How to evaluate this definite integral $\int_0^2(1-x^2)^\frac{1}{3}~dx$ A student asked me to help him calculate this definite integral
$$\int_0^2(1-x^2)^\frac{1}{3}~dx$$
Although I have tried almost all the methods I have learned, I can not still do with it. I have tried the change of variable $x=\sec t$ , and the method of integral by parts. Can anyone help me?
 A: No, no one can help your student calculate this definite integral. It is too horribly nasty by far. Tell him to wait till he gets to grad school and see if he's still interested in awful integrals then.
A: Let $x = 1 -u^3$, we can rewrite the integral $\mathcal{I}$ as
$$\mathcal{I} = \int_0^2\sqrt[3]{1-x^2}dx 
= 3\sqrt[3]{2}\int_{-1}^1 u^3 \left(1 - \frac{u^3}{2}\right)^{\frac13} du\\
$$
Since the power series expansion of $\left(1 - \frac{u^3}{2}\right)^{\frac13}$ at $u = 0$ has radius of convergence $> 1$, we can expand it inside the integral sign and evaluate 
the expansion term by term.  We have
$$\begin{align}
\mathcal{I} \stackrel{[\color{blue}{1}]}{=}& 3\sqrt[3]{2}\int_{-1}^1 u^3 \sum_{k=0}^{\infty} \frac{\left(-\frac13\right)_k}{k!}\left(\frac{u^3}{2}\right)^k du\\
=&3\sqrt[3]{2}\sum_{k=0}^\infty\frac{(-\frac13)_k}{k!}\frac{1}{3k+4}\left[\left(\frac12\right)^k - \left(-\frac12\right)^k\right]\\
\stackrel{[\color{blue}{2}]}{=}&\frac{3\sqrt[3]{2}}{4}\sum_{k=0}^\infty\frac{(-\frac13)_k (\frac43)_k}{k!(\frac73)_k}
\left[\left(\frac12\right)^k - \left(-\frac12\right)^k\right]\\
=&\frac{3\sqrt[3]{2}}{4}\left[\,_2F_1(-\frac13,\frac43;\,\frac73;\,\frac12) -\,_2F_1(-\frac13,\frac43;\,\frac73;\,-\frac12)\right]
\end{align}
$$
Throwing the last expression to WA give us
$$\mathcal{I} \sim -0.18490339160722117817295686175099263891533938048269736635284...$$
consistent with what will get if you ask WA to numerically evaluate the original integral.
Notes
$[\color{blue}{1}]$ $(\alpha)_k = \alpha(\alpha+1)\cdots(\alpha+k-1)$ is the rising Pochhammer symbol.
$[\color{blue}{2}]$ We are using the identity $\frac{(\gamma)_k}{(\gamma+1)_k} = \frac{\gamma}{\gamma+k}$ here.
