# Question a58a3

Jan 12, 2018

$\int {e}^{8 \theta} \sin \left(9 \theta\right) d \theta = - \frac{9}{145} {e}^{8 \theta} \cos \left(9 \theta\right) + \frac{8}{145} {e}^{8 \theta} \sin \left(9 \theta\right) + \text{C}$

OR

$\int {e}^{8 \theta} \sin \left(9 \theta\right) d \theta = - \frac{1}{145} {e}^{8 \theta} \left\{9 \cos \left(9 \theta\right) - 8 \sin \left(9 \theta\right)\right\} + \text{C}$

#### Explanation:

Given $\int {e}^{8 \theta} \sin \left(9 \theta\right) d \theta$

We will Apply Integration by Parts (I.B.P) twice and solve the integral algebraically at the end:

$\int$ ${u}_{1}$ ${\mathrm{dv}}_{1} = u v - \int$ $v$ $\mathrm{du}$

Also: We'll let $\int {e}^{8 \theta} \sin \left(9 \theta\right) d \theta = I$

Let color(red)(u=e^(8theta);dv=sin(9theta)

Thus: color(red)(du_1=8e^(8theta)d theta;v_1=-1/9cos(9theta)

I=color(red)(e^(8theta)*-1/9cos(9theta)-int-1/9cos(9theta)*8e^(8theta)d theta

Simplify

I=color(red)(-1/9e^(8theta)cos(9theta)+8/9intcos(9theta)*e^(8theta)d theta

Apply I.B.P again

color(blue)(u_2=e^(8theta);dv_2=cos(9theta)

color(blue)(du_2=8e^(8theta)d theta;v_2=1/9sin(9theta)

I=color(red)(-1/9e^(8theta)cos(9theta)+8/9)color(blue)({e^(8theta)*1/9sin(9theta)-int1/9sin(9theta)*8e^(8theta)d theta}

Simplify

I=color(red)(-1/9e^(8theta)cos(9theta)+8/9)color(blue)({1/9e^(8theta)sin(9theta)-8/9intsin(9theta)*e^(8theta)d theta}

I=color(red)(-1/9e^(8theta)cos(9theta)+)color(blue)(8/81e^(8theta)sin(9theta)-64/81inte^(8theta)sin(9theta) d theta#

Notice how the integral $\int {e}^{8 \theta} \sin \left(9 \theta\right) d \theta$ appears both on the left and right sides and so we can simplify this algebraic expression simply by solving for $\int {e}^{8 \theta} \sin \left(9 \theta\right) d \theta$

Add $\frac{64}{81} \int {e}^{8 \theta} \sin \left(9 \theta\right) d \theta$ to both sides

$\frac{145}{81} \int {e}^{8 \theta} \sin \left(9 \theta\right) d \theta = - \frac{1}{9} {e}^{8 \theta} \cos \left(9 \theta\right) + \frac{8}{81} {e}^{8 \theta} \sin \left(9 \theta\right)$

Divide $\frac{145}{81}$ to both sides

$\int {e}^{8 \theta} \sin \left(9 \theta\right) d \theta = \frac{81}{145} \left\{- \frac{1}{9} {e}^{8 \theta} \cos \left(9 \theta\right) + \frac{8}{81} {e}^{8 \theta} \sin \left(9 \theta\right)\right\}$

Simplify

$\int {e}^{8 \theta} \sin \left(9 \theta\right) d \theta = - \frac{81}{1305} {e}^{8 \theta} \cos \left(9 \theta\right) + \frac{648}{11745} {e}^{8 \theta} \sin \left(9 \theta\right)$

$\int {e}^{8 \theta} \sin \left(9 \theta\right) d \theta = - \frac{9}{145} {e}^{8 \theta} \cos \left(9 \theta\right) + \frac{8}{145} {e}^{8 \theta} \sin \left(9 \theta\right) + \text{C}$

Alternatively, we can factor out $- \frac{1}{145} {e}^{8 \theta}$ to get:

$\int {e}^{8 \theta} \sin \left(9 \theta\right) d \theta = - \frac{1}{145} {e}^{8 \theta} \left\{9 \cos \left(9 \theta\right) - 8 \sin \left(9 \theta\right)\right\} + \text{C}$