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Question

Solve the integral I=π0sin2xdx.

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Solution

I=π0sin2xdx
π0sin2xdx=π0(1cos2x2)dx(sin2x=1cos2x2)
=12[π0dxπ0(cos2x)dx]
=12[xsin2x2]π0
=12[(π0)(sin2πsin0)2]=π2.

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