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 $\mathbb{R} \ni a \neq b \in \mathbb{R}$


2 Answers 2


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:



$$I = \frac{\mathcal{I}}{2} = \frac{\mathcal{I}}{4}$$

  • $\begingroup$ Nice approach (+1) $\endgroup$
    – Robert Z
    Commented Mar 27, 2019 at 11:32
  • $\begingroup$ Nice gymnastics ! and $\to +1$ for sure. $\endgroup$ Commented Mar 27, 2019 at 11:34

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 $I=0$.

  • $\begingroup$ While this is a quite neat way of proving the given equality, the OP asks for a similar pair of substitutions $($similiar to the linked post$)$, namely a pair of linear and non-linear transformation in order to achieve the result. $\endgroup$
    – mrtaurho
    Commented Mar 27, 2019 at 9:22
  • 1
    $\begingroup$ I missed the day when you reached the $100k$ reputation. Congratulations and thanks for posting so good answers so often. Cheers :-) $\endgroup$ Commented Mar 27, 2019 at 9:38
  • $\begingroup$ @ClaudeLeibovici Thanks for your generous comments! $\endgroup$
    – Robert Z
    Commented Mar 27, 2019 at 11:26
  • $\begingroup$ @mrtaurho It seems that OP found the appropriate non-linear transformations $\endgroup$
    – Robert Z
    Commented Mar 27, 2019 at 11:27

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