TITLE: How to evaluate $\int_0^{\pi}\ln(2+\cos^6x)\,\mathrm dx$
QUESTION [6 upvotes]: How to evaluate
$$\int_0^{\pi}\ln(2+\cos^6x)\,\mathrm dx$$
I tried to let
$$I(a)=\int_0^{\pi}\ln(a+\cos^6x)\,\mathrm dx$$
so
$$I'(a)=\int_0^{\pi}\frac{1}{a+\cos^6x}\,\mathrm dx=2\int_0^{\pi/2}\frac{1}{a+\cos^6x}\,\mathrm dx$$
and let $\cos^2x=t$ we get
$$I'(a)=\int_0^1\frac{\mathrm dt}{(a+t^3)\sqrt{t}\sqrt{1-t}}$$
But I don't know how to go further. Any help will be nice!

REPLY [6 votes]: Due to the symmetry of the cosine function, the given integral equals
$$ I=2\int_{0}^{\pi/2}\log\left(2+\cos(x)^6\right)\,dx \stackrel{x\to\arctan t}{=}2\int_{0}^{+\infty}\log\left[2+\left(\frac{1}{1+t^2}\right)^3\right]\frac{dt}{1+t^2} $$
but that is not very promising, so let us switch to a different approach through Taylor series:
$$ I = \pi\log(2)+ 2\sum_{n\geq 1}\frac{(-1)^{n+1}}{n\,2^n}\int_{0}^{\pi/2}\cos(x)^{6n}\,dx =\pi\log(2)-\pi\sum_{n\geq 1}\frac{(-1)^{n}}{n\, 2^{7n}}\binom{6n}{3n}\tag{1}$$
Not really promising either. Further switch: since $\int_{0}^{\pi/2}\log\cos(x)\,dx = -\frac{\pi}{2}\log(2)$, by differentiation under the integral sign it is enough to apply $\int_{0}^{2}(\ldots)\,dA$ to
$$ \int_{0}^{\pi/2}\frac{dx}{A+\cos(x)^6} = \int_{0}^{+\infty}\frac{dt}{A(1+t^2)+\frac{1}{(1+t^2)^2}} $$
that we may compute through the residue theorem, but still involves the roots of a cubic polynomial. Definitely a tough nut to crack without resorting to the hypergeometric (or dilogarithms) machinery. If we set $\beta=2^{1/3}$ and recall that
$$ f(x)=\sum_{n\geq 1}\frac{(-x)^n}{n 4^n}\binom{2n}{n}=\log(4)-\log(1+\sqrt{1+x}) \tag{2}$$
then the RHS of $(1)$ can be evaluated in terms of $f\left(\frac{1}{\beta}\right),\,f\left(\frac{\omega}{\beta}\right)$ and $f\left(\frac{\omega^2}{\beta}\right)$, with $\omega=e^{2\pi i/3}$, due to the discrete Fourier transform. These numbers being "nice" or not is debatable.