How to evaluate $\int_0^{\pi}\ln(2+\cos^6x)\,\mathrm dx$ 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!
 A: 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.
A: We can split the integral using roots of unity to factor $2+x^6$
$$I=\int_0^\pi\ln(\cos^6\theta+2)d\theta=I(ai)+I(-ai)+I(ae^{i\pi/6})+I(ae^{-i\pi/6})+I(ae^{5i\pi/6})+I(ae^{-5i\pi/6})$$
Where $a=\root6\of2$, and $I(z)$ is defined as
$$I(z)=\int_0^\pi\ln(z+\cos\theta)d\theta$$
Take the derivative
$$I'(z)=\int_0^\pi\frac{d\theta}{z+\cos\theta}$$
Substitute $\theta→\pi-\theta$
$$I'(z)=\int_0^\pi\frac{d\theta}{z-\cos\theta}$$
Add $I'(z)$ to $I'(z)$
$$2I'(z)=\int_0^\pi\frac{2z\space d\theta}{z^2-\cos^2\theta}$$
Use symmetry
$$I'(z)=\int_0^{\pi/2}\frac{2z\space d\theta}{z^2-\cos^2\theta}=\int_0^{\pi/2}\frac{2z\sec^2\theta\space d\theta}{z^2\sec^2\theta-1}$$
Substitute $x=\tan\theta$
$$I'(z)=\int_0^\infty\frac{2z\space dx}{z^2(1+x^2)-1}=\int_0^\infty\frac{2z\space dx}{(xz)^2+(z^2-1)}=\frac{\pi}{\sqrt{z^2-1}}$$
Integrate both sides with respect to z, using either a hyperbolic-trigonometric substitution $z=\cosh t$ or a trigonometric substitution $z=\sec \theta$.
$$I(z)=\pi\ln(z+\sqrt{z^2-1})+C=\pi \operatorname{arcosh}z+C$$
Setting $z=1$ we get $I(1)=C$ and
$$I(1)=\int_0^\pi\ln(1+\cos\theta)d\theta=-\pi\ln2$$
(see here)
We get $$I(z)=\pi\ln\left(\frac{z+\sqrt{z^2-1}}{2}\right)=\pi\operatorname{arcosh}z-\pi\ln 2$$
$$I=-6\pi\ln 2+\sum_{k=1}^6\operatorname{arcosh}(\root{6}\of{2}e^{(2k+1)\pi i/6})$$
We can use trigonometric identities to simplify the sum of 6 arccosines, but it's difficult
A: Integrate in real space as follows
\begin{align}
I=&\int_{0}^{\pi}\log\left(2+\cos^6 x\right)\,\overset{t=\tan x }{dx} \\
= &\ 2\int_{0}^{\infty}\frac{\ln [2(1+t^2)^3 +1]-3\ln(1+t^2)}{1+t^2} dt\\
 = &\ 2\int_{0}^{\infty}\frac{\ln2 -3\ln(t^2+1)
+\ln \left(t^2+1+1/{\sqrt[3]2}\right)}{1+t^2} dt +2K
\\
\\
K=& \int_{0}^{\infty}\frac{\ln \left[(t^2+1)^2-\frac1{\sqrt[3]2}(t^2+1) +\frac1{\sqrt[3]4}\right]}{1+t^2} dt\\
=& \int_{0}^{\infty}\frac{\ln \left(t^4+2a^2t^2\cos2\theta +a^4\right)}{1+t^2} dt=\pi \ln\left(1+2a\cos\theta +a^2\right)\\
\\
\end{align}
where $a=\sqrt[4]{1-\frac1{\sqrt[3]2}+\frac1{\sqrt[3]4}}$ and
$\cos2\theta =\frac12\sqrt{3-\sqrt[3]2}$. Substitute $K$ into $I$ and evaluate the rest of $I$ with $\int_0^\infty \frac{\ln(t^2+b^2)}{1+t^2}dt =\pi\ln(1+b) $ to obtain
$$I =-5\pi\ln2 +2\pi \ln \bigg(1+\sqrt{1+\frac1{\sqrt[3]2}}\bigg) +2 \pi \ln\left(1+2a\cos\theta +a^2\right)
$$
