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Question: \(\int _ { - \pi / 2 } ^ { \pi / 2 } \log \left( \frac { 2 - \sin \theta } { 2 + \sin \theta } \righ...

∫−π/2π/2log⁡(2−sin⁡θ2+sin⁡θ)dθ=\int _ { - \pi / 2 } ^ { \pi / 2 } \log \left( \frac { 2 - \sin \theta } { 2 + \sin \theta } \right) d \theta =

A

0

B

1

C

2

D

None of these

Answer

0

Explanation

Solution

Since f(−θ)=log⁡(2−sin⁡θ2+sin⁡θ)−1=−log⁡(2−sin⁡θ2+sin⁡θ)=−f(θ)f ( - \theta ) = \log \left( \frac { 2 - \sin \theta } { 2 + \sin \theta } \right) ^ { - 1 } = - \log \left( \frac { 2 - \sin \theta } { 2 + \sin \theta } \right) = - f ( \theta )

xx.

Therefore, 2∫0π/2log⁡(2−sin⁡θ2+sin⁡θ)dθ=02 \int _ { 0 } ^ { \pi / 2 } \log \left( \frac { 2 - \sin \theta } { 2 + \sin \theta } \right) d \theta = 0 .