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Question

sin(xα)sin(x+α)dx.

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Solution

We are given, I=sin(xα)sin(x+a)dx
I=sin(x+α2α)sin(x+α)dx
[ Now we know that sin(AB)=sinAcosBcosAsinB]
I=[sin(x+α)cos(2α)cos(xα)sin(2α)]sin(x+α)dx
I=cos2αdxintncot(x+α).sin2αdx
I=cos2α.xln|sin(x+α)|.sin2α
I=x.cos2αsin2α.ln|sin(x+α)|+C

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