The value of ${\int }_0^{\pi }\frac{{\mathrm{e}}^{\cos x}\sin x}{\left(1+{\cos }^{2}x\right)\left({\mathrm{e}}^{\cos x}+{\mathrm{e}}^{-\cos x}\right)}\mathrm{d}x$ is equal to

Given $I=\underset{0}{\overset{\pi }{\int }}\frac{\sin x\cdot e^{\cos x}}{\left(1+{\cos }^{2}x\right)\left(e^{\cos x}+e^{-\cos x}\right)}dx ...\left(\mathrm{i}\right)$

Using property of integral $\underset{a}{\overset{b}{\int f}}\left(a+b-x\right)=\underset{a}{\overset{b}{\int }}f\left(x\right)$

we get $I=\underset{0}{\overset{\pi }{\int }}\frac{\sin x\cdot e^{-\cos x}}{\left(1+{\cos }^{2}x\right)\left(e^{-\cos x}+e^{\cos x}\right)}dx ...\left(\mathrm{ii}\right)$

Adding equation $\left(\mathrm{i}\right)$ & $\left(\mathrm{ii}\right)$ we get

$2I=\underset{0}{\overset{\pi }{\int }}\frac{\sin x}{\left(1+{\cos }^{2}x\right)}\cdot \frac{\left(e^{\cos x}+e^{-\cos x}\right)}{e^{\cos x}+e^{-\cos x}}dx$

$2I=\underset{0}{\overset{\pi }{\int }}\frac{\sin x}{1+{\cos }^{2}x}dx$   $\Rightarrow 2I=2\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin x}{1+{\cos }^{2}x}dx$

Let $\cos x=t\Rightarrow -\sin xdx=dt$

$I=-\underset{1}{\overset{0}{\int }}\frac{dt}{1+t^2}\Rightarrow I=-{\left[{\tan }^{-1}t\right]}_1^0=-\left[0-\frac{\pi }{4}\right]$ $=\frac{\pi }{4}$