Problem 1 - Wallis

Consider $I_n$ defined by

$$I_n = \int_0^{\frac{\pi}{2}}\sin^n xdx \quad (n \in \mathbb{N})$$

1.

Show the following: $I_n \frac{n-1}{n} I_{n-2}\quad (n \geq 2)$.

2.

Find a general form of $I_n$.

3.

Compute $J_n = \int_0^{\frac{\pi}{2}}\cos^nxdx\quad (n\in \mathbb{N})$.

1.

$I_n = \int_0^{\frac{\pi}{2}} \sin^n xdx = \left[ -\cos x \sin^{n-1} x \right]_0^{\frac{\pi}{2}} -\int_0^{\frac{\pi}{2}} -\cos x (n-1)\cos x \sin^{n-2} xdx$
$I_n = 0 + (n-1)\int_0^{\frac{\pi}{2}}(1-\cos^2 x)\sin^{n-2} xdx = (n-1)(I_{n-2} - I_n)$
$I_n = \frac{n-1}{n}I_{n-2}$

2.

$I_{2n} = \frac{2n-1}{2n}I_{2n-2} = \frac{(2n-1)(2n-3)(2n-5)\cdots 1}{(2n)(2n-2)(2n-4)\cdots 2}I_0$
$I_0 = \int_0^{\frac{\pi}{2}dx} = \frac{\pi}{2}$
$I_{2n} = \frac{(2n)!}{(2^n n!)^2}\frac{\pi}{2}$
$I_{2n+1} = \frac{2n}{2n+1}I_{2n-1} = \frac{(2n)(2n-2)(2n-4)\cdots 2}{(2n+1)(2n-1)(2n-3)\cdots 1}I_1$
$I_1 = \int_0^{\frac{\pi}{2}} \sin x dx = [-\cos x]_0^{\frac{\pi}{2}} = 1$
$I_{2n+1} = \frac{(2^n n!)^2}{(2n+1)!}$

3.

$J_n = \int_0^{\frac{\pi}{2}}\cos^n dx = - \int_{\frac{\pi}{2}}^0 \left( \cos (\frac{\pi}{2} -t)\right)^n dt = \int_0^{\frac{\pi}{2}} \sin t dt = I_n$