While I do some investigation of the complicated induction work on no. 335 ...

No. 352 in my list (there were a load of easy sine formulas):

$\displaystyle \int \frac {\mathrm d x} {\sin^3 a x}$

which is supposed to work out to:

$\displaystyle -\frac {\cos a x} {2 a \sin^2 a x} + \frac 1 {2 a} \ln \left\vert{\tan \frac {a x} 2}\right\vert$

but is going a bit astray.

I tried Parts, with $\displaystyle u = \frac 1 {\sin a x}$ and got so far:

$\displaystyle -\frac {\cos a x} {a \sin^2 a x} - \int \frac {\cos^2 a x} {\sin^3 a x} \ \mathrm d x$

Hints?