# How do you find the integral of (e^(8x))sin(9x)dx?

Mar 7, 2018

$\setminus \int {e}^{8 x} \sin \left(9 x\right) \mathrm{dx} = \frac{1}{145} {e}^{8 x} \left(8 \sin \left(9 x\right) - 9 \cos \left(9 x\right)\right) + C$

#### Explanation:

(1) $\setminus \int {e}^{8 x} \sin \left(9 x\right) \mathrm{dx}$

We need to use integration by parts.

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

It's useful to remember the acronym LIATE: Log, Inverse Trig, Algebraic, Trig, Exponential for knowing the prioritization of choosing your $u$ value.

Since trigonometric functions take precedence over exponentials, we will use $u = \sin \left(9 x\right)$.

Substitution:
$u = \sin \left(9 x\right)$
$\mathrm{du} = 9 \cos \left(9 x\right) \mathrm{dx}$
$\mathrm{dv} = {e}^{8 x} \mathrm{dx}$
$v = \frac{1}{8} {e}^{8 x}$

(2) $\setminus \int {e}^{8 x} \sin \left(9 x\right) \mathrm{dx} = \frac{1}{8} {e}^{8 x} \sin \left(9 x\right) - \frac{9}{8} \setminus \int {e}^{8 x} \cos \left(9 x\right) \mathrm{dx}$

Since we haven't improved our situation, it looks like another round of integration by parts. Let's use the letters $y$ and $z$, and let $y = \cos \left(9 x\right)$ and integrate the RHS integral by parts.

$\setminus \int y \mathrm{dz} = y z - \setminus \int z \mathrm{dy}$

Substitution:
$y = \cos \left(9 x\right)$
$\mathrm{dy} = - 9 \sin \left(9 x\right) \mathrm{dx}$
$\mathrm{dz} = {e}^{8 x} \mathrm{dx}$
$z = \frac{1}{8} {e}^{8 x}$

(3) $\setminus \int {e}^{8 x} \cos \left(9 x\right) \mathrm{dx} = \frac{1}{8} {e}^{8 x} \cos \left(9 x\right) + \frac{9}{8} \setminus \int {e}^{8 x} \sin \left(9 x\right) \mathrm{dx}$

Substituting (3) $\to$ (2):

(4) $\setminus \int {e}^{8 x} \sin \left(9 x\right) \mathrm{dx} = \frac{1}{8} {e}^{8 x} \sin \left(9 x\right)$$- \frac{9}{8} \left(\frac{1}{8} {e}^{8 x} \cos \left(9 x\right) + \frac{9}{8} \setminus \int {e}^{8 x} \sin \left(9 x\right) \mathrm{dx}\right)$

$\setminus \int {e}^{8 x} \sin \left(9 x\right) \mathrm{dx} = \frac{1}{8} {e}^{8 x} \sin \left(9 x\right) - \frac{9}{64} {e}^{8 x} \cos \left(9 x\right)$$- \frac{81}{64} \setminus \int {e}^{8 x} \sin \left(9 x\right) \mathrm{dx}$

$\frac{145}{64} \setminus \int {e}^{8 x} \sin \left(9 x\right) \mathrm{dx} = \frac{1}{8} {e}^{8 x} \sin \left(9 x\right) - \frac{9}{64} {e}^{8 x} \cos \left(9 x\right)$

$= \frac{64}{145} \left(\frac{1}{8} {e}^{8 x} \sin \left(9 x\right) - \frac{9}{64} {e}^{8 x} \cos \left(9 x\right)\right) + C$

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

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