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$\int_{\pi / 4}^{\pi / 2} \frac{3 d x}{1+e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}\right)}}=$
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The correct answer is:
$\frac{3}{8} \pi$

$\begin{aligned} & =\int_{\pi / 4}^{\pi / 2} \frac{3 d x}{1+e^{\sqrt{8} \sin \left(\frac{\pi}{2}+\frac{\pi}{4}-x-\frac{3 \pi}{8}\right)}} \\ & {\left[\int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right]} \\ & =\int_{\pi / 4}^{\pi / 2} \frac{3 d x}{1+e^{\sqrt{8} \sin \left(\frac{3 \pi}{8}-x\right)}}=\int_{\pi / 4}^{\pi / 2} \frac{3 d x}{1+e^{-\sqrt{8} \sin \left(x-\frac{3 \pi}{8}\right)}}\end{aligned}$

On adding Eq. (i) and Eq. (ii), we get
$\begin{aligned}
& 2 I=3 \int_{\pi / 4}^{\pi / 2} 1 d x=3[x]_{\pi / 4}^{\pi / 2}=3\left[\frac{\pi}{2}-\frac{\pi}{4}\right]=\frac{3 \pi}{4} \\
\therefore \quad I & =\frac{3 \pi}{8}
\end{aligned}$
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