Calculus II - Integration by Parts

Show Mobile Notice

Mobile Notice

You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 7.1 : Integration by Parts

Back to Problem List

6. Evaluate $\displaystyle \int_{0}^{\pi }{{{x^2}\cos \left( {4x} \right)\,dx}}$ .

Show All Steps Hide All Steps

Start Solution

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 or will at least be an integral that will be easier to deal with.

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

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

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

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

We can deal with the limits as we do the integral or we can just do the indefinite integral and then take care of the limits in the last step. We will be using the later way of dealing with the limits for this problem.

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

$\int{{{x^2}\cos \left( {4x} \right)\,dx}} = \frac{1}{4}{x^2}\sin \left( {4x} \right) - \frac{1}{2}\int{{x\sin \left( {4x} \right)\,dx}}$ Show Step 4

Now, the new integral is still not one that we can do with only Calculus I techniques. However, it is one that we can do another integration by parts on and because the power on the $x$ ’s have gone down by one we are heading in the right direction.

So, here are the choices for $u$ and $dv$ for the new integral.

$\begin{align*}u & = x & \hspace{0.5in} & \to & \hspace{0.25in}du & = dx\\ dv & = \sin \left( {4x} \right)\,dx & \hspace{0.5in} & \to & \hspace{0.25in}v & = - \frac{1}{4}\cos \left( {4x} \right)\end{align*}$ Show Step 5

Okay, all we need to do now is plug these new choices of $u$ and $dv$ into the new integral we got in Step 3 and evaluate the integral.

$\begin{align*}\int{{{x^2}\cos \left( {4x} \right)\,dx}} & = \frac{1}{4}{x^2}\sin \left( {4x} \right) - \frac{1}{2}\left[ { - \frac{1}{4}x\cos \left( {4x} \right) + \frac{1}{4}\int{{\cos \left( {4x} \right)\,dx}}} \right]\\ & = \frac{1}{4}{x^2}\sin \left( {4x} \right) - \frac{1}{2}\left[ { - \frac{1}{4}x\cos \left( {4x} \right) + \frac{1}{{16}}\sin \left( {4x} \right)} \right] + c\\ & = \frac{1}{4}{x^2}\sin \left( {4x} \right) + \frac{1}{8}x\cos \left( {4x} \right) - \frac{1}{{32}}\sin \left( {4x} \right) + c\end{align*}$ Show Step 6

The final step is then to take care of the limits.

$\int_{0}^{\pi }{{{x^2}\cos \left( {4x} \right)\,dx}} = \left. {\left( {\frac{1}{4}{x^2}\sin \left( {4x} \right) + \frac{1}{8}x\cos \left( {4x} \right) - \frac{1}{{32}}\sin \left( {4x} \right)} \right)} \right|_0^\pi = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{8}\pi }}$