How to prove that $\int_{-1}^{+1} (1-x^2)^n dx = \frac{2^{2n+1}(n!)^2}{(2n+1)!}$

Question

I'am trying to prove that $$\int_{-1}^{+1} (1-x^2)^n\:\mathrm{d}x = \frac{2^{2n+1}(n!)^2}{(2n+1)!}$$

Here what I have done so far. We know that: \begin{equation} \sin^{2}x + \cos^{2}x = 1 \end{equation} Let $x = \sin\alpha$ so $\mathrm{d}x = \cos\alpha\:\mathrm{d}\alpha$

\begin{equation} x^{2} = \sin^{2}\alpha \end{equation}

\begin{equation} \cos^{2}x = 1 -\sin^{2}x \end{equation}

We have $x = 1 \Rightarrow \alpha = \frac{\pi}{2}$ and $x = -1 \Rightarrow \alpha = - \frac{\pi}{2}$, so we replace the original integral by:

$$\int_{-1}^{1} (x-1^2)^n\:\mathrm{d}x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n+1}\alpha\:\mathrm{d}\alpha = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos^{2n}\alpha) \cos\alpha\:\mathrm{d}\alpha$$

Now integrating by parts, $$u = \cos^{2n}\alpha \quad \Rightarrow\:\mathrm{d}v = \cos\alpha\:\mathrm{d}\alpha \\ \mathrm{d}u = -2n\cos^{2n-1}\alpha\sin\alpha\:\mathrm{d}\alpha \quad \Rightarrow v = \sin\alpha \\ \int udv = uv - \int vdu$$

Then, \begin{align}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos^{2n}\alpha) \cos\alpha\:\mathrm{d}\alpha &= \underbrace{\cos^{2n}\alpha \sin\alpha \Big|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}}_{\text{= 0}} + 2n\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n-1}\alpha \sin^{2} \alpha\:\mathrm{d}\alpha \\[10pt] &= 2n \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n-1}\alpha (1-\cos^2\alpha)\:\mathrm{d}\alpha \\ &= 2n\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n-1}\alpha\:\mathrm{d}\alpha - 2n\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n+1}\alpha\:\mathrm{d}\alpha \end{align}

Now we see that $$(2n+1)\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n+1}\alpha = 2n\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} cos^{2n-1}\alpha\:\mathrm{d}\alpha$$

I am stuck here, it's almost the same integral as the original one. Any suggestions or corrections will be very welcome. Thanks.

Answer 1

The same working with $n$ replaced by $n-1$ and then $n-2$ gives $$(2n-1)\int_{-\pi/2}^{\pi/2}\cos^{2n-1}\alpha\,d\alpha =(2n-2)\int_{-\pi/2}^{\pi/2}\cos^{2n-3}\alpha\,d\alpha$$ and $$(2n-3)\int_{-\pi/2}^{\pi/2}\cos^{2n-3}\alpha\,d\alpha =(2n-4)\int_{-\pi/2}^{\pi/2}\cos^{2n-5}\alpha\,d\alpha\ .$$ Therefore $$\eqalign{ \int_{-\pi/2}^{\pi/2}\cos^{2n+1}\alpha\,d\alpha &=\frac{2n}{2n+1}\int_{-\pi/2}^{\pi/2}\cos^{2n-1}\alpha\,d\alpha\cr &=\frac{2n}{2n+1}\frac{2n-2}{2n-1} \int_{-\pi/2}^{\pi/2}\cos^{2n-3}\alpha\,d\alpha\cr &=\frac{2n}{2n+1}\frac{2n-2}{2n-1}\frac{2n-4}{2n-3} \int_{-\pi/2}^{\pi/2}\cos^{2n-5}\alpha\,d\alpha\ .\cr}$$ Continuing in the same way gives eventually $$\eqalign{\int_{-\pi/2}^{\pi/2}\cos^{2n+1}\alpha\,d\alpha &=\frac{2n}{2n+1}\frac{2n-2}{2n-1}\frac{2n-4}{2n-3}\cdots\frac45\frac23 \int_{-\pi/2}^{\pi/2}\cos\alpha\,d\alpha\cr &=\frac{2n}{2n+1}\frac{2n-2}{2n-1}\frac{2n-4}{2n-3}\cdots\frac45\frac232 \ .\cr}$$ This is the value of the integral: we now need to simplify. First multiply top and bottom by $(2n)(2n-2)(2n-4)\cdots2$, then take out as many factors of $2$ as possible: the integral is $$I=\frac{2^{2n+1}n^2(n-1)^2(n-2)^2\cdots1^2}{(2n+1)(2n)(2n-1)\cdots(3)(2)} =\frac{2^{2n+1}(n!)^2}{(2n+1)!}\ .$$

Comments

• This is the method of integration by reduction formula (look it up).
• You don't actually need the trig substitution as you can integrate $$\int_{-1}^1(1-x^2)^n\,dx =\bigl[x(1-x^2)^n\bigr]_{-1}^1+2n\int_{-1}^1 x^2(1-x^2)^{n-1}\,dx\ ,$$ which leads to the alternative reduction formula $$\int_{-1}^1(1-x^2)^n\,dx =\frac{2n}{2n+1}\int_{-1}^1 (1-x^2)^{n-1}\,dx\ .$$ Then use this in the same way as above.

Answer 2

Consider the integral $\int_{0}^{+1} (1-x^2)^n dx$
Let $I(n)$ be the above integral do the substitution $x=\sin t$. Then $dx=\cos t dt$ and $1-x^2=cos^2 t$.

Also $t=0$ for $x=0$ and $t=\pi/2$ for $x=1$ so
$I(n) = \int_0^{\pi/2} \cos^{2n+1} t \ dt = \int_0^{\pi/2} \cos^{2n} t \ \cos t \ dt$.
To integrate by parts do the substitution

$$u = \cos^{2n} t$$ and $$dv=\cos t dt$$ From where we get
$$du = 2n \cos^{2n-1} t (-\sin t) dt $$ and
$$v=\sin t$$
Now, $uv$ is zero at both limits so

$$I(n) = 2n \int \cos^{2n-1} t \sin^2 t dt = 2n \int \cos^{2n-1} t (1-\cos^2 t) dt = 2n [ I(n-1) - I(n) ]$$ from where we get
$$I(n) = \frac{2n}{2n+1}I(n-1)$$
It is clear that $I(0)=1$ so $$I(n) = \frac{(2n) 2(n-1) 2(n-2) ... 2}{ [ (2n+1) (2n-1) ... 3]}$$ Multiplying both the numerator and the denominator by $(2n) 2(n-1) 2(n-2) ... 2$ the denominator becomes $(2n+1)!$ where as the numerator is the square of $2^n n!$ So, finally $I(n)= \frac{2^{2n} (n!^2)} { (2n+1)!}$
The asked integral is just the double of the computed integral.

Answer 3

If I try to prove it for $n=1$ I get: $$\int_{-1}^1(1-x^2)dx=(x-x^3/3)|_{-1}^{1}=2-2/3=4/3\ne\frac{3!1!^2}{2^3}=\frac{3}{4}$$