Integrate the integrals: `intsinxcos2xsin3xdx`

Explanation

`int sinx cos 2x sin 3x {dx} `
`=frac{1}{2} int(2 sin x cos 2 x) sin 3 x d x`
`
=frac{1}{2} int[sin (x+2 x)+sin (x-2 x)] sin (3 x) d x `
`
=frac{1}{2} int[sin (3 x)-sin (x)] sin (3 x) d x `
`
=frac{1}{2}[int sin ^{2}(3 x) d x-int sin (x) sin (3 x) d x] `
`
=frac{1}{4}[int 2 sin ^{2}(3 x) d x-int 2 sin (x) sin (3 x) d x] `
`
=frac{1}{4}{int[1-cos (6 x)] d x-int[cos (x-3 x)-cos (x+3 x)] d x}`
`
=frac{1}{4}[int 1 d x-int cos (6 x) d x-int cos (2 x) d x+int cos (4 x) d x] `
`
=frac{1}{4}[x-frac{sin (6 x)}{6}-frac{sin (2 x)}{2}+frac{sin (4 x)}{4}]+c `
`
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