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Q.

If ∫sin2⁡x1+sin2⁡xdx=x−Ktan−1⁡(Mtan⁡x)+C then

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a

M=12

b

K=12

c

M=−12

d

K=−12

answer is B.

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Detailed Solution

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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