Question

$calculate \int_{0}^{\frac{\pi}{2}}\frac{x\ln{\cos(x)}}{\tan x}dx$

Answer

$\begin{aligned} I&=\int_{0}^{\frac{\pi}{2}}\frac{x\ln{(\cos x)}}{\tan x}dx=\int_{0}^{\frac{\pi}{2}}x\ln{(\cos x)}d(\ln(\sin x))\\ &=x\ln{(\cos x)}\ln(\sin x)\left|_{0}^{\pi/2}\right.-\int_{0}^{\frac{\pi}{2}}\ln{(\sin x)}d(x\ln{(\cos x)})dx\\ &=-\int_{0}^{\frac{\pi}{2}}\ln{(\sin x)}(\ln{(\cos x)}-x\tan x)dx\\ &=-\int_{0}^{\frac{\pi}{2}}\ln{(\sin x)}\ln{(\cos x)}dx+\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}} \tan x\ln{(\sin x)}dx-\int_{0}^{\frac{\pi}{2}}(\frac{\pi}{2}-x) \tan x\ln{(\sin x)}dx\\ &=-\underbrace{\int_{0}^{\frac{\pi}{2}}\ln{(\sin x)}\ln{(\cos x)}dx}_{I_1}+\frac{\pi}{2}\underbrace{\int_{0}^{\frac{\pi}{2}} \tan x\ln{(\sin x)}dx}_{I_2}-I \\ \because I_1 &=\int_{0}^{\frac{\pi}{2}}\ln{(\sin x)}\ln{(\cos x)}dx\\ &=\int_0^{\pi/2}\left(\ln(2)+\sum_{j=1}^\infty\frac{\cos(2jx)}{j}\right)\left(\ln(2)+\sum_{k=1}^\infty(-1)^k\frac{\cos(2kx)}{k}\right)\,\mathrm{d}x\\ &=\frac\pi2\ln^2(2)+2\ln(2)\int_0^{\pi/2}\sum_{k=1}^\infty\frac{\cos(4kx)}{2k}\,\mathrm{d}x+\int_0^{\pi/2}\sum_{k=1}^\infty(-1)^k\frac{\cos^2(2kx)}{k^2}\,\mathrm{d}x\\ &=\frac\pi2\ln^2(2)+\frac\pi4\sum_{k=1}^\infty\frac{(-1)^k}{k^2}\\ &=\frac\pi2\ln^2(2)-\frac{\pi^3}{48}\\ \because I_2&=\int_{0}^{\frac{\pi}{2}} \tan x\ln{(\sin x)}dx=-\frac{\pi^2}{24}\\ &\xlongequal{t = \cos x}\int_{0}^1\frac{\ln(\sqrt{1-t^2})}tdt\\ &\xlongequal{y=\sqrt{1-t^2}}\int_{0}^1\frac{y\ln y}{1-y^2}\:dy\\ &=\int_{0}^1y\ln y\left(\sum_{n=0}^{\infty}y^{2n}\right)dy\\ &=\sum_{n=0}^{\infty}\int_{0}^1y^{2n+1}\ln y\:dy\\ &=-\sum_{n=0}^{\infty}\frac1{(2n+2)^2}\\ &=-\frac{\pi^2}{24}\\ \therefore I&=\dfrac{-I_1+\dfrac{\pi}{2}I_2}{2}=-\dfrac{\pi}{4}\ln^2 (2) \end{aligned}$