Paul's Online Notes
Paul's Online Notes
Home / Calculus II / Integration Techniques / Integration by Parts
Show Mobile Notice Show All Notes Hide All Notes
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 views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll 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

4. Evaluate \( \displaystyle \int{{6{{\tan }^{ - 1}}\left( {\frac{8}{w}} \right)\,dw}}\) .

Show All Steps Hide All Steps

Hint : Be careful with your choices of \(u\) and \(dv\) here. If you think about it there is really only one way that the choice can be made here and have the problem be workable.
Start Solution

The first step here is to pick \(u\) and \(dv\).

Note that if we choose the inverse tangent for \(dv\) the only way to get \(v\) is to integrate \(dv\) and so we would need to know the answer to get the answer and so that won’t work for us. Therefore, the only real choice for the inverse tangent is to let it be \(u\).

So, here are our choices for \(u\) and \(dv\).

\[u = 6{\tan ^{ - 1}}\left( {\frac{8}{w}} \right)\hspace{0.5in}dv = \,dw\]

Don’t forget the \(dw\)! The differential \(dw\) still needs to be put into the \(dv\) even though there is nothing else left in the integral.

Show Step 2

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

\[\begin{align*}u & = 6{\tan ^{ - 1}}\left( {\frac{8}{w}} \right) & \hspace{0.5in} & \to & \hspace{0.25in}du & = 6\frac{{ - \frac{8}{{{w^2}}}}}{{1 + {{\left( {\frac{8}{w}} \right)}^2}}}dw = 6\frac{{ - \frac{8}{{{w^2}}}}}{{1 + \frac{{64}}{{{w^2}}}}}dw\\ dv & = \,dw & \hspace{0.5in} & \to & \hspace{0.25in}v & = w\end{align*}\] Show Step 3

In order to complete this problem we’ll need to do some rewrite on \(du\) as follows,

\[du = \frac{{ - 48}}{{{w^2} + 64}}dw\]

Plugging \(u\), \(du\), \(v\) and \(dv\) into the Integration by Parts formula gives,

\[ \int{{6{{\tan }^{ - 1}}\left( {\frac{8}{w}} \right)\,dw}} = 6w\tan ^{ - 1}\left( {\frac{8}{w}} \right) + 48\int{{\frac{w}{{{w^2} + 64}}\,dw}}\] Show Step 4

Okay, the new integral we get is easily doable (with the substitution \(u = 64 + {w^2}\) ) and so all we need to do to finish this problem out is do the integral.

\[ \int{{6{{\tan }^{ - 1}}\left( {\frac{8}{w}} \right)\,dw}} = \require{bbox} \bbox[2pt,border:1px solid black]{{6w\tan ^{ - 1}\left( {\frac{8}{w}} \right) + 24\ln \left| {{w^2} + 64} \right| + c}}\]