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

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Question

(D) माना $I = \int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x \quad ...(1)$ गुणधर्म $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$ का

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$\frac{\pi}{2}$

Vedclass · Answer

(D) माना $I = \int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x \quad ...(1)$ गुणधर्म $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$ का उपयोग करने पर,हमें प्राप्त होता है: $I = \int_{0}^{\pi} \frac{e^{\cos(\pi-x)}}{e^{\cos(\pi-x)}+e^{-\cos(\pi-x)}} d x$ चूँकि $\cos(\pi-x) = -\cos x$,इसलिए: $I = \int_{0}^{\pi} \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}} d x \quad ...(2)$ समीकरण $(1)$ और $(2)$ को जोड़ने पर: $2I = \int_{0}^{\pi} \left( \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} + \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}} \right) d x$ $2I = \int_{0}^{\pi} \frac{e^{\cos x} + e^{-\cos x}}{e^{\cos x} + e^{-\cos x}} d x$ $2I = \int_{0}^{\pi} 1 d x$ $2I = [x]_{0}^{\pi} = \pi - 0 = \pi$ $I = \frac{\pi}{2}$

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

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