$\int_{0}^{\pi} \frac{e^{\cos x}}{\left(e^{\cos x}+e^{-\cos x)}\right.} d x=$
$\int_{0}^{\pi} \frac{e^{\cos x}}{\left(e^{\cos x}+e^{-\cos x)}\right.} d x=$
- $\frac{-\pi}{2}$
- $-\pi$
- $\pi$
- $\frac{\pi}{2}$
Solution
$\begin{aligned}
\text {Let } I &=\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x} d x}...(1) \\
&=\int_{0}^{\pi} \frac{e^{\cos (\pi-x)}}{e^{\cos (\pi-x)}+e^{-\cos (\pi-x)}} d x \\
&=\int_{0}^{\pi} \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x} d x}...(2)
\end{aligned}$
Adding equation (1) \& (2), we get
$2 I=\int_{0}^{\pi} 1 d x=[x]_{0}^{\pi} \Rightarrow I=\frac{\pi}{2}$
Asked in: MHT CET 2020 (20 Oct Shift 1)
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