# Thread: How do you solve this integral?

1. ## How do you solve this integral?

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

thanks

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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