# Problem of the Week #80 - October 7th, 2013

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#### Chris L T521

##### Well-known member
Staff member
Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: For $n\geq 0$, show that
$\int_0^1 (1-x^2)^n\,dx = \frac{2^{2n}(n!)^2}{(2n+1)!}.$

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Hint
:
Start by showing that if $I_n$ denotes the integral, then
$I_{k+1}=\frac{2k+2}{2k+3}I_k.$

#### Chris L T521

##### Well-known member
Staff member
This week's question was correctly answered by anemone and MarkFL. You can find both of their solutions below.

anemone's solution:
We're asked to prove that $$\displaystyle \int_0^1(1-x^2)^n dx=\frac{2^{2n}(n!)^2}{(2n+1)!}$$.

First, let's examine the LHS expression, the definite integral, $$\displaystyle \int_0^1(1-x^2)^n dx$$,

Using the following trigonometric substitution,

$$\displaystyle x=\sin \theta$$ $$\displaystyle \rightarrow dx=\cos \theta d\theta$$

The integral is now

$$\displaystyle \int_0^{\frac{\pi}{2}} \cos^{2n} \theta \cos \theta d\theta=\int_0^{\frac{\pi}{2}} \cos^{2n+1} \theta d\theta$$

Using the integration by parts with the following substitution:

$$\displaystyle u=\cos^{2n} \theta\;\;\rightarrow\;\;\frac{du}{d\theta}=2n (\cos^{2n-1} \theta)(-\sin \theta)$$

$$\displaystyle \frac{dv}{d\theta}=\cos \theta\;\;\rightarrow\;\;v=\sin \theta$$

The integral is then

$$\displaystyle \int_0^{\frac{\pi}{2}} \cos^{2n+1} \theta d\theta=\left[(\cos^{2n} \theta)(\sin \theta) \right]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} 2n (\cos^{2n-1} \theta)(-\sin \theta)(\sin \theta)d \theta$$

$$\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=(0)+2n\int_0^{\frac{\pi}{2}} (\sin^2 \theta)(\cos^{2n-1} \theta)d \theta$$

$$\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=2n\int_0^{\frac{\pi}{2}} (1-\cos^2 \theta)(\cos^{2n-1} \theta)d \theta$$

$$\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=2n\int_0^{\frac{\pi}{2}} (\cos^{2n-1} \theta)d \theta-2n\int_0^{\frac{\pi}{2}} (\cos^{2n+1} \theta)d \theta$$

$$\displaystyle (2n+1)\int_0^{\frac{\pi}{2}} \cos^{2n+1} \theta d\theta =2n\int_0^{\frac{\pi}{2}} (\cos^{2n-1} \theta)d \theta$$

$$\displaystyle \int_0^{\frac{\pi}{2}} \cos^{2n+1} \theta d\theta =\frac{2n}{2n+1}\int_0^{\frac{\pi}{2}} (\cos^{2n-1} \theta)d \theta$$

If we let $$\displaystyle I_n=\int_0^1(1-x^2)^n dx=\int_0^{\frac{\pi}{2}} \cos^{2n+1} \theta d\theta$$, we now have

$$\displaystyle I_n=\frac{2n}{2n+1}I_{n-1}$$

$$\displaystyle I_n=\frac{2n}{2n+1}I_{n-1}$$

$$\displaystyle \;\;\;\;=\frac{2n}{2n+1}\cdot\frac{2(n-1)}{2(n-1)+1}I_{n-2}$$

$$\displaystyle \;\;\;\;=\frac{2n}{2n+1}\cdot\frac{2(n-1)}{2(n-1)+1}\cdot\frac{2(n-2)}{2(n-2)+1}I_{n-3}$$

$$\displaystyle \;\;\;\;=\frac{2n}{2n+1}\cdot\frac{2(n-1)}{2(n-1)+1}\cdot\frac{2(n-2)}{2(n-2)+1}\cdots\frac{6}{7}\cdot\frac{4}{5}\cdot\frac{2}{3}$$

$$\displaystyle \;\;\;\;=\left(\frac{2n}{2n+1}\cdot\frac{2(n-1)}{2(n-1)+1}\cdot\frac{2(n-2)}{2(n-2)+1}\cdots\frac{6}{7}\cdot\frac{4}{5}\cdot\frac{2}{3} \right)\left(\frac{2n}{2n}\cdot\frac{2(n-1)}{2(n-1)}\cdot\frac{2(n-2)}{2(n-2)}\cdots\frac{6}{6}\cdot\frac{4}{4}\cdot\frac{2}{2} \right)$$

$$\displaystyle \;\;\;\;=\frac{((2n)(2(n-1))(2(n-2))\cdots(6)(4)(2))^2}{(2n+1)(2n))\cdots(6)(5)(4)(3)(2)(1)}$$

$$\displaystyle \;\;\;\;=\frac{(2^n(n)(n-1)(n-2)\cdots(3)(2)(1))^2}{(2n+1)!}$$

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

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

and we're done!

MarkFL's solution:
Let:

$$\displaystyle I_n=\int_0^1\left(1-x^2 \right)^n\,dx$$

Apply integration by parts, where:

$$\displaystyle u=\left(1-x^2 \right)^n\,\therefore\,du=n\left(1-x^2 \right)^{n-1}(-2x)\,dx$$

$$\displaystyle dv=dx\,\therefore\,v=x$$

And we may state:

$$\displaystyle I_n=\left[x\left(1-x^2 \right)^n \right]_0^1+2n\int_0^1 x^2\left(1-x^2 \right)^{n-1}\,dx$$

$$\displaystyle I_n=0-2n\int_0^1 \left(1-x^2-1 \right)\left(1-x^2 \right)^{n-1}\,dx$$

$$\displaystyle I_n=-2n\left(\int_0^1 \left(1-x^2 \right)^{n}\,dx-\int_0^1 \left(1-x^2 \right)^{n-1}\,dx \right)$$

Now, using the definition of the definite integral we may write:

$$\displaystyle I_n=2n\left(I_{n-1}-I_n \right)$$

Solving for $I_n$, we find:

$$\displaystyle (2n+1)I_n=2nI_{n-1}$$

$$\displaystyle I_n=\frac{2n}{2n+1}I_{n-1}$$

Now, we should observe that:

$$\displaystyle I_0=\int_0^1\left(1-x^2 \right)^0\,dx=1$$

Iterating all the way down to $n=0$, we have:

$$\displaystyle I_n=\frac{(2n)(2(n-1)(2(n-2))\cdots6\cdot4\cdot2}{(2n+1)(2n-1)(2n-3)\cdots7\cdot5\cdot3}\cdot1$$

Multiplying by $$\displaystyle 1=\frac{(2n)(2(n-1)(2(n-2))\cdots6\cdot4\cdot2\cdot1}{(2n)(2(n-1)(2(n-2))\cdots6\cdot4\cdot2\cdot1}$$ we have:

$$\displaystyle I_n=\frac{\left((2n)(2(n-1)(2(n-2))\cdots6\cdot4\cdot2\cdot1 \right)^2}{(2n+1)(2n)(2n-1)(2n-2)(2n-3)(2n-4)\cdots7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$$

$$\displaystyle I_n=\frac{\left(2^n\cdot n! \right)^2}{(2n+1)!}$$

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

Shown as desired.

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