex] is also part of the integrand.<\/li>\n<li>Substitute [latex]u=g(x)[\/latex] and [latex]du={g}^{\\prime }(x)dx[\/latex] into the integral.<\/li>\n<li>We should now be able to evaluate the integral with respect to [latex]u[\/latex]. If the integral can\u2019t be evaluated we need to go back and select a different expression to use as [latex]u[\/latex].<\/li>\n<li>Evaluate the integral in terms of [latex]u[\/latex].<\/li>\n<li>Write the result in terms of [latex]x[\/latex] and the expression [latex]g(x).[\/latex]<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p>Use substitution to find the antiderivative of [latex]\\displaystyle\\int 6x{(3{x}^{2}+4)}^{4}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572587719\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572587719\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573569711\">The first step is to choose an expression for [latex]u[\/latex]. We choose [latex]u=3{x}^{2}+4.[\/latex] because then [latex]du=6xdx.,[\/latex] and we already have <em>du<\/em> in the integrand. Write the integral in terms of [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170570995689\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int 6x{(3{x}^{2}+4)}^{4}dx=\\displaystyle\\int {u}^{4}du.[\/latex]<\/div>\n<p id=\"fs-id1170573398341\">Remember that <em>du<\/em> is the derivative of the expression chosen for [latex]u[\/latex], regardless of what is inside the integrand. Now we can evaluate the integral with respect to [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170571000121\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} {\\displaystyle\\int {u}^{4}du}\\hfill & =\\frac{{u}^{5}}{5}+C\\hfill \\\\ \\\\ \\\\ & =\\frac{{(3{x}^{2}+4)}^{5}}{5}+C.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170573208347\"><strong>Analysis<\/strong><\/p>\n<p id=\"fs-id1170573569153\">We can check our answer by taking the derivative of the result of integration. We should obtain the integrand. Picking a value for <em>C<\/em> of 1, we let [latex]y=\\frac{1}{5}{(3{x}^{2}+4)}^{5}+1.[\/latex] We have<\/p>\n<div id=\"fs-id1170573336352\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\frac{1}{5}{(3{x}^{2}+4)}^{5}+1,[\/latex]<\/div>\n<p id=\"fs-id1170573411738\">so<\/p>\n<div id=\"fs-id1170573408431\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\hfill {y}^{\\prime }& =(\\frac{1}{5})5{(3{x}^{2}+4)}^{4}6x\\hfill \\\\ & =6x{(3{x}^{2}+4)}^{4}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170573371003\">This is exactly the expression we started with inside the integrand.<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Ak_y3lsBNfE?controls=0&amp;start=52&amp;end=163&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>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\/5.5Substitution52to163_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.5 Substitution&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<p id=\"fs-id1170571121618\">Sometimes we need to adjust the constants in our integral if they don\u2019t match up exactly with the expressions we are substituting.<\/p>\n<section class=\"textbox proTip\">\n<p>As long as you select a [latex]g(x)[\/latex] for [latex]u[\/latex] such that&nbsp;<em>a multiple<\/em> of&nbsp;[latex]g'(x)[\/latex] exists in the integrand, it will work! In other words, make sure the exponents work &#8211; don&#8217;t worry about the constants.<\/p>\n<p>For instance, in the example below, if we select [latex]{u={z}^{2}-5}[\/latex], [latex]g'(x)={2z}[\/latex]. Although [latex]g'(x)={2z}[\/latex] doesn&#8217;t appear in the integrand, [latex]z[\/latex] does. Substitution can work here!&nbsp;<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Use substitution to find the antiderivative of [latex]\\displaystyle\\int z\\sqrt{{z}^{2}-5}dz.[\/latex]<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572587720\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572587720\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573391210\">Rewrite the integral as [latex]\\displaystyle\\int z{({z}^{2}-5)}^{1\\text{\/}2}dz.[\/latex] Let [latex]u={z}^{2}-5[\/latex] and [latex]du=2zdz.[\/latex]<\/p>\n<p>Now we have a problem because [latex]du=2zdz[\/latex] and the original expression has only [latex]zdz.[\/latex] We have to alter our expression for <em>du<\/em> or the integral in [latex]u[\/latex] will be twice as large as it should be. If we multiply both sides of the <em>du<\/em> equation by [latex]\\frac{1}{2}.[\/latex] we can solve this problem. Thus,<\/p>\n<div id=\"fs-id1170573430996\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\hfill u& ={z}^{2}-5\\hfill \\\\ \\hfill du& =2zdz\\hfill \\\\ \\hfill \\frac{1}{2}du& =\\frac{1}{2}(2z)dz=zdz.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170570976225\">Write the integral in terms of [latex]u[\/latex], but pull the [latex]\\frac{1}{2}[\/latex] outside the integration symbol:<\/p>\n<div id=\"fs-id1170573533836\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int z{({z}^{2}-5)}^{1\\text{\/}2}dz=\\frac{1}{2}\\displaystyle\\int {u}^{1\\text{\/}2}du.[\/latex]<\/div>\n<p id=\"fs-id1170573586341\">Integrate the expression in [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170570994415\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\frac{1}{2} {\\displaystyle\\int {u}^{1\\text{\/}2}du}\\hfill & =(\\frac{1}{2})\\frac{{u}^{3\\text{\/}2}}{\\frac{3}{2}}+C\\hfill \\\\ \\\\ & =(\\frac{1}{2})(\\frac{2}{3}){u}^{3\\text{\/}2}+C\\hfill \\\\ & =\\frac{1}{3}{u}^{3\\text{\/}2}+C\\hfill \\\\ & =\\frac{1}{3}{({z}^{2}-5)}^{3\\text{\/}2}+C.\\hfill \\end{array}[\/latex]&nbsp; &nbsp;<\/div>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Ak_y3lsBNfE?controls=0&amp;start=165&amp;end=266&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>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\/5.5Substitution165to266_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.5 Substitution&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Use substitution to evaluate the integral [latex]\\displaystyle\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt.[\/latex]<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572587721\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572587721\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"qfs-id1170573437523\" class=\"hidden-answer\">\n<p id=\"fs-id1170573437523\">We know the derivative of [latex]\\cos t[\/latex] is [latex]\\text{\u2212} \\sin t,[\/latex] so we set [latex]u= \\cos t.[\/latex] Then [latex]du=\\text{\u2212} \\sin tdt.[\/latex] Substituting into the integral, we have<\/p>\n<div id=\"fs-id1170573337947\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt=\\text{\u2212}\\displaystyle\\int \\frac{du}{{u}^{3}}.[\/latex]<\/div>\n<p id=\"fs-id1170573364147\">Evaluating the integral, we get<\/p>\n<div id=\"fs-id1170571136365\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ \\text{\u2212}{\\displaystyle\\int \\frac{du}{{u}^{3}}}\\hfill & =\\text{\u2212} {\\displaystyle\\int {u}^{-3}du}\\hfill \\\\ & =\\text{\u2212}(-\\frac{1}{2}){u}^{-2}+C.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170573409376\">Putting the answer back in terms of [latex]t[\/latex], we get<\/p>\n<div id=\"fs-id1170573587515\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc} {\\displaystyle\\int \\frac{ \\sin t}{{ \\cos }^{3}t}dt}\\hfill & =\\frac{1}{2{u}^{2}}+C\\hfill \\\\ \\\\ & =\\dfrac{1}{2{ \\cos }^{2}t}+C.\\hfill \\end{array}[\/latex] <\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<p id=\"fs-id1170573413569\">Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a constant. We need to eliminate all the expressions within the integrand that are in terms of the original variable. When we are done, [latex]u[\/latex] should be the only variable in the integrand.<\/p>\n<p>In some cases, this means solving for the original variable in terms of [latex]u[\/latex]. This technique should become clear in the next example.<\/p>\n<section class=\"textbox example\">\n<p>Use substitution to find the antiderivative of [latex]\\displaystyle\\int \\frac{x}{\\sqrt{x-1}}dx.[\/latex]<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572587722\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572587722\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573435928\">If we let [latex]u=x-1,[\/latex] then [latex]du=dx.[\/latex] But this does not account for the [latex]x[\/latex] in the numerator of the integrand. We need to express [latex]x[\/latex] in terms of [latex]u[\/latex]. If [latex]u=x-1,[\/latex] then [latex]x=u+1.[\/latex] Now we can rewrite the integral in terms of [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170573333822\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} {\\displaystyle\\int \\frac{x}{\\sqrt{x-1}}dx}\\hfill & = {\\displaystyle\\int \\frac{u+1}{\\sqrt{u}}du}\\hfill \\\\ \\\\ & = {\\displaystyle\\int \\sqrt{u}+\\frac{1}{\\sqrt{u}}du}\\hfill \\\\ & = {\\displaystyle\\int ({u}^{1\\text{\/}2}+{u}^{-1\\text{\/}2})du}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170573544033\">Then we integrate in the usual way, replace [latex]u[\/latex] with the original expression, and factor and simplify the result. Thus,<\/p>\n<div id=\"fs-id1170570997027\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc} {\\displaystyle\\int ({u}^{1\\text{\/}2}+{u}^{-1\\text{\/}2})du}\\hfill & =\\frac{2}{3}{u}^{3\\text{\/}2}+2{u}^{1\\text{\/}2}+C\\hfill \\\\ \\\\ & =\\frac{2}{3}{(x-1)}^{3\\text{\/}2}+2{(x-1)}^{1\\text{\/}2}+C\\hfill \\\\ & ={(x-1)}^{1\\text{\/}2}\\left[\\frac{2}{3}(x-1)+2\\right]+C\\hfill \\\\ & ={(x-1)}^{1\\text{\/}2}(\\frac{2}{3}x-\\frac{2}{3}+\\frac{6}{3})\\hfill \\\\ & ={(x-1)}^{1\\text{\/}2}(\\frac{2}{3}x+\\frac{4}{3})\\hfill \\\\ & =\\frac{2}{3}{(x-1)}^{1\\text{\/}2}(x+2)+C.\\hfill \\end{array}[\/latex] <\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288435\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288435&theme=lumen&iframe_resize_id=ohm288435&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":5,"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\":\"5.5 Substitution\",\"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":"5.5 Substitution","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\/404"}],"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\/404\/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\/404\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=404"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=404"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=404"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=404"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}