# Need help with yet another integral...

#### 123qwerty

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!

#### fysmat

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.

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#### v8archie

Math Team
In other words, by distributing over $(1-x^2)$ you will form a reduction formula.

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#### 123qwerty

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...)

#### 123qwerty

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