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Question

Solve: dxcosx1+cos2x+sin2x;(0<x<π4)

A
2+cotx
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B
tanx+1
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C
2+2tanx
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D
2+2cotx
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Solution

The correct option is D 2+2tanx
Consider, I=dxcosx1+cos2x+sin2x

I=dxcosxsin2x+cos2x+cos2xsin2x+2sinxcosx

I=dxcosx2cos2x+2sinxcosx

I=sec2xdx2+2tanx
Put tanx=t
Then sec2xdx=dt
Therefore, I=dt2+2t

Thus 2+2tanx+C

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