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If sin2x1+sin2xdx=xKtan1(Mtanx)+C then 

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a
M=12
b
K=12
c
M=−12
d
K=−12

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detailed solution

Correct option is B

sin2⁡x1+sin2⁡x=1−11+sin2⁡x=1−sec2⁡x1+2tan2⁡xSo ∫sin2⁡x1+sin2⁡x=x−12tan−1⁡(2tan⁡x)+C.


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