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$\frac{\pi^2}{2}$
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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