The value of $\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^2 x}{1+2^x} d x$ is :
The value of $\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^2 x}{1+2^x} d x$ is :
-
$\pi$
-
$\frac{\pi}{2}$
-
$4 \pi$
-
$\frac{\pi}{4}$
Solution
$\mathrm{I}=\int_{\pi / 2}^{\pi / 2} \frac{\sin ^2 x}{1+2^x} d x$
$
\Rightarrow \mathrm{I}=\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^2 x}{1+2^{-x}} d x \text {, by replacing } x \text { by }
$
$
\begin{aligned}
& \left(\frac{\pi}{2}-\frac{\pi}{2}-x\right) \\
\Rightarrow \quad \mathrm{I} & =\int_{-\pi / 2}^{\pi / 2} \frac{2^x \cdot \sin ^2 x}{1+2^x} d x
\end{aligned}
$
Adding equations (i) and (ii), we get
$
\begin{array}{rl}
2 & I=\int_{-\pi / 2}^{\pi / 2} \sin ^2 x d x=\frac{1}{2} \int_{-\pi / 2}^{\pi / 2}(1-\cos 2 x) d x \\
\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 \mathrm{I} & =\frac{1}{4}\left[\frac{\pi}{2}+\frac{\pi}{2}\right]=\frac{\pi}{4}
\end{array}
$
Asked in: JEE Main 2013 (23 Apr Online)
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