I want to calculate $$ \int_0^{\pi/2}\tan(x)\ln(\sin(x))\,dx. $$
I tried taking $\cos x$ as $t$ and converting the whole expression in $t$ and then integrating by parts. That didn't help. Integrating by parts directly also doesn't help. I'm guessing there must be a different, better approach to solve this.
Can someone help?
\begin{align} I&=\int^{\frac\pi2}_0\tan(x)\ln(\sin(x))dx \\ &=\int_{0}^1\frac{\ln(\sqrt{1-t^2})}tdt\tag{$t=\cos{x}$} \\ &=\int_{0}^1\frac{y\ln y}{1-y^2}\:dy\tag{$y=\sqrt{1-t^2}$} \\ &=\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\tag1 \\ &=-\sum_{n=0}^{\infty}\frac1{(2n+2)^2} \\ &=-\frac{\pi^2}{24} \end{align} Note in $(1)$ $$ \int_{0}^1y^{2n+1}\ln y\:dy=\frac{y^{2n+2}\ln y}{2n+2}\bigg |_0^1-\frac1{2n+2}\int_{0}^1y^{2n+1}\:dy=-\frac1{(2n+2)^2} $$
Hint: $$I=\int^{\frac\pi2}_0\tan(x)\ln(\sin(x))dx$$
Put $\cos x=t; -\sin xdx=dt$
$$=-\int^{0}_1\frac{\ln(\sqrt{1-t^2})}tdt$$
$$=\frac12\int^1_0\frac{2t\ln(\sqrt{1-t^2})}{t^2}dt$$
Put $1-t^2=u;-2tdt=du$ $$\frac12\int^1_0\frac{\ln(\sqrt u)}{u-1}du$$
$$\frac12\int^1_0\frac{\ln(\sqrt u)}{u-1}du=\frac12[-\frac{\text{Li}_2(1-u)}{2}]^1_0=-\frac{\pi^2}{24}$$
