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Question: \(\int _ { - 1 / 2 } ^ { 1 / 2 } ( \cos x ) \left[ \log \left( \frac { 1 - x } { 1 + x } \right) \ri...

∫−1/21/2(cos⁡x)[log⁡(1−x1+x)]dx=\int _ { - 1 / 2 } ^ { 1 / 2 } ( \cos x ) \left[ \log \left( \frac { 1 - x } { 1 + x } \right) \right] d x =

A

0

B

1

C

e1/2e ^ { 1 / 2 }

D

2e1/22 e ^ { 1 / 2 }

Answer

0

Explanation

Solution

I=∫−1/21/2(cos⁡x)[log⁡(1−x1+x)]dxI = \int _ { - 1 / 2 } ^ { 1 / 2 } ( \cos x ) \left[ \log \left( \frac { 1 - x } { 1 + x } \right) \right] d x …..(i)

∴ I=∫−1/21/2cos⁡(−x)[log⁡(1+x1−x)]dxI = \int _ { - 1 / 2 } ^ { 1 / 2 } \cos ( - x ) \left[ \log \left( \frac { 1 + x } { 1 - x } \right) \right] d x

⇒ I=−∫−1/21/2cos⁡x[log⁡(1−x1+x)]dxI = - \int _ { - 1 / 2 } ^ { 1 / 2 } \cos x \left[ \log \left( \frac { 1 - x } { 1 + x } \right) \right] d x …..(ii)

Adding (i) and (ii), we get

2I=∫−1/21/2cos⁡x[log⁡(1−x1+x)]dx−∫−1/21/2cos⁡x[log⁡(1−x1+x)]dx2 I = \int _ { - 1 / 2 } ^ { 1 / 2 } \cos x \left[ \log \left( \frac { 1 - x } { 1 + x } \right) \right] d x - \int _ { - 1 / 2 } ^ { 1 / 2 } \cos x \left[ \log \left( \frac { 1 - x } { 1 + x } \right) \right] d x

or 2I=02 I = 0or I = 0.