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.
 A: For any $R\geq 1$,
$$(R+e^{i\theta})(R+e^{-i\theta}) = (R^2+1)+2R\cos\theta \tag{1}$$
$$ 1+\frac{2R}{R^2+1}\cos\theta = \frac{R^2}{R^2+1}\left(1+\frac{e^{i\theta}}{R}\right)\left(1+\frac{e^{-i\theta}}{R}\right)\tag{2}$$
$$ \log\left(1+\tfrac{2R}{R^2+1}\cos\theta\right) = \log\left(\tfrac{R^2}{R^2+1}\right)+2\sum_{n\geq 1}\frac{(-1)^{n+1}\cos(n\theta)}{n R^n}\tag{3} $$
and for any $n\in\mathbb{N}^+$ we have $\int_{0}^{\pi}\cos(n\theta)\,d\theta=0$, therefore
$$ \int_{0}^{\pi}\log\left(1+\tfrac{2R}{R^2+1}\cos\theta\right)\,d\theta = \pi\log\left(\tfrac{R^2}{R^2+1}\right).\tag{4}$$
Now it is enough to enforce the substitution $\frac{2R}{R^2+1}=\kappa$.
A: Hint: by the binomial theorem, $$\sum_{n\ge 0}\frac{(-\frac{1}{2})(-\frac{3}{2})\cdots(\frac{1}{2}-n)}{n!}(-k^2)^n=(1-k^2)^{-1/2}.$$
A: 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!
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[10px,#ffd]{{1 \over 2}\cdot{1 \over 2}\, k^{2} +
{1 \over 2}\cdot{3 \over 4}\, k^{4} +
{1 \over 2}\cdot{3 \over 4}\cdot{5 \over 6}\, k^{6} + \cdots} =
\sum_{n = 1}^{\infty}{1 \over 2}
\pars{\prod_{j = 1}^{n}{2j - 1 \over 2j}}k^{2n}
\\[5mm] = &\
{1 \over 2}\sum_{n = 1}^{\infty}
\bracks{\prod_{j = 1}^{n}\pars{j - 1/2} \over \prod_{m = 1}^{n}m}k^{2n} =
{1 \over 2}\sum_{n = 1}^{\infty}
{\pars{1/2}^{\overline{n}} \over n!}\,k^{2n}
\\[5mm] = &\
{1 \over 2}\sum_{n = 1}^{\infty}
{\Gamma\pars{1/2 + n}/\Gamma\pars{1/2} \over n!}\, k^{2n} =
{1 \over 2}\sum_{n = 1}^{\infty}
{\pars{n - 1/2}! \over n!\pars{-1/2}!}\,k^{2n} =
{1 \over 2}\sum_{n = 1}^{\infty}
{n - 1/2 \choose n}k^{2n}
\\[5mm] = &\
{1 \over 2}\sum_{n = 1}^{\infty}
\bracks{{-n + 1/2 + n - 1\choose n}\pars{-1}^{n}}k^{2n} =
{1 \over 2}\sum_{n = 1}^{\infty}{-1/2 \choose n}\pars{-k^{2}}^{n}
\\[5mm] = &\
{1 \over 2}\braces{\bracks{1 + \pars{-k^{2}}}^{-1/2} - 1} =
\bbx{{1 \over 2}\pars{{1 \over \root{1 - k^{2}}} - 1}}
\end{align}
