Integrate $\int\frac{Cx}{(\sin x^2)^2}\,\mathrm dx.$ Have been a doing a reduction of order ODE problem and this integral comes up at the last step. Not sure how to go about integrating it. The answers give $\cos x^2$ as the answer. Here's the original question:
Verify that $u_1=\sin x^2$ is a solution to the equation $$xu''-u'+4x^3u=0$$ and use reduction of order to find a second, linearly independent solution.
I've called the second solution $v$ and as far as I can tell, everything is good with my previous working. The only remaining bit is to integrate $$v'=\frac{Cx}{(\sin x^2)^2}\Leftrightarrow v=\int\frac{Cx}{(\sin x^2)^2}\,\mathrm dx.$$
Integration by parts didn't really help. I think there might be a substitution that I'm missing/forgetting.
Thanks.

In view of the $x$ sitting on top, the substitution $u=x^2$ is natural. So let $u=x^2$. Then $x\,dx=\frac{du}{2}$.
We end up with
$$\int \frac{C}{2\sin^2 u}\,du.$$
One way to continue is to rewrite as 
$$\int \frac{C}{2}\csc^2 u\,du.$$
This may not be quite familiar, though integrating its close relative $\sec^2 u$ is familiar. One can verify that our integral is $-\frac{C}{2}\cot u+D$. 
