general solution to finite integrals of the form $\int_{-\sqrt{a}}^{\sqrt{a}} (a-x^2)^b dx$? Recently I was working with higher dimensional spheres and I found the value of this integral:
$$\int_{-\sqrt{a}}^{\sqrt{a}} (a-x^2)^b dx.$$
Although in a way I was working backward. I made this integral from a relationship I already managed to establish. I was wondering if it is possible to find the value of this integral using any integration methods, as in if you were just given this integral would you be able to find its value. I tried some integral calculators online, that didn't give any results. I tried doing it myself and didn't find anything either, however, integration isn't really my area.
 A: 
I was wondering if it is possible to find the value of this integral

If you are Looking For an answer, I have it,(From Mathematica)

Conditional expression Just Meant that there are extra condition involved and that has been Mentioned.
A: The simplest approach would be to use integration by parts, which is also used for deriving the Wallis Product for similar integrals.
Let $I(b) = \int_{-\sqrt{a}}^{\sqrt{a}} (a-x^2)^b dx$, $v'=1$ and $u=(a-x^2)^b$, then $\frac{du}{dx}=-2bx(a-x^2)^{b-1}$. $I(0)=\int_{-\sqrt{a}}^{\sqrt{a}}  dx=2\sqrt{a}$.
$$I(b) = [x(a-x^2)^b]_{-\sqrt{a}}^{\sqrt{a}} - \int_{-\sqrt{a}}^{\sqrt{a}} x(-2bx)(a-x^2)^{b-1} dx$$
$$I(b) = - 2b\int_{-\sqrt{a}}^{\sqrt{a}} (a-x^2)^b + 2ab\int_{-\sqrt{a}}^{\sqrt{a}} (a-x^2)^{b-1} dx$$
$$I(b) = - 2bI(b) + 2abI(b-1)$$
$$I(b) = \frac{2ab}{2b+1}I(b-1)$$
$$I(b) = \frac{2ab}{2b+1}.\frac{2a(b-1)}{2b-1}...\frac{2a(2)}{2(2)+1}\frac{2a(1)}{2(1)+1} I(0)$$
A: The substitution $$x = \sqrt{a}(2u-1), \quad dx = 2 \sqrt{a} \, du,$$ results in the integral $$(2 \sqrt{a})^{2b+1} \int_{u=0}^1 u^b (1-u)^b \, du.$$  This is proportional to a beta integral, whose value is $$(2 \sqrt{a})^{2b+1} \frac{\Gamma(b+1)^2}{\Gamma(2b+2)}.$$  When $b \in \mathbb Z^+$, this is expressible in factorials as $$(2 \sqrt{a})^{2b+1} \frac{(b!)^2}{(2b+1)!} = \frac{(2 \sqrt{a})^{2b+1}}{(b+1) \binom{2b+1}{b}}.$$
A: If you enjoy hypergeometric functions, assuming $a>0$ and $b>0$
$$\int (a-x^2)^b\, dx=a^b\,x\,\, _2F_1\left(\frac{1}{2},-b;\frac{3}{2};\frac{x^2}{a}\right)$$
$$\int_{-t}^t (a-x^2)^b\, dx=2  a^b\,t \, _2F_1\left(\frac{1}{2},-b;\frac{3}{2};\frac{t^2}{a}\right)$$ If $t=\sqrt a$, this leads to the result already given in answers.
