Beta function solution for $\int_0^2(16-x^2)^{-3/2}\,dx$

Question

$$\int_0^2(16-x^2)^{-3/2}\,dx$$ I need to solve this integral by using the beta function. So firstly I substituted $x=4t$, and my limits change to $[0,1/2]$. Then I substited $t^2=u$, $t=\sqrt u$. And my limits change again to $[0,1/4]$. Now I have $$\int_0^{1/4}u^{1/2-1}(1-u)^{-1/2-1}\,du$$ but I need positive values of power minus one in order to get beta function. And limits should be 0 to 1. Did I made mistake somewhere or I need another substitution of $u$?

Answer 1

You may notice that the integral $$ \mathcal{J}=\int_{0}^{1/4}\sqrt{\frac{u}{1-u}}\frac{du}{u(1-u)}$$ greatly simplifies after the substitution $\frac{u}{1-u}=v$, i.e. $u=\frac{v}{1+v}$, $du=\frac{dv}{(1+v)^2}$: $$ \mathcal{J}=\int_{0}^{1/3}\frac{dv}{\sqrt{v}}\stackrel{v=z/3}{=}\frac{1}{\sqrt{3}}\int_{0}^{1}\frac{dz}{\sqrt{z}}=\frac{B\left(1,\tfrac{1}{2}\right)}{\sqrt{3}}=\frac{2}{\sqrt{3}}. $$

Answer 2

substitute $$x=4\sin(t)$$ then we get $$dx=4\cos(t)dt$$