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 $

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 $\frac{2R}{R^2+1}=\kappa$.

Hint: by the binomial theorem, $$\sum_{n\ge 0}\frac{(-\frac{1}{2})(-\frac{3}{2})\cdots(\frac{1}{2}-n)}{n!}(-k^2)^n=(1-k^2)^{-1/2}.$$

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

$$\int_0^\pi \log(1+k\cos(x))\,dx=\pi\log\left(\frac{1+\sqrt{1-k^2}}{2}\right)$$

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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!