Calculus II - Integration by Parts

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Section 7.1 : Integration by Parts

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

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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.

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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.

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