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 If I=(cotxtanx)dx, then I equals 

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a
2log⁡(tan⁡x−cot⁡x)+C
b
2log⁡|sin⁡x+cos⁡x+sin⁡2x|+C
c
2log⁡|sin⁡x−cos⁡x+2sin⁡xcos⁡x|+C
d
2log⁡|sin⁡(x+π/4)+2sin⁡xcos⁡x|+C

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detailed solution

Correct option is B

I=∫cos⁡x−sin⁡xcos⁡xsin⁡xdx Put sin⁡x+cos⁡x=t, so that 2sin⁡xcos⁡x=t2−1∴I=2∫dtt2−1=2log⁡t+t2−1+C=2log⁡|sin⁡x+cos⁡x+sin⁡2x|+C


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(tanx+cotx)dx is equal to


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