Primitive of Sine of a x over x squared

Theorem

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

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:

Also see

 * Primitive of $\dfrac {\cos a x} x$
 * Primitive of $\dfrac {\cos a x} {x^2}$