Check that for every natural $n \ge 1$, has:

Question

I am studying integrals by parts, and in the book I'm studying has an exercise in integrals by parts chapter, I do not even know where to start and I do not know how to proceed ... The exercise says:

Check that for every natural $n \geq 1$, has:

$a)$ $$ {\int_{0}^{1} (1-x^2)^n dx = \frac{2n}{2n+1} \int_{0}^{1}(1 -x^2)^{n-1} dx} $$

$b)$ $$ \int_{0}^{1} (1-x^2)^n dx = \frac{2^{2n}(n!)^2}{(2n+1)!} $$

I appreciate if someone help me with this exercise!

Answer 1

$$\int_0^1(1-x^2)^ndx=x(1-x^2)^n\bigg|_0^1-\int_0^1xd[(1-x^2)^n]$$ $$=-\int_0^1xn(1-x^2)^{n-1}(-2x)dx$$ $$=2n\int_0^1x^2(1-x^2)^{n-1}dx$$ $$=-2n\int_0^1(1-x^2)^ndx+2n\int_0^1(1-x^2)^{n-1}dx$$ Hence $$(2n+1)\int_0^1(1-x^2)^ndx=2n\int_0^1(1-x^2)^{n-1}dx$$ $$\int_0^1(1-x^2)^ndx=\frac{2n}{2n+1}\int_0^1(1-x^2)^{n-1}dx$$

For the second part $$\int_0^1(1-x^2)^ndx=\frac{2n}{2n+1}\int_0^1(1-x^2)^{n-1}dx$$ $$=\frac{2n}{2n+1}\frac{2(n-1)}{2(n-1)+1}\int_0^1(1-x^2)^{n-2}dx$$ $$=\frac{2n}{2n+1}\frac{2(n-1)}{2(n-1)+1}\frac{2(n-2)}{2(n-2)+1}\int_0^1(1-x^2)^{n-3}dx$$ $$\cdots$$ $$=\frac{2n(2n-2)(2n-4)\cdots 2}{(2n+1)(2n-1)(2n-3)\cdots 3}\int_0^1(1-x^2)^0dx$$ $$=\frac{[2n(2n-2)(2n-4)\cdots 2]^2}{(2n+1)!}$$ $$=\frac{2^{2n}(n!)^2}{(2n+1)!}$$

Answer 2

Let $S_n$ be the above integral, and write $S_n = \displaystyle \int_{0}^1 (1-x^2)^{n-1}(1-x^2)dx = S_{n-1} - \displaystyle \int_{0}^1 x^2(1-x^2)^{n-1}dx= S_{n-1} + \dfrac{1}{2n}\displaystyle \int_{0}^1 xd((1-x^2)^n)= S_{n-1} - \dfrac{S_n}{2n}$. From this part $a)$ follows.

For part $b)$, use induction on $n$ and note that $S_n = \dfrac{2nS_{n-1}}{2n+1}$, the answer is around the corner !.