Primitive of Reciprocal of Cube of Sine of a x

Theorem

 * $\displaystyle \int \frac {\mathrm d x} {\sin^3 a x} = -\frac {\cos a x} {2 a \sin^2 a x} + \frac 1 {2 a} \ln \left\vert{\tan \frac {a x} 2}\right\vert + C$

Proof
With a view to expressing the primitive in the form:
 * $\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$

let:

and let:

Then: