Improper Integral Calculation with Lots of Constants

Solution 1:

The idea is to rewrite it this way : Let's keep $4\sqrt{m}$ away for a moment :

\begin{align} \int_{0}^{\sqrt{2E/k}}{\frac{1}{\sqrt{2E-kx^2}}}dx &=\int_{0}^{\sqrt{2E/k}}{\frac{1}{\sqrt{2E}}\frac{1}{\sqrt{1-kx^2/2E}}}dx\\ &=\frac{1}{\sqrt{2E}}\int_{0}^{\sqrt{2E/k}}{\frac{1}{\sqrt{1-[(\sqrt{k/2E})x]^2}}dx} \end{align} If we dentoe $\alpha=\sqrt{k/2E}$ then we'll have $$\frac{1}{\sqrt{2E}}\int_{0}^{1/\alpha}{\frac{1}{\sqrt{1-(\alpha x)^2}}dx}$$

Which is the $Arcsin$ primitive : $$\frac{1}{\sqrt{2E}}\int_{0}^{1/\alpha}{\frac{1}{\sqrt{1-(\alpha x)^2}}dx}=\frac{1}{\sqrt{2E}}\Bigg[\frac{1}{\alpha}arcsin(\alpha x)\Bigg]_0^\frac{1}{\alpha}$$ Which yields : $$\frac{1}{\sqrt{2E}}\Bigg[\frac{1}{\alpha}arcsin(\alpha x)\Bigg]_0^\frac{1}{\alpha}=\frac{1}{\sqrt{2E}}\Bigg[\frac{1}{\alpha}arcsin(1)-\frac{1}{\alpha}arcsin(0)\Bigg]$$

Since $arcsin(1)=\frac{\pi}{2}$ and $arcsin(0)=0$, and by replacing $\alpha$ with $\sqrt{k/2E}$, we get $$\int_{0}^{\sqrt{2E/k}}{\frac{1}{\sqrt{2E-kx^2}}}dx =\frac{1}{\sqrt{2E}}\frac{1}{\sqrt{k/2E}}\frac{\pi}{2}=\frac{1}{\sqrt{2E}}\sqrt{\frac{2E}{k}}\frac{\pi}{2}$$ We are left with $$\frac{\pi}{2\sqrt{k}}$$ By multiplying by the $4\sqrt{m}$ left at the begining, we finally get : $$T=4\sqrt{m}\frac{\pi}{2\sqrt{k}}=2\pi \sqrt{\frac{m}{k}}$$