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(tanx+cotx)dx  is equal to

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a
2sin−1(sinx−cosx)+c
b
2sin−1(sinx+cosx)+c
c
2tan−1(sinx−cosx)+c
d
2tan−1(sinx+cosx)+c

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detailed solution

Correct option is A

∫(tanx+cotx)dx ∫(sinxcosx+cosxsinx)dx ∫(sinx+cosxsinxcosx)dx ∫2(sinx+cosx)2sinxcosxdx 2∫sinx+cosx1−(sinx−cosx)2dx Put sinx−cosx=t ⇒(cosx+sinx)dx=dt =2∫dt1−t2 =2sin−1t+C =2sin−1(sinx−cosx)+C


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