Prove that...(integration)

　∫sin x sin 3x dx
= ∫(- 1/2)(cos (x + 3x) - cos (x - 3x)) dx
= ∫(- 1/2)(cos 4x - cos (- 2x)) dx
= ∫((cos 2x)/2 - (cos 4x)/2) dx
= (sin 2x)/4 - (sin 4x)/8 + C
= (sin x cos x)/2 - (sin 2x cos 2x)/4 + C
= (sin x cos x)/2 - (sin x cos x (1 - 2 sin² x))/2 + C
= (sin x cos x)/2 - (sin x cos x)/2 + (2 sin³ x cos x)/2 + C
= sin³ x cos x + C

其實這沒甚麼啦，不過就是product to sum, integrate，再計form而已。

use "cmopound angle" ... product to sum formula:


　integral { sin x sin 3x }

= integral { 0.5 ( cos 2x - cos 4x ) }

= 0.5 { 0.5 sin 2x - 0.25 sin 4x }

= 0.5 { sin x cos x - 0.5 sin 2x cos 2x }

= 0.5 { sin x cos x - sin x cos x ( 1 - 2 sin*x ) }

= sin^x cos x