This week's question was correctly answered by anemone and MarkFL. You can find both of their solutions below.
anemone's solution: [sp]We're asked to prove that [math]\int_0^1(1-x^2)^n dx=\frac{2^{2n}(n!)^2}{(2n+1)!}[/math].
First, let's examine the LHS expression, the definite integral, [math]\int_0^1(1-x^2)^n dx[/math],
Using the following trigonometric substitution,
[math]x=\sin \theta[/math] [math]\rightarrow dx=\cos \theta d\theta[/math]
The integral is now
[math]\int_0^{\frac{\pi}{2}} \cos^{2n} \theta \cos \theta d\theta=\int_0^{\frac{\pi}{2}} \cos^{2n+1} \theta d\theta[/math]
Using the integration by parts with the following substitution:
[math]u=\cos^{2n} \theta\;\;\rightarrow\;\;\frac{du}{d\theta}=2n (\cos^{2n-1} \theta)(-\sin \theta)[/math]
[math]\frac{dv}{d\theta}=\cos \theta\;\;\rightarrow\;\;v=\sin \theta[/math]
The integral is then
[math]\int_0^{\frac{\pi}{2}} \cos^{2n+1} \theta d\theta=\left[(\cos^{2n} \theta)(\sin \theta) \right]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} 2n (\cos^{2n-1} \theta)(-\sin \theta)(\sin \theta)d \theta[/math]
[math]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=(0)+2n\int_0^{\frac{\pi}{2}} (\sin^2 \theta)(\cos^{2n-1} \theta)d \theta[/math]
[math]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=2n\int_0^{\frac{\pi}{2}} (1-\cos^2 \theta)(\cos^{2n-1} \theta)d \theta[/math]
[math]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=2n\int_0^{\frac{\pi}{2}} (\cos^{2n-1} \theta)d \theta-2n\int_0^{\frac{\pi}{2}} (\cos^{2n+1} \theta)d \theta[/math]
[math](2n+1)\int_0^{\frac{\pi}{2}} \cos^{2n+1} \theta d\theta =2n\int_0^{\frac{\pi}{2}} (\cos^{2n-1} \theta)d \theta[/math]
[math]\int_0^{\frac{\pi}{2}} \cos^{2n+1} \theta d\theta =\frac{2n}{2n+1}\int_0^{\frac{\pi}{2}} (\cos^{2n-1} \theta)d \theta[/math]
If we let [math]I_n=\int_0^1(1-x^2)^n dx=\int_0^{\frac{\pi}{2}} \cos^{2n+1} \theta d\theta[/math], we now have
[math]I_n=\frac{2n}{2n+1}I_{n-1}[/math]
[math]I_n=\frac{2n}{2n+1}I_{n-1}[/math]
[math]\;\;\;\;=\frac{2n}{2n+1}\cdot\frac{2(n-1)}{2(n-1)+1}I_{n-2}[/math]
[math]\;\;\;\;=\frac{2n}{2n+1}\cdot\frac{2(n-1)}{2(n-1)+1}\cdot\frac{2(n-2)}{2(n-2)+1}I_{n-3}[/math]
[math]\;\;\;\;=\frac{2n}{2n+1}\cdot\frac{2(n-1)}{2(n-1)+1}\cdot\frac{2(n-2)}{2(n-2)+1}\cdots\frac{6}{7}\cdot\frac{4}{5}\cdot\frac{2}{3}[/math]
[math]\;\;\;\;=\left(\frac{2n}{2n+1}\cdot\frac{2(n-1)}{2(n-1)+1}\cdot\frac{2(n-2)}{2(n-2)+1}\cdots\frac{6}{7}\cdot\frac{4}{5}\cdot\frac{2}{3} \right)\left(\frac{2n}{2n}\cdot\frac{2(n-1)}{2(n-1)}\cdot\frac{2(n-2)}{2(n-2)}\cdots\frac{6}{6}\cdot\frac{4}{4}\cdot\frac{2}{2} \right)[/math]
[math]\;\;\;\;=\frac{((2n)(2(n-1))(2(n-2))\cdots(6)(4)(2))^2}{(2n+1)(2n))\cdots(6)(5)(4)(3)(2)(1)}[/math]
[math]\;\;\;\;=\frac{(2^n(n)(n-1)(n-2)\cdots(3)(2)(1))^2}{(2n+1)!}[/math]
[math]\;\;\;\;=\frac{(2^n(n!))^2}{(2n+1)!}[/math]
[math]\;\;\;\;=\frac{2^{2n}(n!)^2}{(2n+1)!}[/math]
and we're done![/sp]
MarkFL's solution: [sp]Let:
$$I_n=\int_0^1\left(1-x^2 \right)^n\,dx$$
Apply integration by parts, where:
$$u=\left(1-x^2 \right)^n\,\therefore\,du=n\left(1-x^2 \right)^{n-1}(-2x)\,dx$$
$$dv=dx\,\therefore\,v=x$$
And we may state:
$$I_n=\left[x\left(1-x^2 \right)^n \right]_0^1+2n\int_0^1 x^2\left(1-x^2 \right)^{n-1}\,dx$$
$$I_n=0-2n\int_0^1 \left(1-x^2-1 \right)\left(1-x^2 \right)^{n-1}\,dx$$
$$I_n=-2n\left(\int_0^1 \left(1-x^2 \right)^{n}\,dx-\int_0^1 \left(1-x^2 \right)^{n-1}\,dx \right)$$
Now, using the definition of the definite integral we may write:
$$I_n=2n\left(I_{n-1}-I_n \right)$$
Solving for $I_n$, we find:
$$(2n+1)I_n=2nI_{n-1}$$
$$I_n=\frac{2n}{2n+1}I_{n-1}$$
Now, we should observe that:
$$I_0=\int_0^1\left(1-x^2 \right)^0\,dx=1$$
Iterating all the way down to $n=0$, we have:
$$I_n=\frac{(2n)(2(n-1)(2(n-2))\cdots6\cdot4\cdot2}{(2n+1)(2n-1)(2n-3)\cdots7\cdot5\cdot3}\cdot1$$
Multiplying by $$1=\frac{(2n)(2(n-1)(2(n-2))\cdots6\cdot4\cdot2\cdot1}{(2n)(2(n-1)(2(n-2))\cdots6\cdot4\cdot2\cdot1}$$ we have:
$$I_n=\frac{\left((2n)(2(n-1)(2(n-2))\cdots6\cdot4\cdot2\cdot1 \right)^2}{(2n+1)(2n)(2n-1)(2n-2)(2n-3)(2n-4)\cdots7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$$
$$I_n=\frac{\left(2^n\cdot n! \right)^2}{(2n+1)!}$$
$$I_n=\frac{2^{2n}(n!)^2}{(2n+1)!}$$
Shown as desired.[/sp]
