<s>[INST] series sum $\frac{1}{2}\cdot \frac{1}{2}k^2+\frac{1}{2}\cdot \frac{3}{4}\cdot k^4+\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot k^6+\cdots $ Finding value of $\displaystyle \int^{\pi}_{0}\ln(1+k\cos x)dx$ for $0<k<1$
what I try
Let $\displaystyle I =\int^{\pi}_{0}\ln(1+k\cos x)dx$ 
put $\displaystyle x\rightarrow \frac{\pi}{2}-x$
$\displaystyle I=\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\ln(1-k\sin x)dx$
$\displaystyle I =\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\bigg[k\sin x-\frac{k^2\sin^2 x}{2}+\frac{k^3\sin^3 x}{3}-\cdots \bigg]dx$
$\displaystyle I =-2\int^{\frac{\pi}{2}}_{0}\bigg[\frac{k^2\sin^2 x}{2}+\frac{k^4\sin^4 x}{4}+\cdots \bigg]dx$
$\displaystyle I =-\pi\bigg[\frac{1}{2}\cdot \frac{1}{2}k^2+\frac{1}{2}\cdot \frac{3}{4}\cdot k^4+\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot k^6+\cdots \bigg]$
How can I find sum of that series. Help me please.
 [/INST] METHODOLOGY $1$: Feynman's Trick
Let $I(a,k)$ be the integral given by
$$I(a,k)=\int_0^\pi \log(a+k\cos(x))\,dx\tag1$$
Differentiating $(1)$ with respect to $a$ reveals
$$\begin{align}
\frac{\partial I(a,k)}{\partial a}&=\int_0^\pi \frac{1}{a+k\cos(x)}\,dx\\\\
&=\frac{\pi}{\sqrt{a^2-k^2}}\tag2
\end{align}$$
Integrating $(2)$ with respect to $a$ and using $I(a,0)=\pi \log(a)$ yields
$$I(a,k)=\pi \log(a+\sqrt{a^2-k^2})-\pi\log(2)\tag3$$
Setting $a=1$ in $(3)$, we obtain the coveted result
$$\int_0^\pi \log(1+k\cos(x))\,dx=\pi\log\left(\frac{1+\sqrt{1-k^2}}{2}\right)$$

METHODOLOGY $2$: Series Evaluation
Enforcing the substitution $x\mapsto \pi/2-x$ in $(1)$, we have 
$$\begin{align}
I(1,k)&=\int_{-\pi/2}^{\pi/2}\log(1+k\sin(x))\,dx\\\\
&=\int_{-\pi/2}^{\pi/2}\sum_{n=1}^\infty \frac{(-1)^{n-1}k^n\sin^n(x)}{n}\,dx\\\\
&=-\sum_{n=1}^\infty \frac{k^{2n}}{n}\int_0^{\pi/2}\sin^{2n}(x)\,dx\\\\
&=-\pi\int_0^k \sum_{n=1}^\infty \frac{(2n-1)!!}{(2n)!!}\,x^{2n-1}\,dx\\\\
&=-\pi\int_0^k \frac1x \sum_{n=1}^\infty \binom{-1/2}{n}(-x^2)^n\,dx\\\\
&=-\pi \int_0^k \frac1x\left(\frac1{\sqrt{1-x^2}}-1\right)\,dx\\\\
&=\pi \log\left(\frac{1+\sqrt{1-k^2}}{2}\right)
\end{align}$$
as expected!
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