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Question

Evaluate: sinxcosx5/2dx

A
23tan3/2x+c
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B
23tan2/3x+c
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C
32tan3/2x+c
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D
32tan2/3x+c
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Solution

The correct option is D 23tan3/2x+c
Let I=sinxcos5/2xdx=tanxsec2xdx
Put tanx=tsec2xdx=dt
I=tdt=2t3/23=23tan3/2x

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