Integrate $x\sqrt{1-x^3}$ I used integration by parts, with
$$u = x, du = dx$$
$$dv = \sqrt{1-x^3}dx, v = ??$$
Then to find v, I would need to integrate $\sqrt{1-x^3}dx$. I immediately thought to use trig substitution, but with the $x^3$, it didn't work out.
How would I continue this method if it is correct, or is there another way to approach solving this question?
 A: This has no elementary antiderivative. It can be written as an Incomplete Beta Function:
$$
\begin{align}
\int_0^xt\sqrt{1-t^3}\,\mathrm{d}t
&=\frac13\int_0^{x^3}t^{-1/3}\left(1-t\right)^{1/2}\,\mathrm{d}t\\[6pt]
&=\frac13\operatorname{B}\left(x^3;\frac23,\frac32\right)
\end{align}
$$
If $x=1$, we can write this as a ratio of Gamma functions:
$$
\begin{align}
\frac13\operatorname{B}\left(1;\frac23,\frac32\right)
&=\frac13\operatorname{B}\left(\frac23,\frac32\right)\\
&=\frac13\frac{\Gamma\left(\frac23\right)\Gamma\left(\frac32\right)}{\Gamma\left(\frac{13}6\right)}\\
&=\frac{6\sqrt\pi}7\frac{\Gamma\left(\frac23\right)}{\Gamma\left(\frac16\right)}
\end{align}
$$
A: As said in comments,  this is a very difficult integral, the solution of which involving elliptic integrals (which are not the most funny things to handle).
One solution could be to expand the integrand as a Taylor series (or use the general binomial theorem) using $$\sqrt{1+y}
=(1+y)^{1/2}
= \sum_{k=0}^{\infty} \binom{1/2}{k}y^k$$ Replacing $y$ by $-x^3$, this gives $$\sqrt{1-x^3}=\sum_{k=0}^{\infty} \binom{1/2}{k}(-1)^k x^{3k}$$ So $$x\sqrt{1-x^3}=\sum_{k=0}^{\infty} \binom{1/2}{k}(-1)^k x^{3k+1}$$ and integrating $$\int x\sqrt{1-x^3}\,dx=\sum_{k=0}^{\infty} \binom{1/2}{k}(-1)^k \int x^{3k+1}\,dx=\sum_{k=0}^{\infty} \binom{1/2}{k}(-1)^k \frac {x^{3k+2}}{3k+2}$$ 
Edit
As robjohn commented, it is worth to remember that $$\binom{1/2}{k}=\frac{(-1)^{k-1}}{4^k(2k-1)}\binom{2k}{k}$$ All of this makes $$\int x\sqrt{1-x^3}\,dx=-\sum_{k=0}^{\infty} \frac{ \binom{2 k}{k} }{4^k(2 k-1) (3 k+2)}x^{3 k+2}$$ We could also write $$\binom{1/2}{k}=\frac{\sqrt{\pi }}{2 \,\Gamma \left(\frac{3}{2}-k\right)\,k!}$$
