Integral of $\log(\sin(x)) \tan(x)$ I would like to see a direct proof of the integral $$\int_0^{\pi/2} \log(\sin(x)) \tan(x) \, \mathrm{d}x = -\frac{\pi^2}{24}.$$ I arrived at this integral while trying different ways to evaluate $\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + ...$ and found this value using the known sum $\frac{\pi^2}{6}.$ It would be better to have a direct way to evaluate this.
 A: Note that
$$\begin{align}\int_0^{\pi/2} \log(\sin x) \tan x  \, \mathrm{d}x &= \frac1{2}\int_0^{\pi/2} \log(\sin^2 x ) \tan x  \, \mathrm{d}x \\ &= \frac1{2}\int_0^{\pi/2} \frac{\log(\sin^2 x ) \sin x}{\cos x}  \, \mathrm{d}x \\ &= \frac1{2}\int_0^{\pi/2} \frac{\log(1-\cos^2 x ) \cos x \sin x}{\cos^2 x}  \, \mathrm{d}x  \end{align}.$$
Changing variables to $u = \cos^2 x$ we get 
$$\begin{align}\int_0^{\pi/2} \log(\sin x) \tan x  \, \mathrm{d}x &= \frac1{4}\int_0^{1} \frac{\log(1-u)}{u}  \mathrm{d}u \end{align}.$$
Now integrate the Taylor series expansion of the integrand termwise to finish.
$$\begin{align}\int_0^{\pi/2} \log(\sin x) \tan x  \, \mathrm{d}x &= -\frac1{4}\int_0^{1} \sum_{k=1}^{\infty}\frac{u^{k-1}}{k}  \mathrm{d}u \\ &= -\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^2}  \end{align}.$$
A: Hint: Use the transformation $$\sin x=e^{-z}$$ to get $$I=\int_{0}^{\infty}\frac{ze^{-z} }{\sqrt{1-e^{-2z}}}dz$$ Now, expand $(1-e^{-2z})^{-1/2}$ and use the Gamma and zeta functions to evaluate the integral.
A: We know that, for positive values of a and b, $\displaystyle\int_0^\tfrac\pi2\sin^{2a-1}x~\cos^{2b-1}x~dx~=~\frac12~B\big(a,b\big)$, 
where  B is the beta function, implying that our original integral can be written in terms 
of $~\displaystyle\lim_{b\to0}B^{(1,0)}(1,b)=-\zeta(2)=-\dfrac{\pi^2}6$, which follows from the well-known relation between 
the beta and $\Gamma$ functions, as well as the latter's connection to the digamma, trigamma, and 
polygamma functions.
