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

if you feel that this integral is easy, just post hints.

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But can't this be done without knowledge of Beta or Gamma function? Answer $2\pi 3^{-3/2}$ I guess. – GEdgar Mar 20 '13 at 20:48
Okay, but just wait a moment, maybe someone has an idea, this problem is from "Klub Prepa". – aziiri Mar 20 '13 at 20:53
@GEdgar Yes, it can be done, see my answer below. – Artem Mar 21 '13 at 2:15
up vote 2 down vote accepted

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.

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I like it. It really is elementary. – GEdgar Mar 21 '13 at 3:30

Substitute $u = 1-x^3$ and recall the definition of a beta function:

$$B(m,n) = \int_0^1 du\: u^{m-1} (1-u)^{n-1} = \frac{\Gamma(m) \Gamma(n)}{\Gamma(m+n)}$$

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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.

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thank you, but this was intended for Analysis II, I think there is a solution without the beta function. – aziiri Mar 20 '13 at 20:45
@aziiri I learn beta function on analysis II :) – Cortizol Mar 20 '13 at 20:47
in our university, Analysis II is in the second semester of the first year (LMD system), Beta function is not used at this level. – aziiri Mar 20 '13 at 20:50
@aziiri: See the added. – Mhenni Benghorbal Mar 20 '13 at 20:56
So, you don't have to know Gamma function, but [for your definition ${n \choose k}=\frac {n!}{(n-k)!k!}$] you do have to know $(1/3)!$ ... – GEdgar Mar 20 '13 at 21:07

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