<s>[INST] 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.
 [/INST] 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.
$$\begin{align}\int_0^{\pi/2} \log(\sin x) \tan x  \, \mathrm{d}x &= -\frac1{4}\int_0^{1} \sum_{k=1}^{\infty}\frac{u^{k-1}}{k}  \mathrm{d}u \\ &= -\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^2}  \end{align}.$$
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