Calculus II - Integration by Parts

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Section 7.1 : Integration by Parts

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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 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.

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