The value of

Question:

The value of $\int_{-\pi / 2}^{\pi / 2} \frac{1}{1+e^{\sin x}} d x$ is:

  1. (1) $\frac{\pi}{4}$

  2. (2) $\pi$

  3. (3) $\frac{\pi}{2}$

  4. (4) $\frac{3 \pi}{2}$


Correct Option: , 3

Solution:

$I=\int_{-\pi / 2}^{\pi / 2} \frac{1}{1+e^{\sin x}} d x$

$=\int_{-\pi / 2}^{0} \frac{1}{1+e^{\sin x}} d x+\int_{0}^{\pi / 2} \frac{1}{1+e^{\sin x}} d x$

$=\int_{0}^{\pi / 2}\left(\frac{1}{1+e^{\sin x}}+\frac{1}{1+e^{-\sin x}}\right) d x$

$=\int_{0}^{\pi / 2} \frac{1+e^{\sin x}}{1+e^{\sin x}} d x=\frac{\pi}{2}$

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