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I want to evaluate $$ \int_0^1 ( 1 - x^2)^{10} dx $$ One way I can do this is by expanding out $(1 - x^2)^{10}$ term by term, but is there a better way to do this?

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  • $\begingroup$ Yes, by coming up with the reduction formula! $\endgroup$
    – user209663
    Commented Jun 12, 2019 at 21:31

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This answer is exactly what the asker is NOT asking for. However, I believe it will be useful to show the asker that expanding out "term-by-term" is not actually that hard.

We have

\begin{align} \int_0^1 (1-x^2)^{10}dx &= \int_0^1 \sum_{i=0}^{10} C(10,i)(-x^2)^idx \\ % & = \sum_{i=0}^{10} (-1)^iC(10,i)\int_0^1 x^{2i}dx\\ % &= \sum_{i=0}^{10} \frac{(-1)^iC(10,i)}{2i+1}, \end{align}

where $$C(n,m) = \frac{n!}{m!(n-m)!}.$$

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An alternate approach is through the beta function. We start by u-substitution and try to make our way towards an integral form of the beta function:

$$\begin{align*} \int_0^1(1-x^2)^{10}\,\mathrm dx &= \frac{1}{2} \int_0^1 (1-u)^{10}u^{-1/2}\,\mathrm du,\qquad x=\sqrt{u} \\ &= \frac{1}{2} \int_0^1 (1-u)^{11-1}u^{1/2-1}\,\mathrm du\\ &= \frac{1}{2}\cdot\frac{\Gamma(11)\Gamma(1/2)}{\Gamma(11+1/2)}\\ &= \frac{1}{2}\cdot\frac{10!\sqrt{\pi}}{\frac{22!}{4^{11}\cdot11!}\sqrt{\pi}}\\ &= \frac{262144}{969969}. \end{align*}$$

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    $\begingroup$ Upvoted. There's a special place in my heart for using special functions to solve problems that look like they only admit basic methods :-) $\endgroup$
    – JMJ
    Commented Jun 12, 2019 at 22:10
  • $\begingroup$ what formula did you use to calculate $\Gamma(11.5)$? $\endgroup$
    – Henry Lee
    Commented Jun 14, 2019 at 23:34
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    $\begingroup$ @Henry Lee It's a rearrangement of the Legendre duplication formula. $\endgroup$
    – dxdydz
    Commented Jun 14, 2019 at 23:36
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    $\begingroup$ @Henry Lee If you like that, it gets even better. The duplication formula is a special case of that one. $\endgroup$
    – dxdydz
    Commented Jun 14, 2019 at 23:46
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    $\begingroup$ So it has a general formula! These are the kinds of equations I love finding but never bump into doing my usual work. @dxdydz thank you so much $\endgroup$
    – Henry Lee
    Commented Jun 14, 2019 at 23:49
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Similar to dxdydz's answer, let $x=\sin(t)$ $$\int_0^1 ( 1 - x^2)^{10}\, dx=\int_0^{\frac \pi 2}\cos^{21}(t)\,dt$$ and remember that $$\int_0^{\frac \pi 2}\cos^{n}(t)\,dt=\frac{\sqrt{\pi }}2 \frac{ \Gamma \left(\frac{n+1}{2}\right)}{ \Gamma \left(\frac{n+2}{2}\right)}$$

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When you encounter this type of problem, the first thing to do is to come up with a reduction formula. Let us find the reduction formula for $\int(k^2 - x^2)^n dx $.

Using integration by parts: $u = (k-x^2)^n $ and $dv = dx$ then $du = n(k^2 - x^2)^{n-1}(-2x)dx $ and $v =x$

Thus,

$$\int(k^2 - x^2)^n dx = x(k^2 - x^2)^{n-1} + 2n \int x^2(k^2 - x^2)^{n-1} \\ = x(k^2 - x^2)^{n-1} + 2n \int \bigg[ k^2(k^2 - x^2)^{n-1} - (k^2 - x^2)^{n} \bigg] dx \\ = x(k^2 - x^2)^{n-1} + 2n \int k^2(k^2 - x^2)^{n-1} dx - 2n \int (k^2 - x^2)^{n} dx $$ From here you can see that: $$2n\int(k^2 - x^2)^n dx + \int(k^2 - x^2)^n dx = x(k^2 - x^2)^{n-1} + 2n \int k^2(k^2 - x^2)^{n-1} dx - 2n \int (k^2 - x^2)^{n} dx $$

Thus, $$\int(k^2 - x^2)^n dx = \frac{x(k^2 - x^2)^{n-1}}{2n+1} + \frac{2nk^2}{2n+1} \int (k^2 - x^2)^{n-1} dx$$

For your problem, $k=1$ and $n=10$. Can you take it from here?

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  • $\begingroup$ wow! I knew there must be a shorter way. Thank you! $\endgroup$
    – user680412
    Commented Jun 12, 2019 at 21:45
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    $\begingroup$ Is this really a shorter way? Evaluating all of the terms of the recursion seems like just as much effort as simply using the binomial expansion. $\endgroup$
    – JMJ
    Commented Jun 12, 2019 at 21:55
  • $\begingroup$ @MathStudent No problem. $\endgroup$
    – user209663
    Commented Jun 12, 2019 at 21:58
  • $\begingroup$ @SZN Note that this is a definite integral... look at the recursion.... what happen to the first part? This is similar to $\int_0^{\pi} sin^n(x) dx$. You also get a recursion... but the trick is the bound of the integral. It makes it easier to evaluate $\endgroup$
    – user209663
    Commented Jun 12, 2019 at 21:58
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    $\begingroup$ @user209663 what you say is perfectly fine. But by using the binomial expansion I can evaluate the integral in just a few lines without any recursion or argument about bounds. $\endgroup$
    – JMJ
    Commented Jun 12, 2019 at 22:06

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