# Q. `int` `sinx sin2x sin3x dx` `=?`

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`I = int sin x* sin 2x* sin 3x dx`

`= (1/2) int [ 2 sin 2x* sin x ] sin 3x dx`

`=(1/2) int [ cos ( 2x-x) - cos ( 2x+x) ] sin 3x dx`

`= (1/2) int ( cos x - cos 3x ) sin 3x dx`

`=(1/4) int [ ( 2 sin 3x cos x ) - ( 2 sin 3x cos 3x ) ] dx`

`=(1/4) int [{ sin ( 3x + x ) + sin ( 3x - x ) } - sin (2*3x ) ] dx`

`=(1/4)int∫ [ ( sin 4x + sin 2x ) - ( sin 6x ) ] dx`

`= (1/4)[( - cos 4x ) / 4] +[( - cos 2x ) / 2] - [( - cos 6x ) / 6 ] + C`

`=( -1/16) cos 4x - (1/8) cos 2x + (1/24) cos 6x + C .`

=>answer.

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