The value of the integral _ - / 2 ^ / 2 ^2 x 1+e^x ,d x is - Vidyadip

MCQ

The value of the integral $\int \limits_{-\pi / 2}^{\pi / 2} \frac{\sin ^2 x}{1+e^x}\,d x$ is

A
$\frac{\pi}{6}$
✓
$\frac{\pi}{4}$
C
$\frac{\pi}{2}$
D
$\frac{\pi^2}{2}$

✓

Answer

Correct option: B.

$\frac{\pi}{4}$

b
(b)

We have, $\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^2 x}{1+e^x} d x$

$+\int_0^{\pi / 2}\left(\frac{\sin ^2 x}{1+e^x}+\frac{\sin ^2(-x)}{1+e^{-x}}\right) d x$

$=\int_0^{\pi / 2}\left(\frac{\sin ^2 x}{1+e^x}+\frac{e^x \sin ^2 x}{1+e^x}\right) d x$

$=\int_0^{\pi / 2} \sin ^2 x d x=\int_0^{\pi / 2}\left(\frac{1-\frac{\cos 2 x}{2}}{2}\right) d x$

$=\frac{1}{2}\left(x-\frac{\sin 2 x}{2}\right)_0^{\pi / 2}=\frac{\pi}{4}$

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