The value of the integral

${\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2}\frac{dx}{(1+e^x)({\sin }^{6}x + {\cos }^{6}x)}$ is equal to

$$\begin{gathered} I = {\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2}\frac{dx}{(1+e^x)({\sin }^{6}x + {\cos }^{6}x)} \\ = {\int }_0^{\mathrm{\pi }/2}\left(\frac{1}{(1+e^x)({\sin }^{6}x + {\cos }^{6}x)} + \frac{1}{(1+e^{-x})({\sin }^{6}x + {\cos }^{6}x)}\right)dx \\ = {\int }_0^{\mathrm{\pi }/2}\frac{dx}{({\sin }^{6}x + {\cos }^{6}x)} \\ = {\int }_0^{\mathrm{\pi }/2}\frac{dx}{{\sin }^{4}x - {\sin }^{2}x {\cos }^{2}x + {\cos }^{4}x} \\ = {\int }_0^{\mathrm{\pi }/2}\frac{se{c}^{4}x}{{\tan }^{4}x - {\tan }^{2}x + 1}dx \\ = {\int }_0^{\mathrm{\pi }/2}\frac{(1 + {\tan }^{2}x)\left(se{c}^{2}x\right)}{{\tan }^{4}x - {\tan }^{2}x + 1}dx \\ Let \tan x = t \Rightarrow se{c}^{2}x dx = dt \\ = {\int }_0^{\infty }\frac{(1 + t^2)dt}{t^4-t^2+1} = {\int }_0^{\infty }\frac{(1 + {\displaystyle \frac{1}{t^2}})}{t^2 -1 + {\displaystyle \frac{1}{t^2}}}dt \\ = {\int }_0^{\infty }\frac{(1 + \frac{1}{t^2})}{{(t - {\displaystyle \frac{1}{t}})}^2 + 1}dt \\ = {{\left[{\tan }^{-1}\left(t - \frac{1}{t}\right)\right]}^{\infty }}_0 \\ = \frac{\mathrm{\pi }}{2} + \frac{\mathrm{\pi }}{2} \\ = \mathrm{\pi } \end{gathered}$$