Q.

∫−π/2π/2cos x1+ex dx=

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a

0

b

–1

c

1

d

2π

answer is C.

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Detailed Solution

I=∫−π/2π/2Cos x1+exdx     =∫−π/2π/2Cos x1+e−xdx     since ∫-aafxdx= ∫-aaf-xdx     =∫−π/2π/2ex.Cos x1+exdx 2I=∫−π/2π/2cosx+ex.Cos x1+exdx=∫−π/2π/2cosx dx =22I=2
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