Prove the following equality: integral from -1 to 1

Prove the following equality: $\int_{-1}^1 (1-x^2)^n \, dx = \frac{2^{2n+1} \cdot (n!)^2}{(2n+1)!}$.

We will prove this formula using a trigonometric substitution and relating it to the Wallis integrals.

Step 1: Use the substitution $x = \sin\theta$

Then $dx = \cos\theta \, d\theta$, and $1 - x^2 = \cos^2\theta$.

Step 2: Change the limits

When $x=-1$: $\theta=-\frac\pi2$. When $x=1$: $\theta=\frac\pi2$.

Step 3: Transform the integral

\[\int_{-1}^1 (1-x^2)^n \, dx = \int_{-\pi/2}^{\pi/2} \cos^{2n}\theta \cdot \cos\theta \, d\theta = \int_{-\pi/2}^{\pi/2} \cos^{2n+1}\theta \, d\theta\]

Step 4: Use symmetry

Since $\cos^{2n+1}\theta$ is even (odd power, but cosine is even):

\[\int_{-\pi/2}^{\pi/2} \cos^{2n+1}\theta \, d\theta = 2\int_0^{\pi/2} \cos^{2n+1}\theta \, d\theta\]

Step 5: Apply the reduction formula for odd powers

Let $I_m = \int_0^{\pi/2} \cos^m\theta \, d\theta$. For odd $m=2n+1$: $I_{2n+1} = \frac{2n}{2n+1} I_{2n-1}$. Applying this recursively, with base case $I_1 = \int_0^{\pi/2}\cos\theta\,d\theta = 1$:

\[I_{2n+1} = \frac{2n \cdot (2n-2) \cdots 4 \cdot 2}{(2n+1)(2n-1)\cdots 5 \cdot 3} \cdot 1 = \frac{(2n)!!}{(2n+1)!!}\]

where $(2n)!! = 2\cdot4\cdots(2n) = 2^n n!$.

Step 6: Express in terms of factorials

Since $(2n)!! = 2^n \cdot n!$, and $(2n+1)!! = \frac{(2n+1)!}{(2n)!!} = \frac{(2n+1)!}{2^n\cdot n!}$:

\[I_{2n+1} = \frac{2^n \cdot n!}{\frac{(2n+1)!}{2^n \cdot n!}} = \frac{2^{2n} \cdot (n!)^2}{(2n+1)!}\]

Step 7: Complete the calculation

\[\int_{-1}^1 (1-x^2)^n \, dx = 2 \cdot I_{2n+1} = \frac{2^{2n+1} \cdot (n!)^2}{(2n+1)!}\]

Therefore:

\[\boxed{\int_{-1}^1 (1-x^2)^n \, dx = \frac{2^{2n+1} \cdot (n!)^2}{(2n+1)!}}\]

Verification with examples

Example 1 ($n=0$): $\int_{-1}^1 1 \, dx = 2$. Formula: $\frac{2^1 \cdot 1}{1!} = 2$.

Example 2 ($n=1$): $\int_{-1}^1 (1-x^2) \, dx = \frac{4}{3}$ (computed directly in Problem 6). Formula: $\frac{2^3\cdot1}{3!} = \frac{8}{6} = \frac{4}{3}$.

Example 3 ($n=2$): Formula: $\int_{-1}^1 (1-x^2)^2 \, dx = \frac{2^5\cdot4}{5!} = \frac{128}{120} = \frac{16}{15}$.

Connection to Beta function

This result is related to the Beta function $B(p,q) = \int_0^1 t^{p-1}(1-t)^{q-1}\,dt = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$; for integer arguments $\Gamma(n+1)=n!$, and our integral can be expressed using the Beta function, giving an alternative proof.