MCQ
The value of $\int \limits_{-\pi / 2}^{\pi / 2} \frac{1}{1+ e ^{\sin x}} d x$
  • A
    $\pi$
  • B
    $\frac{3 \pi}{2}$
  • C
    $\frac{\pi}{4}$
  • $\frac{\pi}{2}$

Answer

Correct option: D.
$\frac{\pi}{2}$
d
$I=\int_{-\pi / 2}^{\pi / 2} \frac{1}{1+e^{\sin x}} d x$

Apply King property

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

$\begin{array}{l}2 I =\int_{-\pi / 2}^{\pi / 2} d x =\pi \\I =\frac{\pi}{2}\end{array}$

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