The value of int limits_{0}^{pi} frac{e^{cos | vedclass

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(C) Let $I = \int \limits_{0}^{\pi} \frac{e^{\cos x} \sin x}{(1+\cos^2 x)(e^{\cos x}+e^{-\cos x})} dx$  $(1)$Using the property $\int_{0}^{a} f(x) dx

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$\frac{\pi}{4}$

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(C) Let $I = \int \limits_{0}^{\pi} \frac{e^{\cos x} \sin x}{(1+\cos^2 x)(e^{\cos x}+e^{-\cos x})} dx$  $(1)$Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:$I = \int \limits_{0}^{\pi} \frac{e^{\cos(\pi-x)} \sin(\pi-x)}{(1+\cos^2(\pi-x))(e^{\cos(\pi-x)}+e^{-\cos(\pi-x)})} dx$$I = \int \limits_{0}^{\pi} \frac{e^{-\cos x} \sin x}{(1+\cos^2 x)(e^{-\cos x}+e^{\cos x})} dx$  $(2)$Adding $(1)$ and $(2)$:$2I = \int \limits_{0}^{\pi} \frac{(e^{\cos x} + e^{-\cos x}) \sin x}{(1+\cos^2 x)(e^{\cos x}+e^{-\cos x})} dx$$2I = \int \limits_{0}^{\pi} \frac{\sin x}{1+\cos^2 x} dx$Since $\frac{\sin x}{1+\cos^2 x}$ is symmetric about $x = \frac{\pi}{2}$,$2I = 2 \int \limits_{0}^{\pi/2} \frac{\sin x}{1+\cos^2 x} dx$$I = \int \limits_{0}^{\pi/2} \frac{\sin x}{1+\cos^2 x} dx$Let $\cos x = t$,then $-\sin x dx = dt$. When $x=0, t=1$; when $x=\pi/2, t=0$.$I = -\int \limits_{1}^{0} \frac{dt}{1+t^2} = \int \limits_{0}^{1} \frac{dt}{1+t^2} = [	an^{-1} t]_{0}^{1} = 	an^{-1}(1) - 	an^{-1}(0) = \frac{\pi}{4}$

The value of $\int \limits_{0}^{\pi} \frac{e^{\cos x} \sin x}{\left(1+\cos ^{2} x\right)\left(e^{\cos x}+e^{-\cos x}\right)} d x$ is equal to

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