# How do you integrate int e^-xsin4x by integration by parts method?

Dec 24, 2016

The answer is $= \frac{- {e}^{- x} \left(\sin 4 x + 4 \cos 4 x\right)}{17} + C$

#### Explanation:

We use the integration by parts

$\int u v ' \mathrm{dx} = u v - \int u ' v \mathrm{dx}$

Here,

$u = \sin 4 x$, $\implies$, $u ' = 4 \cos 4 x$

$v ' = {e}^{- x}$, $\implies$, $v = - {e}^{- x}$

$\int {e}^{- x} \sin 4 x \mathrm{dx} = - {e}^{- x} \sin 4 x - \int 4 \cdot - {e}^{- x} \cos 4 x \mathrm{dx}$

$= - {e}^{- x} \sin 4 x + 4 \int {e}^{- x} \cos 4 x \mathrm{dx}$

For the integral $\int {e}^{- x} \cos 4 x \mathrm{dx}$, we apply the integration by parts a second time

$u = \cos 4 x$, $\implies$, $u ' = - 4 \sin 4 x$

$v ' = {e}^{- x}$, $\implies$, $v = - {e}^{- x}$

$\int {e}^{- x} \cos 4 x \mathrm{dx} = - {e}^{- x} \cos 4 x - 4 \int {e}^{- x} \sin 4 x \mathrm{dx}$

Putting it all together

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

$= - {e}^{- x} \sin 4 x - 4 {e}^{- x} \cos 4 x - 16 \int {e}^{- x} \sin 4 x \mathrm{dx}$

Therefore,

$17 \int {e}^{- x} \sin 4 x \mathrm{dx} = - {e}^{- x} \left(\sin 4 x + 4 \cos 4 x\right)$

$\int {e}^{- x} \sin 4 x \mathrm{dx} = \frac{- {e}^{- x} \left(\sin 4 x + 4 \cos 4 x\right)}{17} + C$