Question:
The value of $\int_{-\pi / 2}^{\pi / 2} \frac{1}{1+e^{\sin x}} d x$ is
Correct Option: , 4
Solution:
$I=\int_{-\pi / 2}^{\pi / 2} \frac{1}{1+e^{\sin x}} d x$ $\ldots(1)$
Apply King property
$I=\int_{-z / 2}^{\pi / 2} \frac{1}{1+e^{-\sin x}} d x=\int_{-\pi / 2}^{\pi / 2} \frac{e^{\sin x}}{1+e^{\sin x} x} d x \ldots(2)$
Add (1) & (2)
$2 \mathbf{I}=\int_{-\pi / 2}^{\pi / 2} \mathrm{~d} \mathrm{x}=\pi$
$I=\frac{\pi}{2}$
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