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Question

π20sin2x.cosx dx

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Solution

We have,

π20sin2xcosxdx

π202sinxcosxcosxdx

2π20sinxcos2xdx


Let

cosx=t

sinxdx=dt


Therefore,

201t2dt


On integrating and we get,

=2[t33]10

Then,

=2[03133]

=23


Hence, this is the answer.


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