# How do you integrate int xsin(10x) by integration by parts method?

##### 1 Answer
Jan 29, 2017

$\frac{x}{10} \cos 10 x + \frac{1}{100} \sin 10 x + C$

#### Explanation:

It is important that the integration by parts formula be memorised

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

when using the IBP the choice of $u \text{ & } v '$ is crucial.

in this case

$u = x \implies u ' = 1$

$v ' = \sin 10 x \implies v = - \frac{1}{10} \cos 10 x$

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

becomes

$I = - \frac{x}{10} \cos 10 x - \left(\int \frac{1}{10} \cos 10 x \mathrm{dx}\right)$

$I = - \frac{x}{10} \cos 10 x + \frac{1}{100} \sin 10 x + C$