# How do you find the indefinite integral of ∫sin 4xe^sin2x dx?

Apr 13, 2018

$I = {e}^{\sin \left(2 x\right)} \left(\sin \left(2 x\right) - 1\right) + C$

#### Explanation:

We want to solve

$I = \int \sin \left(4 x\right) {e}^{\sin \left(2 x\right)} \mathrm{dx}$

Using the trig identity color(blue)(sin(2a)=2cos(a)sin(a)

$I = 2 \int \cos \left(2 x\right) \sin \left(2 x\right) {e}^{\sin \left(2 x\right)} \mathrm{dx}$

Make a substitution $u = \sin \left(2 x\right) \implies \mathrm{du} = 2 \cos \left(2 x\right) \mathrm{dx}$

$I = \int u {e}^{u} \mathrm{du}$

$I = u {e}^{u} - \int {e}^{u} \mathrm{du}$

$\textcolor{w h i t e}{I} = u {e}^{u} - {e}^{u} + C$

$\textcolor{w h i t e}{I} = {e}^{u} \left(u - 1\right) + C$

Substitute back $u = \sin \left(2 x\right)$

$I = {e}^{\sin \left(2 x\right)} \left(\sin \left(2 x\right) - 1\right) + C$

Apr 13, 2018

$= \sin 2 x \setminus {e}^{\sin 2 x} - {e}^{\sin 2 x} + C$

#### Explanation:

$\int \setminus \sin 4 x \setminus {e}^{\sin 2 x} \setminus \mathrm{dx}$

$= 2 \int \setminus \sin 2 x \setminus \cos 2 x \setminus {e}^{\sin 2 x} \setminus \mathrm{dx}$

$= \int \setminus \sin 2 x \setminus d \left({e}^{\sin 2 x}\right)$

$= \sin 2 x \setminus {e}^{\sin 2 x} - \int \setminus d \left(\sin 2 x\right) {e}^{\sin 2 x}$

$= \sin 2 x \setminus {e}^{\sin 2 x} - 2 \int \cos 2 x \setminus {e}^{\sin 2 x} \setminus \mathrm{dx}$

$= \sin 2 x \setminus {e}^{\sin 2 x} - \int d \left({e}^{\sin 2 x}\right)$

$= \sin 2 x \setminus {e}^{\sin 2 x} - {e}^{\sin 2 x} + C$