Primitive of Sine of a x over x squared

Theorem

 * $\ds \int \frac {\sin a x \rd x} {x^2} = -\frac {\sin a x} x + a \int \frac {\cos a x \rd x} x$

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

let:

and let:

Then:

Also see

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


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