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11+sinx+cosxdx

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a
12log⁡tan⁡x2+1+C
b
log⁡tan⁡(x+1)+C
c
log⁡tan⁡x2+1+C
d
None of these

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detailed solution

Correct option is C

Putting sin⁡x=2tan⁡x/21+tan2⁡x/2 and cos⁡x=1−tan2⁡x/21+tan2⁡x/2 ,  I=∫11+sin⁡x+cos⁡xdx=∫11+2tan⁡x/21+tan2⁡x/2+1−tan2⁡x/21+tan2⁡x/2dx=∫1+tan2⁡x/21+tan2⁡x/2+2tan⁡x/2+1−tan2⁡x/2dx=∫sec2⁡x/22+2tan⁡x/2dx Putting tan⁡x2=t and 12sec2⁡x2dx=dt or sec2⁡x2dx=2dt, we get I=∫2dt2+2t=∫1t+1dt=log⁡|t+1|+C=log⁡tan⁡x2+1+C


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