Section 7.1 : Integration by Parts
5. Evaluate \( \displaystyle \int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}}\) .
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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 2Next, 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 3Plugging \(u\), \(du\), \(v\) 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 4We’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 5Plugging 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 6To 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.
\[ \int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{32}}{{65}}{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right) + \frac{4}{{65}}{{\bf{e}}^{2z}}\sin \left( {\frac{1}{4}z} \right) + c}}\]