Calculus II - Integration by Parts

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Section 7.1 : Integration by Parts

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1. Evaluate $\displaystyle \int{{4x\cos \left( {2 - 3x} \right)\,dx}}$ .

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The first step here is to pick $u$ and $dv$ . We want to choose $u$ and $dv$ so that when we compute $du$ and $v$ and plugging everything into the Integration by Parts formula the new integral we get is one that we can do.

With that in mind it looks like the following choices for $u$ and $dv$ should work for us.

$u = 4x\hspace{0.5in}dv = \cos \left( {2 - 3x} \right)\,dx$ Show Step 2

Next, we need to compute $du$ (by differentiating $u$ ) and $v$ (by integrating $dv$ ).

$\begin{align*}u & = 4x & \hspace{0.5in} & \to & \hspace{0.25in}du & = 4dx\\ dv & = \cos \left( {2 - 3x} \right)\,dx & \hspace{0.25in} & \to & \hspace{0.25in}v & = - \frac{1}{3}\sin \left( {2 - 3x} \right)\end{align*}$ Show Step 3

Plugging $u$ , $du$ , $v$ and $dv$ into the Integration by Parts formula gives,

$\begin{align*}\int{{4x\cos \left( {2 - 3x} \right)\,dx}} & = \left( {4x} \right)\left( { - \frac{1}{3}\sin \left( {2 - 3x} \right)} \right) - \int{{ - \frac{4}{3}\sin \left( {2 - 3x} \right)\,dx}}\\ & = - \frac{4}{3}x\sin \left( {2 - 3x} \right) + \frac{4}{3}\int{{\sin \left( {2 - 3x} \right)\,dx}}\end{align*}$ Show Step 4

Okay, the new integral we get is easily doable and so all we need to do to finish this problem out is do the integral.

$\int{{4x\cos \left( {2 - 3x} \right)\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{4}{3}x\sin \left( {2 - 3x} \right) + \frac{4}{9}\cos \left( {2 - 3x} \right) + c}}$