Need help with yet another integral...

Dec 2012
850
310
Hong Kong
Let \(\displaystyle \alpha = \int_0^1 (1-x^2)^{10} dx \) and \(\displaystyle \beta = \int_0^1 (1-x^2)^9 dx \)

Then express \(\displaystyle \alpha\) in terms of \(\displaystyle \beta\).

(Hint: Write \(\displaystyle (1 - x^2)^{10} = (1-x^2)^9(1-x^2)\) and use integration by parts.)

I feel dumb for not being able to do this...

This is what I've got:

\(\displaystyle
\begin{align*}
\alpha &= \int_0^1 (1-x^2)^9 (1-x^2) dx \\
&= \int_0^{2/3} (1-x^2)^9 d(x-\frac{x^3}{3}) \\
&= (1-x^2)^9(x-\frac{x^3}{3})|_0^{2/3} - \int_0^{2/3} (x-\frac{x^3}{3}) d(1-x^2)^9 \\
&= (1-x^2)^9(x-\frac{x^3}{3})|_0^{2/3} - \int_1^{\sqrt{1 - (2/3)^{1/9}}} 9(-2x)(1-x^2)^8 (x-\frac{x^3}{3}) dx
\end{align*}
\)

At this point, the problem has become such an intractable mess that I have no idea how to proceed (thoguh I've attempted several times). Is there something wrong I've done here? Thanks!
 
Dec 2013
212
72
some subspace
Hint: Multiply first, use integration by parts once you see something useful. You should first find something which is directly \(\displaystyle \beta\), and something which turns out to give \(\displaystyle \alpha\) once you have integrated by parts. Then you can write an equation and solve \(\displaystyle \alpha\).

A side note: I don't understand how you've determined the limits of the integration. Integration by parts doesn't affect the limits.
 
Last edited by a moderator:
  • Like
Reactions: 1 person

v8archie

Math Team
Dec 2013
7,709
2,677
Colombia
In other words, by distributing over $(1-x^2)$ you will form a reduction formula.
 
  • Like
Reactions: 1 person
Dec 2012
850
310
Hong Kong
Ah, I see now. Thanks guys! I knew I'd done something wrong. (The distribution thing is something I've done with trigonometric integrals before, funny that I didn't see it this time.)

As for the limits - according to my understanding, isn't that how it's supposed to work? I only changed the limits when I used substitution, not when I integrated by parts. (I'm not completely sure, because when I solve definite integrals, I generally solve the indefinite one first, *then* write in the limits and use the second fundamental theorem...)
 
Dec 2012
850
310
Hong Kong
Never mind about the limits. I looked through the formal statement of the integration by parts theorem and I get it now. The udv/vdu thing can be a bit misleading :p