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If $a>0$, then $\int_{-\pi}^\pi \frac{\sin ^2 x}{1+a^x} d x$ is equal to
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Verified Answer
The correct answer is:
$\frac{\pi}{2}$
Let $l=\int_{-\pi}^\pi \frac{\sin ^2 x}{1+a^x} d x$
Put $x=-x$, we get
$\begin{aligned} & I=-\int_\pi^{-\pi} \frac{\sin ^2(-x)}{1+a^{-x}} d x \\ \Rightarrow & I=\int_{-\pi}^\pi a^x \frac{\sin ^2 x}{1+a^x} d x\end{aligned}$
On adding Fqs. (i) and (ii), we get
$\begin{aligned} 2 I & =\int_{-\pi}^\pi \frac{\left(1+a^x\right) \sin ^2 x}{\left(1+a^x\right)} d x \\ & =\int_{-\pi}^\pi \sin ^2 x d x \\ & =2 \int_0^\pi \sin ^2 x d x=\int_0^\pi(1-\cos 2 x) d x \\ & =\left[x+\frac{\sin 2 x}{2}\right]_0^\pi \\ & =\left(\pi+\frac{\sin 2 \pi}{2}-0-0\right) \\ \Rightarrow \quad 2 I & =\pi \\ \Rightarrow \quad I & =\frac{\pi}{2}\end{aligned}$
Put $x=-x$, we get
$\begin{aligned} & I=-\int_\pi^{-\pi} \frac{\sin ^2(-x)}{1+a^{-x}} d x \\ \Rightarrow & I=\int_{-\pi}^\pi a^x \frac{\sin ^2 x}{1+a^x} d x\end{aligned}$
On adding Fqs. (i) and (ii), we get
$\begin{aligned} 2 I & =\int_{-\pi}^\pi \frac{\left(1+a^x\right) \sin ^2 x}{\left(1+a^x\right)} d x \\ & =\int_{-\pi}^\pi \sin ^2 x d x \\ & =2 \int_0^\pi \sin ^2 x d x=\int_0^\pi(1-\cos 2 x) d x \\ & =\left[x+\frac{\sin 2 x}{2}\right]_0^\pi \\ & =\left(\pi+\frac{\sin 2 \pi}{2}-0-0\right) \\ \Rightarrow \quad 2 I & =\pi \\ \Rightarrow \quad I & =\frac{\pi}{2}\end{aligned}$
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