# How do you integrate int x csc^2x by integration by parts method?

##### 1 Answer
Apr 1, 2018

The answer is $= - x \cot x + \ln \left(| \sin x |\right) + C$

#### Explanation:

$\text{ Reminder }$

$\int \cot x \mathrm{dx} = \ln \left(| \sin x |\right) + C$

Perform an integration by parts

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

$u = x$, $\implies$, $u ' = 1$

$v ' = {\csc}^{2} x$, $\implies$, $v = - \cot x$

Therefore,

$\int x {\sec}^{2} x \mathrm{dx} = - x \cot x + \int \cot x \mathrm{dx}$

$= - x \cot x + \ln \left(| \sin x |\right) + C$