On some notes I found the following equation:
$$\sqrt{2ml^2}*4\int_0^{\theta_{\max}}{\sqrt{E-\frac{mgl\theta^2}{2}}d\theta} = \frac{\sqrt{2ml^2}}{2\pi} \frac{\sqrt{mgl^2}}{2}*4\theta^2_{\max} \int_0^1{\sqrt{1-x^2}dx}$$
EDIT: The only clues are:
- $\theta$ is an angular variable
- there was performed a variable change $x=\theta/\theta_{\max}$
- $E=mgl\theta^2_{\max}/2$
Someone could offer a (possibly) step by step answer?
First of all note that your RHS can be simplified to $$\frac{\sqrt{2g}}{\pi}\cdot 4ml^2\theta_{max}^2 \int_0^1 \sqrt{1-x^2}dx $$ Now $0 \leq \theta \leq \theta_{max}$ so $\frac{0}{\theta_{max}} \leq \frac{\theta}{\theta_{max}} \leq \frac{\theta_{max}}{\theta_{\max}}$, i.e. $0 \leq x \leq 1$.
We have also $\theta=\theta_{max}x$ and $dx=d \left(\frac{\theta}{\theta_{max}} \right)=\frac{1}{\theta_{max}}d\theta$, i.e. $d\theta=\theta_{max}dx$
Now $$\sqrt{2ml^2}\cdot 4\int_0^{\theta_{max}}{\sqrt{E-\frac{mgl\theta^2}{2}}d\theta} = \sqrt{2ml^2}\cdot 4 \int_0^1 \sqrt{E-\frac{mglx^2 \theta_{max}^2}{2}}\theta_{max}dx=\\=\sqrt{2ml^2}\cdot 4 \int_0^1 \sqrt{\frac{mgl\theta_{max}^2}{2}-\frac{mglx^2 \theta_{max}^2}{2}}\theta_{max}dx=\sqrt{2ml^2}\cdot 4 \int_0^1 \sqrt{\frac{mgl\theta_{max}^2}{2}(1-x^2)}\theta_{max}dx=\\=\sqrt{2ml^2}\cdot 4\theta_{max} \sqrt{\frac{mgl\theta_{max}^2}{2}}\int_0^1 \sqrt{1-x^2}dx=\sqrt{2ml^2}\cdot 4\theta_{max}^2 \sqrt{\frac{mgl}{2}}\int_0^1 \sqrt{1-x^2}dx=\\=\sqrt{\frac{2m^2gl^3}{2}}\cdot 4\theta_{max}^2\int_0^1 \sqrt{1-x^2}dx=\sqrt{gl}\cdot 4ml\theta_{max}^2\int_0^1 \sqrt{1-x^2}dx$$
If we assume $\frac{\sqrt{2}}{\pi}=\sqrt{l}$ then we have the RHS but without any other assumptions I cannot do more..