# Math Help - integral question..

1. ## integral question..

$\int \frac{x}{\sin ^2x}dx=-arcotx -\int -arccot x dx$

u=x u'=1
$v'=\frac{1}{\sin ^2x}dx$

how to solve integral of arc cot??

2. Originally Posted by transgalactic
$\int \frac{x}{\sin ^2x}dx=-arcotx -\int -arccot x dx$

u=x u'=1
$v'=\frac{1}{\sin ^2x}dx$

how to solve integral of arc cot??
$\int u \, dv = uv - \int v \, du$.

$u = x \Rightarrow du = dx$

$dv = \frac{1}{\sin^2 x} \, dx \Rightarrow v = - \cot x$.

So your integral becomes $-x \cot x + \int \cot x \, dx$, which is simple to find.

3. thanks