How to Integrate this function $\int(1-x^2)^ndx$

Question

The actual question was to find the area bounded by the curve $y=\int_{-1}^{1}(1-x^2)^ndx$ and coordinate axes. But I haven't came across these type of problems with power $n$.

Answer 1

By parts,

$$I_n=\int(1-x^2)^ndx=x(1-x^2)^n+2n\int x^2(1-x^2)^{n-1}dx.$$

But $x^2=1-(1-x^2)$, so that

$$I_n=x(1-x^2)^n+2nI_{n-1}-2nI_n,$$ and

$$I_n=\frac{x(1-x^2)^n}{2n+1}+\frac{2n}{2n+1}I_{n-1}.$$

With $I_0=x$, you can compute for any $n$.

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When evaluating the definite integral, the first term vanishes and

$$I_n=\frac{2(2n)!!}{(2n+1)!!}.$$

Answer 2

Using the binomial theorem, $(1-x^2)^n=\sum_{k=0}^n\binom{n}{k}(-1)^kx^{2k}$ and therefore

$$\int (1-x^2)^n\,dx=\sum_{k=0}^n\binom{n}{k}(-1)^k\frac{x^{2k+1}}{2k+1}+C$$

and

$$\int_{-1}^1 (1-x^2)^n\,dx=\sum_{k=0}^n\binom{n}{k}(-1)^k\frac{2}{2k+1}$$

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ALTERNATIVE APPROACH TO EVALUATE THE DEFINITE INTEGRAL:

First, note that $\int_{-1}^1 (1-x^2)^n\,dx=2\int_0^1 (1-x^2)^n\,dx$. Next, enforce the substitution $x\to \sqrt{x}$ to obtain

$$\begin{align} \int_{-1}^1 (1-x^2)^n\,dx&=2\int_0^1 (1-x)^n x^{-1/2}\,dx\\\\ &=2\,B(n+1,1/2)\\\\ &=2\frac{\Gamma(n+1)\Gamma(1/2)}{\Gamma(n+3/2)}\\\\ &=2\frac{n!\Gamma(1/2)}{(n+1/2)!\Gamma(1/2)}\\\\ &=2\frac{n!}{\frac32\,\frac52\,\frac72\,\cdots\,\frac{2n+1}{2}}\\\\ &=2\frac{2^n\,n!}{(2n+1)!!}\\\\ &=2\frac{(2n)!!}{(2n+1)!!} \end{align}$$

as expected!

Answer 3

This integral appears in the normalization of the Legendre Polynomials. $n$ is assumed to be a natural number. Here is the computation. First we make the change of variables $x = cos(t)$ and use pithagorean identity to get:

$$ I = \int_{-1}^1(1-x^2)^ndx = \int_{\pi}^0[\sin^2(t)]^n (-\sin t)dt = -\int_{\pi}^0\sin^{2n+1}(t) dt $$

Now we integrate by parts $n$ times, noting that the square bracket vanishes at both limits each time, getting: $$I = - \frac{2n}{2n+1} \frac{2n-2}{2n-1} \frac{2n-4}{2n-3} ... \frac{2}{3} \int_{\pi}^0\sin(t) dt = \frac{(2n)!!}{(2n+1)!!} · 2 $$ Now we can express the result in terms of factorials insetead of double factorials using:

$$ (2n)!! = 2^n n! $$ $$ (2n+1)!! = \frac{(2n + 1)!}{n! 2^n} $$

We finally get:

$$ I = \frac{2^{2n+1}(n!)^2}{(2n+1)!} $$

Answer 4

Let $$f(n)=\int_{-1}^1{\left(1-x^2\right)^n}\,\mathrm{d}x$$ It may be easier to calculate $f\left(n-\frac12\right)$ rather than $f(n)$.
With the substitution $x=\sin\theta$, we have $$\begin{align}f\left(n-\frac12\right)&=\int_{-1}^1{\left(1-x^2\right)^{n-\frac12}}\,\mathrm{d}x\\&=\int_{-1}^1{\left(1-x^2\right)^{\frac{2n-1}2}}\,\mathrm{d}x\\&=\int_{-\frac\pi2}^{\frac\pi2}\cos^{2n}\theta\,\mathrm{d}\theta\\&=\int_{-\frac\pi2}^{\frac\pi2}\frac1{2^{2n-1}}\sum_{r=1}^{n-1}{2n\choose r}\cos(\left(n-r\right)\theta)+\frac1{2^{2n}}{2n\choose n}\,\mathrm{d}\theta\\&=\frac\pi{2^{2n}}{2n\choose n}\\&=\frac{\pi(2n)!}{2^{2n}(n!)^2}\end{align}$$ From this we can conclude that $$f(n)=\frac{\pi(2n+1)!}{2^{2n+1}\left(\left(n+\frac12\right)!\right)^2}$$ Evaluating $\left(n+\frac12\right)!$, $$\begin{align}\left(n+\frac12\right)!&=\left(-\frac12\right)!\prod_{r=0}^n\frac{2r+1}2\\&=\frac1{2^{n+1}}\sqrt\pi\prod_{r=0}^n(2r+1)\\&=\frac{\sqrt\pi(2n+1)!}{2^{2n+1}n!}\end{align}$$ Therefore we can conclude that $$\begin{align}\int_{-1}^1\left(1-x^2\right)^n\,\mathrm{d}x&=f(n)\\&=\frac{\pi(2n+1)!}{2^{2n+1}}\cdot\frac{2^{4n+2}(n!)^2}{\pi((2n+1)!)^2}\\&=\frac{2^{2n+1}(n!)^2}{(2n+1)!}\end{align}$$ This is indeed equal to the $\frac{2(2n)!!}{(2n+1)!!}$ from the above solutions.