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Question

Evaluate π0[sin2π2Cos2x2]dx.

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Solution

π0[sin2π2cosx2]dx
=sin2π2dxcos2x2dx
1cos2π/22dx1+cos2π/22dx
=12cosπ2dx12+cosπ2dx
=x2sinπ2x2+sinπ4
=sinx4sinπ4+C

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