$\displaystyle \int \frac{x}{\sin ^2x}dx=-arcotx -\int -arccot x dx$
u=x u'=1
$\displaystyle v'=\frac{1}{\sin ^2x}dx$
how to solve integral of arc cot??
$\displaystyle \int \frac{x}{\sin ^2x}dx=-arcotx -\int -arccot x dx$
u=x u'=1
$\displaystyle v'=\frac{1}{\sin ^2x}dx$
how to solve integral of arc cot??