# Math Help - How do you solve this integral?

1. ## How do you solve this integral?

$\int{x*cotx^2 dx}$

thanks

2. Originally Posted by mojo0716
$\int{x*cotx^2 dx}$

thanks
$\int{x*cotx^2 dx}=\int{\frac{x \cos{x^2}}{\sin{x^2}}\,dx}$

Let $u=\sin{x^2}$ which gives $\frac{du}{dx}=2x\cos{x^2}$.

So $\int{\frac{x \cos{x^2}}{\sin{x^2}}\,dx}=\frac{1}{2}\int{\frac{2 x \cos{x^2}}{\sin{x^2}}\,dx}$

$=\frac{1}{2}\int{\frac{1}{u}\frac{du}{dx}\,dx}$

$=\frac{1}{2}\int{\frac{1}{u}\,du}$

$=\frac{1}{2}[\ln|u|+c]$

$=\frac{1}{2}\ln|\sin{x^2}|+C$

3. Originally Posted by Prove It
$\int{x*cotx^2 dx}=\int{\frac{x \cos{x^2}}{\sin{x^2}}\,dx}$

Let $u=\sin{x^2}$ which gives $\frac{du}{dx}=2x\cos{x^2}$.

So $\int{\frac{x \cos{x^2}}{\sin{x^2}}\,dx}=\frac{1}{2}\int{\frac{2 x \cos{x^2}}{\sin{x^2}}\,dx}$

$=\frac{1}{2}\int{\frac{1}{u}\frac{du}{dx}\,dx}$

$=\frac{1}{2}\int{\frac{1}{u}\,du}$

$=\frac{1}{2}[\ln|u|+c]$

$=\frac{1}{2}\ln|\sin{x^2}|+C$
ah thanks!