MCQ
The value of $\int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}\,x}}{{1 + {2^x}}}dx} $ is
- A$ \pi $
- B$\frac{\pi}{2}$
- C$4 \pi $
- ✓$\frac{\pi}{4}$
✓
Answer
Correct option: D.
$\frac{\pi}{4}$
d
(d) $I = \int\limits_{\pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} $ $......(1)$
(d) $I = \int\limits_{\pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} $ $......(1)$
$ \Rightarrow I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 - {2^{ - x}}}}dx} $ by replacing $x$ by
$\left(\frac{\pi}{2}-\frac{\pi}{2}-x\right)$
$ \Rightarrow I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{2^x}.{{\sin }^2}x}}{{1 + {2^x}}}dx} $
Adding equations $(i)$ and $(ii),$ we get
$2I = \int\limits_{ - \pi /2}^{\pi /2} {{{\sin }^2}xdx} $
$ = \frac{1}{2}\int\limits_{ - \pi /2}^{\pi /2} {\left( {1 - \cos 2x} \right)dx} $
$\Rightarrow \quad \mathrm{I}=\frac{1}{4}\left[x+\frac{\sin 2 x}{2}\right]_{-\pi / 2}^{\pi / 2}$
$=\frac{1}{4}\left[\left(\frac{\pi}{2}+\frac{\sin \pi}{2}\right)-\left(-\frac{\pi}{2}+\frac{\sin (-\pi)}{2}\right)\right]$
$\Rightarrow \quad 1=\frac{1}{4}\left[\frac{\pi}{2}+\frac{\pi}{2}\right]=\frac{\pi}{4}$
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