$\int_0^{\pi/2} \frac{\log \cos x}{\tan x }dx$

Evaluate the integral:

$$\int_{0}^{\pi/2} \frac{\log \cos x}{\tan x}\, {\rm d}x$$


Solution

\begin{align*} \int_{0}^{\pi/2} \frac{\log \cos x}{\tan x}\, {\rm d}x &=\cancelto{0}{\left [ \log \sin x \log \cos x \right ]_0^{\pi/2}} + \int_{0}^{\pi/2} \tan x \log \sin x \, {\rm d}x \\
&=\int_{0}^{\pi/2} \frac{\sin x}{\cos x} \log \sqrt{1-\cos^2 x}\, {\rm d}x \\
 &\!\!\!\!\overset{u=\cos x}{=\! =\! =\! =\!} \frac{1}{2} \int_{0}^{1} \frac{\log \left ( 1-u^2 \right )}{u}\, {\rm d}u\\
&=-\frac{1}{2} \int_{0}^{1} \frac{1}{u} \sum_{n=1}^{\infty} \frac{u^{2n}}{n} \, {\rm d}u \\
&= -\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{1}u^{2n-1} \, {\rm d}u \\
&=- \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^2} \\
 &=- \frac{\pi^2}{24} \end{align*}