The value of the integral _-/2^/2(dx)/((1+e^x)(^6x+^6x)) is equal to:

Question

The value of the integral $\displaystyle\int_{-\pi/2}^{\pi/2}\frac{dx}{(1+e^x)(\sin^6x+\cos^6x)}$ is equal to:

Accepted Answer

By King's rule ($x\to-x$), since $\dfrac1{1+e^x}+\dfrac1{1+e^{-x}}=1$ and $\sin^6x+\cos^6x$ is even:

$$2I = \int_{-\pi/2}^{\pi/2}\frac{dx}{\sin^6x+\cos^6x} = 2\int_0^{\pi/2}\frac{dx}{1-\frac34\sin^22x}$$

Evaluating this standard integral (via $t=\tan x$ or the general $\int_0^{\pi/2}\frac{dx}{a+b\cos^22x}$ formula) gives $2I=2\pi \implies I=\pi$.

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