Primitive of Sine of a x by Sine of b x

Theorem

For $p \ne q$:

$\ds \int \sin a x \sin b x \rd x = \frac {\map \sin {a - b} x} {2 \paren {a - b} } - \frac {\map \sin {a + b} x} {2 \paren {a + b} } + C$

$\ds \int \sin a x \sin b x \rd x$

$=$

$\ds \int \paren {\frac {\map \cos {a x - b x} - \map \cos {a x + b x} } 2} \rd x$

$\ds $

$=$

$\ds \frac 1 2 \int \map \cos {a - b} x \rd x - \frac 1 2 \int \map \cos {a + b} x \rd x$

$\ds $

$=$

$\ds \frac 1 2 \frac {\map \sin {a - b} x} {a - b} - \frac 1 2 \frac {\map \sin {a + b} x} {a + b} + C$

$\ds $

$=$

$\ds \frac {\map \sin {a - b} x} {2 \paren {a - b} } - \frac {\map \sin {a + b} x} {2 \paren {a + b} } + C$

simplifying

$\blacksquare$