Establish a recurrence relation for a given integral

Define

$I_n = \int_0^1 (1-x^2)^n \, dx.$

Prove that this integral satisfies the recurrence relation,

$(2n+1)I_n = 2n I_{n-1}.$

Using this compute $I_2, I_3, I_4$ , and $I_5$ .

Proof. We start by integrating $I_n$ by parts with

$\begin{align*} u &= (1-x^2)^n & du &= -2xn(1-x^2)^{n-1} \, dx \\ dv &= dx & v &= x. \end{align*}$

Therefore we have

$\begin{align*} I_n = \int_0^1 (1-x^2)^n \, dx &= \left( uv - \int v \, du \right) \Bigr \rvert_0^1 \\[9pt] &= x(1-x^2)^n \Bigr \rvert_0^1 + 2n \int_0^1 x^2 (1-x^2)^{n-1} \, dx \\[9pt] &= 2n \int_0^1 x^2 (1-x^2)^{n-1} \, dx \\[9pt] &= 2n \int_0^1 (1-1+x^2)(1-x^2)^{n-1} \, dx \\[9pt] &= 2n \int_0^1 (1-x^2)^{n-1} \, dx - 2n \int_0^1 (1-x^2)(1-x^2)^{n-1} \, dx \\[9pt] &= 2n I_{n-1} - 2n I_n \\[9pt] \implies (2n+1)I_n &= 2n I_{n-1}. \qquad \blacksquare \end{align*}$