# How do you integrate int e^xsinx by parts from [0,1]?

Dec 19, 2016

The answer is $= \frac{1}{2} {e}^{x} \left(\sin x - \cos x\right) + C$

#### Explanation:

The integration by parts is applied 2 times

$\int u v ' = u v - \int u ' v$

Here,

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

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

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

We do the integration by parts a second time

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

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

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

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

Therefore,

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

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

So,

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

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