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 Evaluate ex(1+x)cos2xexdx

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a
tan(xex)+c
b
tan(ex)+c
c
tan(x+ex)+c
d
sec(ex)+c

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detailed solution

Correct option is A

Let z=xex. Then dz=1ex+xexdx=ex1+xdx .  Thus,     ∫ex(1+x)cos2⁡xexdx=∫dzcos2⁡z                                             =∫sec2⁡zdz                                             =tan⁡z+c=tan⁡xex+c


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