TITLE: How to prove $\int_0^\pi\frac{\ln(2+\cos\phi)}{\sqrt{2+\cos\phi}}d\phi=\frac{\ln3}{\sqrt3}K\left(\sqrt{\frac23}\right)$?
QUESTION [16 upvotes]: How can I prove the following conjectured identity?
$$\int_0^\pi\frac{\ln(2+\cos\phi)}{\sqrt{2+\cos\phi}}d\phi\stackrel?=\frac{\ln3}{\sqrt3}K\left(\sqrt{\frac23}\right),\tag1$$
where $K(x)$ is the complete elliptic integral of the 1ˢᵗ kind:
$$K(x)={_2F_1}\left(\frac12,\frac12;\ 1;\ x^2\right)\cdot\frac\pi2.\tag2$$

REPLY [6 votes]: $$I=\int_0^\pi \frac{\ln(2+\cos x)}{\sqrt{2+\cos x}}dx\overset{2+\cos x=t}=\int_1^3\frac{\color{blue}{\ln t}}{\sqrt{t}}\frac{dt}{\sqrt{1-(2-t)^2}}\overset{t\to \frac{3}{t}}=\int_1^3\frac{\color{red}{\ln\left(\frac{3}{t}\right)}}{\sqrt t}\frac{dt}{\sqrt{1-(2-t)^2}}$$
$$\require{cancel}\Rightarrow 2I= \int_1^3\frac{\color{blue}{\cancel{\ln 3}}+\color{red}{\ln 3-\cancel{\ln t}}}{\sqrt t}\frac{dt}{\sqrt{1-(2-t)^2}}\overset{t=2+\cos x}=\ln 3\int_0^\pi\frac{dx}{\sqrt{2+\cos x}}$$
$$\Rightarrow I\overset{x\to 2x}=\frac{\ln 3}{\sqrt 3}\int_0^\frac{\pi}{2}\frac{dx}{\sqrt {\frac{2+\cos(2x)}{3}}}=\frac{\ln 3}{\sqrt 3}\int_0^\frac{\pi}{2} \frac{dx}{\sqrt{1-\frac{2}{3}\sin^2 x}}=\frac{\ln 3}{\sqrt 3}K\left(\sqrt{\frac23}\right)$$