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Question

Evaluate: cos2xcos2αcosxcosαdx.

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Solution

I=(cos2xcos2αcosxcosα)dx
I=⎜ ⎜ ⎜2sin2x+2α2sin2x2α22sinx+α2sinxα2⎟ ⎟ ⎟dx
I=sin(x+α)sin(xα)sinx+α2sinxα2dx
I=2cosx+α2×2cosxα2dx
cos(A+B)cos(AB)=cos2Asin2B
I=(4cos2x2sin2a2)dx
I=4cos2x24sin2a2dx
I=2(1+cosx)dx4xsin2a2
I=2x+2sinx4sin2a2+c

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