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Question

(1+2sinxcos2x)dx.

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Solution

we are given, I=(1+2sinxcos2x)dx

I=(sec2x+2tanx.secx)dx

asweknow,1cosθ=secθ & sinθcosθ=tanθ

I=sec2xdx+2tanx.secxdx

I=tanx+2secx+c

so,(1+2sinxcos2x)dx=tanx+2secx+c

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