Primitive of Cube of Sine Function/Proof 1

Theorem

$\ds \int \sin^3 x \rd x = \frac {\cos^3 x} 3 - \cos x + C$

Proof

From Primitive of $\sin^3 a x$:

$\ds \int \sin^3 a x \rd x = -\frac {\cos a x} a + \frac {\cos^3 a x} {3 a} + C$

The result follows by setting $a = 1$.

$\blacksquare$