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$\int_{0}^{\pi} \frac{e^{\cos x}}{\left(e^{\cos x}+e^{-\cos x)}\right.} d x=$
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2275 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi}{2}$
$$
\text { Let } \begin{aligned}
I &=\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x} d x}...(1) \\
&=\int_{0}^{\pi} \frac{e^{\cos (\pi-x)}}{e^{\cos (\pi-x)}+e^{-\cos (\pi-x)}} d x \\
&=\int_{0}^{\pi} \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x} d x}...(2)
\end{aligned}
$$
Adding equation (1) \& (2), we get
$$
2 I=\int_{0}^{\pi} 1 d x=[x]_{0}^{\pi} \Rightarrow I=\frac{\pi}{2}
$$
\text { Let } \begin{aligned}
I &=\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x} d x}...(1) \\
&=\int_{0}^{\pi} \frac{e^{\cos (\pi-x)}}{e^{\cos (\pi-x)}+e^{-\cos (\pi-x)}} d x \\
&=\int_{0}^{\pi} \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x} d x}...(2)
\end{aligned}
$$
Adding equation (1) \& (2), we get
$$
2 I=\int_{0}^{\pi} 1 d x=[x]_{0}^{\pi} \Rightarrow I=\frac{\pi}{2}
$$
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