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# evaluate:sinx. sin2x. sin3x

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:
Hello Student,

You can solve in many ways which would depend on the options you want.
You can convert all in the form of sinx & cosx.
I am doing it by one way from which you can get the idea.







lily
16 Points
7 years ago
thank you very much sir.
Shivam
13 Points
4 years ago
I = ? sin x. sin 2x. sin 3x dxI= (1/2) ? [ 2 sin 2x. sin x ] sin 3x dxI= (1/2) ? [ cos ( 2x-x) - cos ( 2x+x) ] sin 3x dxI= (1/2) ? ( cos x - cos 3x ) sin 3x dxI= (1/4) ? [ ( 2 sin 3x cos x ) - ( 2 sin 3x cos 3x ) ] dxI= (1/4) ? { [ sin ( 3x + x ) + sin ( 3x - x ) ] - ( sin 2·3x ) } dxI= (1/4) ? [ ( sin 4x + sin 2x ) - ( sin 6x ) ] dxI= (/4) { [ ( - cos 4x ) / 4 ] + [ ( - cos 2x ) / 2 ] - [ ( - cos 6x ) / 6 ] } + CI= ( -1/16) cos 4x - (1/8) cos 2x + (1/24) cos 6x + C