int_{0}^{pi} frac{e^{cos x}}{e^{cos x}+e^{-co | vedclass

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(D) Let $I = \int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x \quad ...(1)$ Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x

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$\frac{\pi}{2}$

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(D) Let $I = \int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x \quad ...(1)$ Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we get: $I = \int_{0}^{\pi} \frac{e^{\cos(\pi-x)}}{e^{\cos(\pi-x)}+e^{-\cos(\pi-x)}} d x$ Since $\cos(\pi-x) = -\cos x$,we have: $I = \int_{0}^{\pi} \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}} d x \quad ...(2)$ Adding equations $(1)$ and $(2)$: $2I = \int_{0}^{\pi} \left( \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} + \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}} \right) d x$ $2I = \int_{0}^{\pi} \frac{e^{\cos x} + e^{-\cos x}}{e^{\cos x} + e^{-\cos x}} d x$ $2I = \int_{0}^{\pi} 1 d x$ $2I = [x]_{0}^{\pi} = \pi - 0 = \pi$ $I = \frac{\pi}{2}$

$\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x=$

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