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(tanx+cotx)dx is equal to

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a
2sin−1⁡(sin⁡x−cos⁡x)+C
b
2sin−1⁡(sin⁡x+cos⁡x)+C
c
2tan−1⁡(sin⁡x−cos⁡x)+C
d
None of the above

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detailed solution

Correct option is A

∫(tan⁡x+cot⁡x)dx=∫sin⁡xcos⁡x+cos⁡xsin⁡xdx=∫sin⁡x+cos⁡xsin⁡xcos⁡xdx=∫2(sin⁡x+cos⁡x)2sin⁡xcos⁡xdx=2∫sin⁡x+cos⁡x1−(sin⁡x−cos⁡x)2dxPut ⁡sin⁡x−cos⁡x=t⇒(cos⁡x+sin⁡x)dx=dt=2∫dt1−t2=2sin−1⁡t+C=2sin−1⁡(sin⁡x−cos⁡x)+C


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