Evaluation of $\int_{0}^1 \frac{1}{x} \log^3{(1-x)}dx =-\frac{\pi^4}{15}$ and $\int_{-\pi}^{\pi} \log(2\cos{\frac{x}{2}}) dx =0$ In the following encyclopedia, http://m.encyclopedia-of-equation.webnode.jp/including-integral/
I found the relations below
\begin{eqnarray}
\int_{0}^1 \frac{1}{x} \log^3{(1-x)}dx &=&-\frac{\pi^4}{15} \tag{1} \\
\int_{-\pi}^{\pi} \log(2\cos{\frac{x}{2}}) dx &=& 0 \tag{2}
\end{eqnarray}
I tried to prove these equation, but I didn't success to prove.
How do you go about evaluating those integrals to obtain the repsective values?
 A: You may write

$$\begin{align}
\int_{0}^1 \frac{1}{x} \log^3{(1-x)}dx&=\int_{0}^1 \frac{1}{1-x} \log^3xdx\\
&=\sum_{n=0}^{\infty}\int_0^1x^n\log^3x \:dx\\
&=-6\sum_{n=0}^{\infty}\frac{1}{(n+1)^4}\\
&=-\frac{\pi ^4}{15}
\end{align}$$

where we have used 
$$
\begin{align}
&\int_0^1x^\alpha\log^3x \:dx=-\frac{3!}{(\alpha+1)^4} \quad (\text{a direct integration by parts)}\\
&\zeta(4)=\frac{\pi ^4}{90} .
\end{align}
$$ (Riemann zeta function at even integers)
Your second integral rewrites, by parity of the integrand and a change of variable,  

$$
\int_{-\pi}^{\pi} \log(2\cos{\frac{x}{2}}) dx=2\int_0^{\pi} \log(2\cos{\frac{x}{2}}) dx=4\int_0^{\pi/2} \log(2\cos x) dx=0
$$ 

a proof of the latter equality may be found here.
A: For the first integral: substitute $1-x=q$, the integral becomes
$$
I_1=\int_0^1 \frac{\log^3(q)}{1-q}
$$
The next substitution is obvious $e^{r}=q$, which yields after some some algebra 
$$
I_1=\int_{0}^{\infty}\frac{r^3}{1-e^{r}}
$$
This integral is now solvable by various means, for example by expanding the denominator into a geometric series, integrate termwise and using the special value $\zeta(4)=\frac{\pi^4}{90}$ to obtain
$$
I_1=-\frac{\pi^4}{15}
$$ 
