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Question

Solve the integral I = π0sin2xdx


A

π

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B

π2

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C

3π2

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D

0

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Solution

The correct option is B

π2


I=sin2xdx=π01cos 2x2dxsin2x=1cos 2x2

=12[π0dxπ0(cos 2x)dx]=12[xsin 2x2]π0

=12[(π0)(sin 2πsin 0)2]π0sin2xdx=π2


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