The value of ∫ _ 0 ^ π e ^ x x (1 + ^ 2 x)( e ^ x + e ^ - x ) dx is...

Given $I = \mathop{∫}_{0}^{\pi}\dfrac{\sin x ⋅ {e}^{\cos x}}{\left(1 + \cos^{2}x\right)\left({e}^{\cos x} + {e}^{- \cos x}\right)}dx . . . \left(i\right)$

Using property of integral $\mathop{∫ f}_{a}^{b}\left(a + b - x\right) = \mathop{∫}_{a}^{b}f\left(x\right)$

we get $I = \mathop{∫}_{0}^{\pi}\dfrac{\sin x ⋅ {e}^{- \cos x}}{\left(1 + \cos^{2}x\right)\left({e}^{- \cos x} + {e}^{\cos x}\right)}dx . . . \left(ii\right)$

Adding equation \left(i\right)& \left(ii\right) we get

$2I = \mathop{∫}_{0}^{\pi}\dfrac{\sin x}{\left(1 + \cos^{2}x\right)} ⋅ \dfrac{\left({e}^{\cos x} + {e}^{- \cos x}\right)}{{e}^{\cos x} + {e}^{- \cos x}}dx$

$2I = \mathop{∫}_{0}^{\pi}\dfrac{\sin x}{1 + \cos^{2}x}dx$ $⇒ 2I = 2\mathop{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{\sin x}{1 + \cos^{2}x}dx$

Let $\cos x = t ⇒ - sinxdx = dt$

$I = - \mathop{∫}_{1}^{0}\dfrac{dt}{1 + {t}^{2}} ⇒ I = - {\left[\tan^{- 1}t\right]}_{1}^{0} = - \left[0 - \dfrac{\pi}{4}\right]$ $= \dfrac{\pi}{4}$