# How do you integrate int sinx*e^-x by integration by parts method?

May 25, 2018

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

#### Explanation:

Integrate by parts:

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

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

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

and then again:

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

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

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

The same integral appears on both sides and we can solve fro it:

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

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