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ln(tanx)sinxcosxdx, is equal to 

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a
12ln⁡(tan⁡x)+C
b
12ln⁡tan2⁡x+C
c
12(ln⁡(tan⁡x))2+C
d
None of these

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detailed solution

Correct option is C

I=∫ln⁡(tan⁡x)sin⁡x⋅cos⁡xdx, let t=ln⁡(tan⁡x)⇒ dtdx=sec2⁡xtan⁡x⇒dt=dxsin⁡x⋅cos⁡x∴ I=∫tdt=12t2+C=12(ln⁡(tan⁡x))2+C


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