The value of the integral ∫ _ - (π)/(2) ^ (π)/(2) dx (1 + e ^ x )(...

Let, $I = {∫}_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}}\dfrac{dx}{\left(1 + {e}^{x}\right)\left({sm}^{6}x + \cos^{6}x\right)} . . . \left(i\right)$

Now we know ${∫}_{a}^{b}f\left(x\right)dx = {∫}_{a}^{b}f\left(a + b - x\right)dx$

$I = {∫}_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}}\dfrac{{e}^{x}dx}{\left(1 + {e}^{x}\right)\left(\sin^{6}x + \cos^{6}x\right)} . . . \left(ii\right)$

Adding \left(i\right)$and \left(ii\right)$

$2I = {∫}_{\dfrac{- \pi}{2}}^{\dfrac{\pi}{2}}\dfrac{\left(1 + {e}^{x}\right)dx}{\left(1 + {e}^{x}\right)\left(\sin^{6}x + \cos^{6}x\right)}$

$I = \dfrac{1}{2} \times 2{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{\left(\sin^{2}x + \cos^{2}x\right)\left(\sin^{4}x + \cos^{4}x - \sin^{2}x\cos^{2}x\right)}$

$\mathop{= ∫}_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{{\left(\sin^{2}x + \cos^{2}x\right)}^{2} - 3\sin^{2}x\cos^{2}x} = \mathop{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{1 - \dfrac{3}{4}\sin^{2}2x}$

$= {∫}_{0}^{\dfrac{\pi}{2}}\dfrac{4\sec^{2}2xdx}{4\left(1 + \tan^{2}2x\right) - 3\tan^{2}2x} = 2{∫}_{0}^{\dfrac{\pi}{4}}\dfrac{4\sec^{2}2x}{4 + \tan^{2}2x}dx$

Let $\tan 2x = t$ and $2\sec^{2}2xdx = dt$

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