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Q.

Evaluate ∫ex(1+x)cos2⁡xexdx

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a

tan(xex)+c

b

tan(ex)+c

c

tan(x+ex)+c

d

sec(ex)+c

answer is A.

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Detailed Solution

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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