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