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Question

If I=(tanx+cotx)dx, then I equals

A
2sin1(sinx+cosx)+C
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B
2cos1(sinxcosx)+C
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C
2sin1(sinxcosx)+C
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D
2cos1(sinx+cosx)+C
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Solution

The correct option is C 2sin1(sinxcosx)+C
Let I=(tanx+cotx)dx=tanx+1tanxdx
Substituting tanx=t2sec2xdx=2tdt
I=t2+1t2.2tt4+1dt=2t2+1t4+1dt=21+1t2t2+1t22+2dt=21+1t2(t1t)2+(2)2dt
Put t1t=u(1+1t2)dt=du
I=2duu2+(2)2=22tan1(u2)+c=2tan1((tanxcotx)2)+c=2sin1(sinxcosx)+C

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