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dxsinx+cosx is equal to

 

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a
log⁡tan⁡(π/4+x/8)+C
b
2log⁡tan⁡(x/2+π/8)+C
c
12log⁡tan⁡x2+π8+C
d
2log⁡tan⁡x4−π8+C

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detailed solution

Correct option is C

∫dxsin⁡x+cos⁡x=12∫dxsin⁡(π/4)sin⁡x+cos⁡(π/4)cos⁡x=12∫dxcos⁡(x−π/4)=12∫sec⁡(x−π/4)dx=12log⁡tan⁡π4+x2−π8+C=12log⁡tan⁡π8+x2+C


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