Prove that $\int_0^1\,\frac{\ln(x)}{\sqrt{1-x^2}}\,\text{d}x=-\frac{\pi}{2}\,\ln(2)$. I have discovered via contour integration that
$$\int_0^\infty\,\frac{\exp(t\,u)}{\exp(u)+1}\,\text{d}u={\text{csc}(\pi\,t)}\,\left(\frac{\pi}{2}-\int_0^{\frac{\pi}{2}}\,\frac{\sin\big((1-2t)\,y\big)}{\sin(y)}\,\text{d}y\right)\tag{*}$$
for all $t\in\mathbb{C}\setminus\mathbb{Z}$ such that $\text{Re}(t)<1$.  By taking $t\to 0$, I deduce that
$$\int_0^\infty\,\frac{1}{\exp(u)+1}\,\text{d}u=\frac{2}{\pi}\,\int_0^{\frac{\pi}{2}}\,y\,\cot(y)\,\text{d}y\,.$$
With a step of integration by parts, I obtain
$$\int_0^\infty\,\frac{1}{\exp(u)+1}\,\text{d}u=-\frac{2}{\pi}\,\int_0^{\frac{\pi}{2}}\,\ln\big(\sin(y)\big)\,\text{d}y\,.$$
Setting $x:=\sin(y)$, I get
$$\int_0^\infty\,\frac{1}{\exp(u)+1}\,\text{d}u=-\frac{2}{\pi}\,\int_0^1\,\frac{\ln(x)}{\sqrt{1-x^2}}\,\text{d}x\,.$$
This shows that
$$\int_0^1\,\frac{\ln(x)}{\sqrt{1-x^2}}\,\text{d}x=-\frac{\pi}{2}\,\int_0^\infty\,\frac{1}{\exp(u)+1}\,\text{d}u\,.$$
The integral $\displaystyle\int_0^\infty\,\frac{1}{\exp(u)+1}\,\text{d}u$ can be easily obtained since
$$\int\,\frac{1}{\exp(u)+1}\,\text{d}u=u-\ln\big(\exp(u)+1\big)+\text{constant}\,.$$
That is, I have
$$\int_0^1\,\frac{\ln(x)}{\sqrt{1-x^2}}\,\text{d}x=-\frac{\pi}{2}\,\ln(2)\,.\tag{#}$$
However, this proof is a very roundabout way to verify the equality above.  Is there a more direct way to prove that (#) is true?  Any technique is appreciated.

A nice consequence of (*) is that
$$\int_0^\infty\,\frac{\sinh(t\,u)}{\exp(u)+1}\,\text{d}u=\frac{\pi}{2}\,\text{csc}(\pi\,t)-\frac{1}{2\,t}$$
for all $t\in\mathbb{C}\setminus\{0\}$ such that $\big|\text{Re}(t)\big|<1$.  This provides a proof that
$$\eta(2r)=\frac{1}{(2r-1)!}\,\int_0^\infty\,\frac{u^{2r-1}}{\exp(u)+1}\,\text{d}u=\frac{\pi^{2r}}{2}\,\Biggl(\left[t^{2r-1}\right]\Big(\text{csc}(t)\Big)\Biggr)$$
for $r=1,2,3,\ldots$.  Here, $\eta$ is the Dirichtlet eta function.  In addition, $[t^k]\big(g(t)\big)$ denotes the coefficient of $t^k$ in the Laurent expansion of $g(t)$ about $t=0$.  This also justifies the well known results that
$$\eta(2r)=\frac{\left(2^{2r-1}-1\right)\,\big|B_{2r}\big|\,\pi^{2r}}{(2r)!}\text{ and }\zeta(2r)=\frac{2^{2r-1}\,\big|B_{2r}\big|\,\pi^{2r}}{(2r)!}$$
for $r=1,2,3,\ldots$, where $\left(B_j\right)_{j\in\mathbb{Z}_{\geq0}}$ is the sequence of Bernoulli numbers and $\zeta$ is the Riemann zeta function.

Similarly,
$$\begin{align}\int_0^\infty\,\frac{\exp(t\,u)-1}{\exp(u)-1}\,\text{d}u&=\ln(2)+2\,\int_0^{\frac{\pi}{2}}\,\frac{\sin\big((1-t)\,y)\,\sin(t\,y)}{\sin(y)}\,\text{d}y\\
&\phantom{aaaaa}-\cot(\pi\,t)\,\left(\frac{\pi}{2}-\int_0^{\frac{\pi}{2}}\,\frac{\sin\big((1-2t)\,y\big)}{\sin(y)}\,\text{d}y\right)\,,\end{align}$$
for all $t\in\mathbb{C}\setminus\mathbb{Z}$ such that $\text{Re}(t)<1$.
This gives
$$\int_0^\infty\,\frac{\sinh(t\,u)}{\exp(u)-1}\,\text{d}u=\frac{1}{2\,t}-\frac{\pi}{2}\,\cot(\pi\,t)$$
for all $t\in\mathbb{C}\setminus\{0\}$ such that $\big|\text{Re}(t)\big|<1$.

Another consequence of (*) is that
$$\int_0^{\frac{\pi}{2}}\,\frac{\sin(k\,y)}{\sin(y)}\,\text{d}y=\frac{\pi}{2}\,\text{sign}(k)$$
for all odd integers $k$.  It is an interesting challenge to determine the integral $\displaystyle \int_0^{\frac{\pi}{2}}\,\frac{\sin(k\,y)}{\sin(y)}\,\text{d}y$ for all even integers $k$.

Riemann sums are enough. For any $n\in\mathbb{N}^+$ we have
$$ \prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right)=\frac{2n}{2^n},\tag{1} $$
hence
$$ \int_{0}^{\pi/2}\log\sin\theta\,d\theta=\frac{1}{2}\int_{0}^{\pi}\log\sin\theta\,d\theta = \frac{\pi}{2}\lim_{n\to +\infty}\frac{\log(2n)-n\log 2}{n} = \color{red}{-\frac{\pi}{2}\log 2}.\tag{2}$$
