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

5. Evaluate \( \displaystyle \int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}}\) .

Show All Steps Hide All Steps

Hint : This is one of the few integration by parts problems where either function can go on \(u\) and \(dv\). Be careful however to not get locked into an endless cycle of integration by parts.
Start Solution

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

In this case we can put the exponential in either the \(u\) or the \(dv\) and the cosine in the other. It is one of the few problems where the choice doesn’t really matter.

For this problem well use the following choices for \(u\) and \(dv\).

\[u = \cos \left( {\frac{1}{4}z} \right)\hspace{0.5in}dv = {{\bf{e}}^{2z}}\,dz\] Show Step 2

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

\[\begin{align*}u & = \cos \left( {\frac{1}{4}z} \right) & \hspace{0.5in} & \to & \hspace{0.25in}du & = - \frac{1}{4}\sin \left( {\frac{1}{4}z} \right)dz\\ dv & = \,{{\bf{e}}^{2z}}dz & \hspace{0.5in} & \to & \hspace{0.25in}v & = \frac{1}{2}{{\bf{e}}^{2z}}\end{align*}\] Show Step 3

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

\[ \int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}} = \frac{1}{2}{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right) + \frac{1}{8}\int{{{{\bf{e}}^{2z}}\sin \left( {\frac{1}{4}z} \right)\,dz}}\] Show Step 4

We’ll now need to do integration by parts again and to do this we’ll use the following choices.

\[\begin{align*}u & = \sin \left( {\frac{1}{4}z} \right) & \hspace{0.5in} & \to & \hspace{0.25in}du & = \frac{1}{4}\cos \left( {\frac{1}{4}z} \right)dz\\ dv & = \,{{\bf{e}}^{2z}}dz & \hspace{0.5in} & \to & \hspace{0.25in}v & = \frac{1}{2}{{\bf{e}}^{2z}}\end{align*}\] Show Step 5

Plugging these into the integral from Step 3 gives,

\[\begin{align*}\int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}} & = \frac{1}{2}{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right) + \frac{1}{8}\left[ {\frac{1}{2}{{\bf{e}}^{2z}}\sin \left( {\frac{1}{4}z} \right) - \frac{1}{8}\int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}}} \right]\\ & = \frac{1}{2}{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right) + \frac{1}{{16}}{{\bf{e}}^{2z}}\sin \left( {\frac{1}{4}z} \right) - \frac{1}{{64}}\int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}}\end{align*}\] Show Step 6

To finish this problem all we need to do is some basic algebraic manipulation to get the identical integrals on the same side of the equal sign.

\[\begin{align*}\int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}} & = \frac{1}{2}{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right) + \frac{1}{{16}}{{\bf{e}}^{2z}}\sin \left( {\frac{1}{4}z} \right) - \frac{1}{{64}}\int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}}\\ \int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}} + \frac{1}{{64}}\int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}} & = \frac{1}{2}{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right) + \frac{1}{{16}}{{\bf{e}}^{2z}}\sin \left( {\frac{1}{4}z} \right)\\ \frac{{65}}{{64}}\int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}} & = \frac{1}{2}{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right) + \frac{1}{{16}}{{\bf{e}}^{2z}}\sin \left( {\frac{1}{4}z} \right)\end{align*}\]

Finally, all we need to do is move the coefficient on the integral over to the right side.

\[ \int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{32}}{{65}}{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right) + \frac{4}{{65}}{{\bf{e}}^{2z}}\sin \left( {\frac{1}{4}z} \right) + c}}\]