Evaluate : $\int_0^1 \frac{dx}{\sqrt[3]{1-x^3}}$. Evaluate : $$ \int_0^1 \frac{\mathrm{d} x}{\sqrt[3]{1-x^3}}.$$

if you feel that this integral is easy, just post hints.
 A: Hint: Make the change of variables $ t=x^3 $ and then use the beta function.
Added: Another way to go is by expanding the integrand in terms of power series as 
$$ \int_0^1 \frac{\mathrm{d} x}{\sqrt[3]{1-x^3}}.= \sum_{k=0}^{\infty}{-\frac{1}{3} \choose k}(-1)^k \int_{0}^{1} x^{3k} dx = \dots, $$
where $ {n \choose k}=\frac {n!}{(n-k)!k!} .$ I think you can proceed now.
A: Substitute $u = 1-x^3$ and recall the definition of a beta function:
$$B(m,n) = \int_0^1 du\: u^{m-1} (1-u)^{n-1} = \frac{\Gamma(m) \Gamma(n)}{\Gamma(m+n)}$$
A: You do not need any beta- or gamma-functions for this integral. I will deal with indefinite integral, you should take care of the limits.
Use the substitution 
$$
1-\frac{1}{x^3}=t^3.
$$
Then, you have
$$
x=\frac{1}{\sqrt[3]{1-t^3}},\quad dx=\frac{t^2\,dt}{(1-t^3)^{4/3}},\quad \sqrt[3]{1-x^3}=-\frac{t}{\sqrt[3]{1-t^3}}.
$$
Plugging everything in, you get
$$
-\int \frac{t}{1-t^3}dt,
$$
which can be integrated using standard methods (e.g., partial fractions).
The big question is how would I know about this substitution? You can find a recipe in this post.
