Section 7.1 : Integration by Parts
1. Evaluate \( \displaystyle \int{{4x\cos \left( {2 - 3x} \right)\,dx}}\) .
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Hint : Remember that we want to pick \(u\) and \(dv\) so that upon computing \(du\) and \(v\) and plugging everything into the Integration by Parts formula the new integral is one that we can do.
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 2Next, 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 3Plugging \(u\), \(du\), \(v\) 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 4Okay, the new integral we get is easily doable and so all we need to do to finish this problem out is do the integral.
\[ \int{{4x\cos \left( {2 - 3x} \right)\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{4}{3}x\sin \left( {2 - 3x} \right) + \frac{4}{9}\cos \left( {2 - 3x} \right) + c}}\]