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      <text>13.4.1.1 Integration by parts</text>
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      <text><![CDATA[<h3><span>Indefinite integration</span><br></h3>

<p>The technique known as <strong>integration by parts</strong> is used to integrate a product of two functions, such as in these two examples:
</p>
<ul>
    <li> \({\displaystyle\int e^{2x}\sin 3x \, \mathrm{d}x}\)</li>
    <li> \({\displaystyle\int_0^1 x^3 e^{-2x} \, \mathrm{d}x}\)</li>
</ul>
<p>Note that in the first example, the integrand is the product of the functions \(e^{2x}\) and \(\sin 3x\), and in the second example the integrand is the product of the functions \(x^3\) and \(e^{-2x}\). Note also that we can change the order of the terms
    in the product if we wish and write
</p>
<ul>
    <li>\({\displaystyle\int (\sin 3x) e^{2x} \, \mathrm{d}x}\)</li>
    <li>\({\displaystyle\int_0^1 e^{-2x}x^3 \, \mathrm{d}x}\)</li>
</ul>
What you must never do is integrate each term in the product separately and then multiply - the integral of a product is not the product of the separate integrals. However, it is often possible to find integrals involving products using the method of
integration by parts - you can think of this as a <em>product rule</em> for integrals.
<p></p>
<p>The integration by parts formula states:</p>

<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <p><strong>Integration by Parts for Indefinite Integrals</strong></p>
    <p>given functions \(f(x)\) and \(g(x)\):</p>

    \({\displaystyle\int f \cdot g \, \mathrm{d}x}\) = \(f\) \(\cdot\) \({\displaystyle\int g \, \mathrm{d}x}\) - \({\displaystyle\int \left(\dfrac{df}{dx} \displaystyle\int g \, \mathrm{d}x \right) \, \mathrm{d}x}\)

    <p>Alternatively, given functions \(u\) and \(v\):</p>
    \({\displaystyle \int u\dfrac{d v}{d x} \, \mathrm{d}x = u.v-\displaystyle \int v\dfrac{d u}{d x} \, \mathrm{d}x}\)
</div>
<p></p>
<p>Study the formula carefully and note the following observations. Firstly, to apply the formula we must be able to differentiate the function \(f\) to find \(\dfrac{d f}{d x}\), and we must be able to integrate the function, \(g\).</p>
<p>Secondly the formula replaces one integral, the one on the left, with a different integral, that on the far right. The intention is that the latter, whilst it may look more complicated in the formula above, is simpler to evaluate. Consider the following
    Example:</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Find the integral of the product of \(x\) with \(\sin x\); that is, find \({\displaystyle \int x\sin x \, \mathrm{d}x}\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>Compare the required integral with the formula for integration by parts: we choose \(f = x\) and \(g = \sin x\)</p>

<p>It follows that \(\dfrac{df}{dx} = 1\) and \({\displaystyle\int g \, \mathrm{d}x = \displaystyle\int \sin x \, \mathrm{d}x} = -\cos x\)</p>
<p>(When integrating \(g\) there is no need to worry about a constant of integration. When you become confident with the method, you may like to think about why this is the case.)</p>
<p>Applying the formula we obtain</p>
\begin{eqnarray*} \int x\sin x \, \mathrm{d} x &amp;=&amp; f\cdot \int g \, \mathrm{d} x -\int \left( {d f\over d x}\cdot\int g \, \mathrm{d} x \right)\,d x \\ &amp;=&amp; x(-\cos x) - \int 1(-\cos x) \, \mathrm{d} x\\ &amp;=&amp; -x\cos x + \int \cos
x \, \mathrm{d} x\\ &amp;=&amp; -x\cos x + \sin x +c \end{eqnarray*}]]></text>
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