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Question

Solve : π20sin2xdx

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Solution


Let I=π20sin2xdx .................. (1)

By applying a0f(x)dx=a0f(ax)dx

I=π20sin2(π2x)dx

I=π20cos2(x)dx .............. (2)

Add (1) and (2) ,

2I=π20(sin2x+cos2x)dx

2I=π201dx

2I=[x]π20

2I=π2

I=π4

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