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<div id="main-column" class="span9"> <article id="integration-by-parts-tutorial" style="margin-top: 16px;">
<h1 class="section-title">Integration by parts (tutorial)</h1>
<ul class="breadcrumb"><li><a title="Home" href="../../../index.html"><i class="fa fa-home"></i></a><span class="divider">/</span></li><li><span class="gray">5. Calculus</span><span class="divider">/</span></li><li><span class="active">Integration by parts (tutorial)</span></li></ul>
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<h2><u>Tutorial</u>: Integration by Parts</h2><p><img alt="" src="../../../ib/mathanalysis/analysis/calculus/images/intparts_img1-1.jpg" style="float: right; width: 250px; height: 144px;">In developing a technique for integrating functions it is important to remember the fact that differentiation and integration are inverse operations. An earlier tutorial demonstrated that the technique of <strong><em>integration by substitution </em></strong>can be developed by <em>reversing</em> the Chain Rule (i.e. the rule for differentiating composite functions). </p><p>Hence, a good starting point for investigating another integration method is to take a look back at some other differentiation technique. Along with the Chain Rule another very useful differentiation rule is the <strong>Product Rule</strong>. This rule for differentiating a function that consists of the product of two functions can be stated formally as follows.</p><p style="text-align: center;"><img alt="" src="../../../ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-parts/int_parts_img2.jpg" style="width: 680px; height: 297px;"></p><p>To begin developing an integration technique for products (i.e. <em>reversing </em>the process of differentiation), it seems reasonable to take the Product Rule and integrate both sides.</p><p style="margin-left: 80px;"><span class="math-tex">\(\frac{d}{{dx}}\left[ {uv} \right] = u\frac{{dv}}{{dx}}\; + \;v\frac{{du}}{{dx}}\)</span> (using the Leibniz form of the Product Rule)</p><p style="margin-left: 80px;"><span class="math-tex">\(\int {\left( {\frac{d}{{dx}}\left[ {uv} \right]} \right)dx = \int {\left( {u\frac{{dv}}{{dx}}} \right)dx\; + \;\int {\left( {v\frac{{du}}{{dx}}} \right)dx} } } \)</span> (integrating both sides)</p><p style="margin-left: 80px;"><span class="math-tex">\(uv = \int {u\;dv\; + \;\int {v\;du} } \)</span> (treating differentials algebraically)</p><p style="margin-left: 80px;"><span class="math-tex">\(\int {u\;dv\; = \;uv - \int {v\;du} } \)</span> (subtracting from both sides)</p><p>This formula expresses the original integral in terms of another ‘new’ integral. Depending on the choices of <em><span class="math-tex">\(u\)</span></em> and <em><span class="math-tex">\(dv\)</span></em>, it may be easier to evaluate the ‘new’ integral, <span class="math-tex">\(\int {v\;du} \)</span>, than the original one. It quickly becomes clear that the choices of<em> <span class="math-tex">\(u\)</span></em> and <em><span class="math-tex">\(dv\)</span></em> (the two ‘parts’ of the integrand) are critical for this particular technique – hence, this integration technique is known as <strong><em>integration by parts</em></strong>.</p><h3>Using integration by parts</h3><p><strong>Example 1</strong>: Evaluate <span class="math-tex">\(\int {x\;\ln x\;dx} \)</span></p><p>Explanation / Solution: </p><p>To re-write this integral in the format required by integration by parts, i.e. <span class="math-tex">\(\int {u\;dv\; = \;uv - \int {v\;du} } \)</span>, there are only two possible choices for<em> <span class="math-tex">\(u\)</span></em> and <em><span class="math-tex">\(dv\)</span></em>. Let’s try the choice: <span class="math-tex">\(u = x\)</span> and <span class="math-tex">\(dv = \ln x\;dx\)</span>. If <span class="math-tex">\(u = x\)</span> then <span class="math-tex">\(du = dx\)</span>; and if <span class="math-tex">\(dv = \ln x\;dx\)</span> then <span class="math-tex">\(v = \int {\ln x\;dx} \)</span>. But what is <span class="math-tex">\(\int {\ln x\;dx} \)</span>? It certainly cannot be found by inspection. The function <span class="math-tex">\(\ln x\)</span> does not have a straightforward anti-derivative. Hence, this choice for <em><span class="math-tex">\(u\)</span></em> and <em><span class="math-tex">\(dv\)</span></em> does not seem to be a good one. Let’s try the other possibility: <span class="math-tex">\(u = \ln x\)</span> and <span class="math-tex">\(dv = x\;dx\)</span>. If <span class="math-tex">\(u = \ln x\)</span> then <span class="math-tex">\(du = \frac{1}{x}dx\)</span>; and if <span class="math-tex">\(dv = x\;dx\)</span> then <span class="math-tex">\(v = \int {x\;dx = \frac{1}{2}{x^2}} \)</span>.</p><p>Substituting into the formula <span class="math-tex">\(\int {u\;dv\; = \;uv - \int {v\;du} } \)</span> gives:</p><p style="margin-left: 40px;"><span class="math-tex">\(\int {x\;\ln x\;dx = \left( {\ln x} \right)\left( {\frac{1}{2}{x^2}} \right) - \int {\frac{1}{2}{x^2}\frac{1}{x}dx = \frac{1}{2}{x^2}\ln x - \frac{1}{2}\int {x\;dx} } } \)</span></p><p>The key question at this stage is whether the ‘new’ integral can be evaluated in a straightforward fashion – i.e. can it be found by inspection? The answer for this example is clearly ‘yes’, since the anti-derivative of <span class="math-tex">\(x\)</span> is <span class="math-tex">\(\frac{1}{2}{x^2}\)</span>. Let’s continue:</p><p style="margin-left: 40px;"><span class="math-tex">\(\int {x\;\ln x\;dx = \frac{1}{2}{x^2}\ln x - \int {\frac{1}{2}{x^2}\frac{1}{x}dx = \frac{1}{2}{x^2}\ln x - \frac{1}{2}\int {x\;dx} } } = \)</span></p><p style="margin-left: 120px;"><span class="math-tex">\( = \frac{1}{2}{x^2}\ln x - \frac{1}{2}\left( {\frac{1}{2}{x^2}} \right) + C\)</span></p><p style="margin-left: 120px;"><span class="math-tex">\( = \frac{1}{2}{x^2}\ln x - \frac{1}{4}{x^2} + C\)</span></p> <p><strong>Exercise 1</strong>:</p><p>Check the result for Example 1 by showing that the derivative of <span class="math-tex">\(\frac{1}{2}{x^2}\ln x - \frac{1}{4}{x^2} + C{\textrm{ }}\)</span> is equal to the original integrand <span class="math-tex">\(x\;\ln x\)</span>.</p> <p><strong>Example 2</strong>: Evaulate <span class="math-tex">\(\int {x\;\sin x\;dx} \)</span>.</p><p>Explanation / Solution:</p><p>The first task is to make choices for the two ‘parts’ <span class="math-tex">\(u\)</span> and<em> </em><span class="math-tex">\(dv\)</span>. Let’s try <span class="math-tex">\(u = \sin x\)</span> and <span class="math-tex">\(dv = x\;dx\)</span>. Then <span class="math-tex">\(du = \cos x\;dx\)</span> and <span class="math-tex">\(v = \int {x\;dx = \frac{1}{2}{x^2}} \)</span>. There was no problem evaluating <span class="math-tex">\(du\)</span> and <span class="math-tex">\(v\)</span> for this set of choices. Now substituting into the formula for integration by parts <span class="math-tex">\(\int {u\;dv\; = \;uv - \int {v\;du} } \)</span> gives:</p><p style="margin-left: 40px;"><span class="math-tex">\(\int {x\sin x\;dx = \frac{1}{2}{x^2}\sin x - \int {\frac{1}{2}{x^2}} } \cos x\;dx\)</span></p><p>The ‘new’ integral, <span class="math-tex">\(\frac{1}{2}\int {{x^2}\cos x\;dx} \)</span>, does not look any friendlier than the original integral. Hence, let’s start over and try the other possible set of choices, i.e. <span class="math-tex">\(u = x\)</span> and <span class="math-tex">\(dv = \sin x\;dx\)</span>. Then <span class="math-tex">\(du = dx\)</span> and <span class="math-tex">\(v = \int {\sin x\;dx = - \cos x\;dx} \)</span>. Substituting into the formula gives:</p><p style="margin-left: 40px;"><span class="math-tex">\(\int {x\sin x\;dx = x\left( { - \cos x} \right) - \int {( - \cos x)\;dx} } = - x\cos x + \int {\cos x\;dx} \)</span></p><p style="margin-left: 320px;"><span class="math-tex">\( = - x\cos x\; + \;\sin x\; + \;C\)</span></p><p><strong>Exercise 2</strong>:</p><p>Check the result for Example 2 by showing that the derivative of <span class="math-tex">\( - x\cos x\; + \;\sin x\; + \;C{\textrm{ }}\)</span> is equal to the original integrand <span class="math-tex">\(x\;\sin x\)</span>.</p> <p>After the first two examples it is even more clear that the key step in this integration technique is the choices of the two parts <span class="math-tex">\(u\)</span> and <span class="math-tex">\(dv\)</span> in the integrand that is being integrated. Given this and taking a closer look at the first two examples, provides support for forming some guidelines for choosing <span class="math-tex">\(u\)</span> and <span class="math-tex">\(dv\)</span> when performing integration by parts.</p><p style="text-align: center;"><img alt="" src="../../../ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-parts/int_parts_img3.jpg" style="width: 650px; height: 159px;"></p> <p>Some integrals require repeated use of the integration by parts formula.</p> <p><strong>Example 3</strong>: Evaluate <span class="math-tex">\(\int {{x^2}\sin x\;dx} \)</span>.</p><p>Explanation / Solution:</p><p>The factors <span class="math-tex">\({x^2}\)</span> and <span class="math-tex">\(\sin x\)</span> are equally easy to integrate. However, the derivative of <span class="math-tex">\({x^2}\)</span> becomes simpler, whereas the derivative of <span class="math-tex">\(\sin x\)</span> does not. Therefore, it is best to choose <span class="math-tex">\(u = {x^2}\)</span> and <span class="math-tex">\(dv = \sin x\;dx\)</span>. Then <span class="math-tex">\(du = 2x\;dx\)</span> and <span class="math-tex">\(v = \int {\sin x\;dx = - \cos x} \)</span>. Substituting into the formula produces the following.</p><p style="margin-left: 40px;"><span class="math-tex">\(\int {{x^2}\sin x\;dx = - {x^2}\cos x + \int {2x\cos x\;dx} } \)</span></p><p>The first use of integration by parts has succeeded in simplifying the original integral, but the ‘new’ integral on the right side still does not fit a basic integration rule. To evaluate that integral, integration by parts can be applied again. This time let <span class="math-tex">\(u = 2x\)</span> and <span class="math-tex">\(dv = \cos x\;dx\)</span>; and then <span class="math-tex">\(du = 2\;dx\)</span> and <span class="math-tex">\(v = \int {\cos x\;dx = \sin x} \)</span>. Applying integration by parts on the ‘new’ integral produces</p><p style="margin-left: 40px;"><span class="math-tex">\(\int {{x^2}\sin x\;dx = - {x^2}\cos x + \int {2x\cos x\;dx} } = - {x^2}\cos x + 2x\sin x - \int {2\sin x\;dx} \)</span></p><p style="margin-left: 320px;"><span class="math-tex">\( = - {x^2}\cos x + 2x\sin x + 2\cos x + C\)</span></p> <p><strong>Exercise 3</strong>: Check the result for Example 3 by showing that the derivative of <span class="math-tex">\( - {x^2}\cos x\; + \;2x\sin x\; + \;2\cos x + C\)</span> is equal to the original integrand <span class="math-tex">\({x^2}\;\sin x\)</span>.</p> <p>When making repeated applications of integration by parts, it is important not to interchange the substitutions in successive applications. For instance, in Example 3, the first substitution was <span class="math-tex">\(u = {x^2}\)</span> and .<span class="math-tex">\(dv = \sin x\;dx\)</span>. If, in the second application, the substitutions were switched so that <span class="math-tex">\(u = \cos x\)</span> and <span class="math-tex">\(dv = 2x\;dx\)</span>, the following would have been obtained</p><p style="margin-left: 40px;"><span class="math-tex">\(\int {{x^2}\sin x\;dx = - {x^2}\cos x + \int {2x\cos x\;dx} } \)</span></p><p style="margin-left: 120px;"><span class="math-tex">\( = - {x^2}\cos x + {x^2}\cos x + \int {{x^2}\sin x\;dx} \)</span></p><p style="margin-left: 120px;"><span class="math-tex">\( = \int {{x^2}\sin x\;dx} \)</span></p><p>thus, undoing the previous integration and returning to the <em>original</em> integral.</p><p>When making repeated applications of integration by parts, it is also important to watch for the appearance of a <em>constant multiple</em> of the original integral. This is when the original integral appears on the right side so that it can be added to both sides producing a constant multiple of the original integral on the left side. Remember this when working on the next exercise.</p><p><strong><u>Exercise 4</u>:</strong> Evaluate <span class="math-tex">\(\int {{e^x}\cos 2x\;dx} \)</span></p><p>Hint: Let <span class="math-tex">\(u = \cos 2x\)</span> and <span class="math-tex">\(dv = {e^x}dx\)</span> in the first substitution; and let <span class="math-tex">\(u = \sin 2x\)</span> and <span class="math-tex">\(dv = {e^x}dx\)</span> in the second substitution.</p><p><strong><u>Exercise 5</u>:</strong> Evaluate <span class="math-tex">\(\int {\ln x\;dx} \)</span></p><p>This integral was a stumbling block in Example 1. At first glance it appears that the integrand does not have two parts – but it does.</p><p><strong><u>Exercises 6 – 11</u>:</strong> Use integration by parts to evaluate each indefinite integral.</p><p>6. <span class="math-tex">\(\int {{e^x}\sin x\;dx} \)</span> 7. <span class="math-tex">\(\int {\frac{{\ln x}}{{{x^2}}}dx} \)</span><br> </p><p>8. <span class="math-tex">\(\int {x\sqrt {x - 1} \;dx} \)</span> 9. <span class="math-tex">\(\int {\left( {{x^2} - 1} \right){e^x}dx} \)</span><br> </p><p>10. <span class="math-tex">\(\int {{x^2}\ln x\;dx} \)</span> 11. <span class="math-tex">\(\int {{x^2}\cos x\;dx} \)</span> </p><p><strong><u>Exercise 12</u>:</strong> Use the technique of integration by substitution to evaluate the integral in 8. above.</p><p>Hint: Let <span class="math-tex">\(u = x - 1\)</span> and then consider what is <em>x</em> in terms of <em>u</em>.</p><p><strong><u>Exercise 13</u>:</strong></p><p>(a) Evaluate the integral <em> <span class="math-tex">\(\int {{x^n}{e^x}\;dx} \)</span> </em>for <span class="math-tex">\(n = \)</span>0, 1, 2, and 3. </p><p>(b) Use the results to formulate a general rule for <span class="math-tex">\(\int {{x^n}{e^x}\;dx} \)</span> for any positive integer <em>n</em>.</p><p>(c) Use mathematical induction to prove your general formula for <span class="math-tex">\(\int {{x^n}{e^x}\;dx} \)</span><em> .</em></p><p><strong><u>Exercise 14</u>:</strong></p><p>Use integration by parts to verify the formula (<em>n</em> is a positive integer).</p><p><span class="math-tex">\(\int {{x^n}\sin x\;dx = - {x^n}\cos x + n\int {{x^{n - 1}}\cos x\;dx} } \)</span></p><script>document.querySelectorAll('.tib-teacher-only').forEach(e => e.remove());</script>
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