How to solve $\int_{-4}^{4}x^2\sqrt{16-x^2 }dx$ with beta function

Question

I need to solve the integral with use of the beta function.

I have tried derationalization of the square root, got nowhere. I tried some substitutions, but the thing is I never got the bounds right.

I would need help solving this.

Answer 1

Notice that integrated function $f(x)$ is even. Thus this integral equals to $2\int_0^4 f(x) dx$. Then make variable replacement $t=\frac{x^2}{16}$.

Answer 2

$$I=\int_{-4}^{4}x^2\sqrt{16-x^2 }dx=2\int_{0}^{4}x^2\sqrt{16-x^2 }dx=8\int_0^4x^2\sqrt{1-\frac{x^2}{16}}dx$$ Now make substitution $u=\frac{x^2}{16}$, $du=\frac18xdx$. The $u$ limits are from $0$ to $1$.

Answer 3

Let $x=4u$ to transform the integral to $$16^2\int_{-1}^1u^2\sqrt{1-u^2}\,du=2\cdot16^2\int_0^1u^2\sqrt{1-u^2}\,du$$ Then $v=u^2,du=\frac1{2\sqrt v}\,dv$: $$=16^2\int_0^1v^{1-1/2}(1-v)^{1/2}\,dv=16^2\mathrm B(3/2,3/2)=32\pi$$