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cosxsinx1+sin2xdx is equal to

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a
sin⁡x+cos⁡x+C
b
−(sin⁡x+cos⁡x)+C
c
−1(sin⁡x+cos⁡x)+C
d
None of the above

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detailed solution

Correct option is C

Let⁡I=∫cos⁡x−sin⁡x1+sin⁡2xdx=∫cos⁡x−sin⁡xsin2⁡x+cos2⁡x+2sin⁡xcos⁡xdx ∵sin2⁡x+cos2⁡x=1;sin⁡2x=2sin⁡xcos⁡x=∫cos⁡x−sin⁡x(sin⁡x+cos⁡x)2dxLet cos⁡x+sin⁡x=t⇒−sin⁡x+cos⁡x=dtdx⇒ dx=dt(cos⁡x−sin⁡x)∴ I=∫cos⁡x−sin⁡xt2⋅dt(cos⁡x−sin⁡x)=∫1t2dt=∫t−2dt=t−2+1−2+1+C=−1cos⁡x+sin⁡x+C


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