find integral of (sinx sin2x sin3x)?

Asked by Aparna | 16th Mar, 2013, 11:58: AM

Expert Answer:

I = ? sin x. sin 2x. sin 3x dx

I= (1/2) ? [ 2 sin 2x. sin x ] sin 3x dx

I= (1/2) ? [ cos ( 2x-x) - cos ( 2x+x) ] sin 3x dx

I= (1/2) ? ( cos x - cos 3x ) sin 3x dx

I= (1/4) ? [ ( 2 sin 3x cos x ) - ( 2 sin 3x cos 3x ) ] dx

I= (1/4) ? { [ sin ( 3x + x ) + sin ( 3x - x ) ] - ( sin 2·3x ) } dx

I= (1/4) ? [ ( sin 4x + sin 2x ) - ( sin 6x ) ] dx

I= (/4) { [ ( - cos 4x ) / 4 ] + [ ( - cos 2x ) / 2 ] - [ ( - cos 6x ) / 6 ] } + C

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

Answered by  | 16th Mar, 2013, 07:52: PM

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