The value of the definite integral _-/4^/4dx(1+e^x x)(^4x+^4x) is equal to:

Question

The value of the definite integral $\displaystyle\int_{-\pi/4}^{\pi/4}\frac{dx}{(1+e^{x\cos x})(\sin^4x+\cos^4x)}$ is equal to:

Accepted Answer

Let $I=\displaystyle\int_{-\pi/4}^{\pi/4}\frac{dx}{(1+e^{x\cos x})(\sin^4x+\cos^4x)}\ \ldots(1)$

Using $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$ (i.e. $x\to-x$):
$$I=\int_{-\pi/4}^{\pi/4}\frac{dx}{(1+e^{-x\cos x})(\sin^4x+\cos^4x)}\ \ldots(2)$$

Adding (1) and (2): $2I=\displaystyle\int_{-\pi/4}^{\pi/4}\frac{dx}{\sin^4x+\cos^4x}=2\int_0^{\pi/4}\frac{dx}{\sin^4x+\cos^4x}$

Dividing numerator and denominator by $\cos^4x$: $I=\displaystyle\int_0^{\pi/4}\frac{(1+\tan^2x)\sec^2x}{\tan^4x+1}dx$. Writing $\tan^4x+1=(\tan x-\tfrac1{\tan x})^2+2$ (dividing by $\tan^2x$) and substituting $t=\tan x-\dfrac1{\tan x}$:
$$I=\int_{-\infty}^0\frac{dt}{t^2+2}=\frac{1}{\sqrt2}\tan^{-1}\frac{t}{\sqrt2}\Big|_{-\infty}^0=0-\frac{1}{\sqrt2}\left(-\frac{\pi}{2}\right)=\frac{\pi}{2\sqrt2}$$

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