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ex1+sinx1+cosxdx is equal to 

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a
exsec⁡x2+C
b
ex1+cos⁡x+C
c
extan⁡x2+C
d
None of these

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detailed solution

Correct option is C

∫ex11+cos⁡x+sin⁡x1+cos⁡xdx =∫ex12cos2⁡x2+2sin⁡x2cos⁡x22cos2⁡x2dx ∵cos⁡2x=2cos2⁡x−1 and sin⁡2x=2sin⁡xcos⁡x =∫exsec2⁡x22+tan⁡x2dx =∫extan⁡x2+12sec2⁡x2dx Let    f(x)=tan⁡x2⇒f′(x)=sec2⁡x22 ∴∫extan⁡x2+sec2⁡x22dx=extan⁡x2+C                                                           ∵∫exf(x)+f′(x)dx=exf(x)


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