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Question: \(\int _ { - \pi / 2 } ^ { \pi / 2 } \frac { \cos x } { 1 + e ^ { x } } d x =\)...

∫−π/2π/2cos⁡x1+exdx=\int _ { - \pi / 2 } ^ { \pi / 2 } \frac { \cos x } { 1 + e ^ { x } } d x =

A

1

B

0

C

−1- 1

D

None of these

Answer

1

Explanation

Solution

I=∫−π/2π/2cos⁡x1+exdx=∫−π/20cos⁡x1+exdx+∫0π/2cos⁡x1+exdxI = \int _ { - \pi / 2 } ^ { \pi / 2 } \frac { \cos x } { 1 + e ^ { x } } d x = \int _ { - \pi / 2 } ^ { 0 } \frac { \cos x } { 1 + e ^ { x } } d x + \int _ { 0 } ^ { \pi / 2 } \frac { \cos x } { 1 + e ^ { x } } d x…..(i)

Putting x=−tx = - t in ∫−π/20cos⁡x1+exdx\int _ { - \pi / 2 } ^ { 0 } \frac { \cos x } { 1 + e ^ { x } } d x, we get

I=∫−π/20cos⁡x1+exdx=∫0π/2excos⁡x1+exdxI = \int _ { - \pi / 2 } ^ { 0 } \frac { \cos x } { 1 + e ^ { x } } d x = \int _ { 0 } ^ { \pi / 2 } \frac { e ^ { x } \cos x } { 1 + e ^ { x } } d x

I=∫0π/2excos⁡x1+exdx+∫0π/2cos⁡x1+exdxI = \int _ { 0 } ^ { \pi / 2 } \frac { e ^ { x } \cos x } { 1 + e ^ { x } } d x + \int _ { 0 } ^ { \pi / 2 } \frac { \cos x } { 1 + e ^ { x } } d x

=∫0π/2(1+ex)cos⁡xdx(1+ex)= \int _ { 0 } ^ { \pi / 2 } \frac { \left( 1 + e ^ { x } \right) \cos x d x } { \left( 1 + e ^ { x } \right) } =∫0π/2cos⁡xdx=[sin⁡x]0π/2=1= \int _ { 0 } ^ { \pi / 2 } \cos x d x = [ \sin x ] _ { 0 } ^ { \pi / 2 } = 1 .