Prove $\int_0^2 \frac{\log{x}}{\sqrt{4-x^2}}\text{d}x=0$ without integrating Inspired by this post:
Prove that $\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}}~dx=0$ (without trigonometric substitution)
I was wondering if there exists a similar pair of substitutions for the following integral:
$$I = \int_0^2 \frac{\log{x}}{\sqrt{4-x^2}}\text{d}x$$
More precisely:
Does there exist a linear and non-linear transformation (or possibly two non-linear transformations) $u(x), v(x)$ such that by applying them to the above integral, we can obtain $aI = bI$ where $\mathbb{R} \ni a \neq b \in \mathbb{R}$
 A: I have clearly overthought this.
Substitute $x \mapsto x^2$ to transform $I$ into a quarter of the integral linked.
Nesting the transformations, we find the appropriate non-linear transformations for this integral are:
$$u\mapsto \sqrt{4-x^2}, u \mapsto x^2 - 2$$
Resulting in:
$$I=\frac{1}{2}\int_0^2\frac{\log{(4-u^2)}}{\sqrt{4-u^2}}\text{d}u$$
$$I=\frac{1}{4}\int_{-2}^2\frac{\log{(2+u)}}{\sqrt{4-u^2}}\text{d}u=\frac{1}{4}\int_0^2\frac{\log{(4-u^2)}}{\sqrt{4-u^2}}\text{d}u$$
$$I = \frac{\mathcal{I}}{2} = \frac{\mathcal{I}}{4}$$
A: I found a way of proving that $I=2I$ without integrating, although it is not similar to the one explained in the linked post.
Let $x=2\sin(t)$ then 
$$\begin{align}I&=\int_0^{\pi/2}\log(2\sin(t))\,dt\\
&=\int_0^{\pi/2}\log(4\sin(t/2)\cos(t/2))\,dt\\
&=\int_0^{\pi/2}\log(2\sin(t/2))\,dt+\int_0^{\pi/2}\log(2\cos(t/2))\,dt\\
&=\int_0^{\pi/2}\log(2\sin(t/2))\,dt+\int_{\pi/2}^{\pi}\log(2\sin(s/2))\,ds\\
&=\int_0^{\pi}\log(2\sin(s/2))\,ds=2\int_0^{\pi/2}\log(2\sin(u))\,du=2I
\end{align}$$
where $s=\pi-t$ and $u=s/2$. Hence $I=0$.
