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Can someone tell me how to evaluate this integral please?

$$\int_{-1}^1 (1-x^2)^k, k \in \mathbb{N}$$

I tried using the substitution x = sin(t), which would allow me to express this as:

$$\int_{-1}^1 cos^{2k+1}(t) dt$$

but this doesn't really help. Any other tricks?

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    $\begingroup$ Use Newton's Binomial Theorem to expand $(1-x^2)^{k}$ into a polynomial, and then integrate each summand, which is of the form $x^{n}$. $\endgroup$ – avs May 31 '17 at 21:48
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    $\begingroup$ Note: the integral would actually become $$\int_{-\pi}^\pi\cos^{2k+1}(t)\,dt$$ under that substitution. $\endgroup$ – John Doe May 31 '17 at 21:50
  • $\begingroup$ $\int_{-1}^{1} (1-x^2)^n dx = \frac{2^{2n+1}(n!)^2}{(2n+1)!}$ ... This is a duplicate from several weeks ago ... still delightful result ?! $\endgroup$ – Donald Splutterwit May 31 '17 at 21:53
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Here is a standard route using integration by parts, $$\begin{align} I_k&=\int_{-1}^{1}(1-x^2)^kdx \\\\&=\left[x(1-x^2)^k\right]_{-1}^{1}+2k\int_{-1}^{1}x^2(1-x^2)^{k-1}dx \\\\&=0+2k\int_{-1}^{1}\left[(1-(1-x^2))(1-x^2)^{k-1}\right]dx \\\\&=2kI_{k-1}-2kI_{k} \end{align} $$ then, with $I_0=2,\,I_1=\frac43,$ one gets $$ I_{k}=\frac{2k}{2k+1}\cdot I_{k-1}, \quad k\ge1, $$ finally

$$ I_k=\int_{-1}^{1} (1-x^2)^k dx = \frac{2^{2k+1}(k!)^2}{(2k+1)!}, \quad k\ge1. $$

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Enforcing the substitution $x\to x^{1/2}$ reveals

$$\begin{align} \int_0^1 (1-x^2)^k\,dx&=2\int_0^1(1-x^2)^k\,dx\\\\ &=\int_0^1 x^{-1/2}(1-x)^k\,dx\\\\ &=B(1/2,k+1)\\\\ &=\frac{\Gamma(1/2)\Gamma(k+1)}{\Gamma(k+3/2)}\\\\ &=\frac{\sqrt{\pi}\,k!}{(k+1/2)\Gamma(k+1/2)}\\\\ &=\frac{\sqrt{\pi}\,k!\Gamma(k)}{(k+1/2)2^{1-2k}\sqrt{\pi}\Gamma(2k)}\\\\ &=2\frac{4^k(k!)^2}{(2k+1)!} \end{align}$$

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When we expand $$(1-x^2)^k$$ into a binomial, we get $$_kC_0-_kC_1x^2+_kC_2x^4-...+_kC_kx^{2k}(-1)^k$$ and when we integrate termwise, we get $$_kC_0x-\frac{1}{3}{_k}C_1x^3+\frac{1}{5}{_k}C_2x^5-...+\frac{1}{2k}{_k}C_kx^{2k}(-1)^k$$ Now, since we are integrating from $-1$ to $1$, and since the function is symmetric about the $y$-axis, it is the same as twice the integral from $0$ to $1$, which is easily evaluated: $$2\bigg({_k}C_0-\frac{1}{3}{_k}C_1+\frac{1}{5}{_k}C_2-...+\frac{1}{2k+1}{_k}C_k(-1)^{k}\bigg)$$ And with the help of Wolfram Alpha (thanks, Wolfram!) we find this to be $$\frac{2(2k)!!}{(2k+1)!!}$$

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  • $\begingroup$ Did you mean to write $!!$? This doesn't agree with other answers $\endgroup$ – John Doe May 31 '17 at 22:03
  • $\begingroup$ @JohnDoe .... double factorial notation ... mathworld.wolfram.com/DoubleFactorial.html $\endgroup$ – Donald Splutterwit May 31 '17 at 22:21
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    $\begingroup$ @DonaldSplutterwit Ohh, I see - I have never seen this before, thanks $\endgroup$ – John Doe May 31 '17 at 22:36
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This is simply a Beta function. Using the substitution $t=x^2$, this reduces to

$$2\int_0^1 \frac{1}{2\sqrt{t}} (1-t)^k\ dt$$

$$\int_0^1 t^{-\frac{1}{2}} (1-t)^k$$

$$B\left(\frac{1}{2},k+1\right)$$

which can be expressed in a lot of different ways (see the linked Wikipedia page).

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