Evaluate $\int_0^{\pi/2}\sin^nxdx$

Question

Prove
$$
\int_0^{\pi/2}\sin^nx.dx=\int_0^{\pi/2}\cos^nx.dx=\begin{cases}
\dfrac{(n-1)(n-3)….2}{n(n-2)….1}\quad\text{if $n$ is odd}\\
\dfrac{(n-1)(n-3)….1}{n(n-2)….2}\quad\text{if $n$ is even}
\end{cases}
$$

$$
I_n=\int_0^{\pi/2}\sin^nx.dx=\int_0^{\pi/2}\sin^{n-1}x.\sin x.dx\\
=\sin^{n-1}x.(-\cos x)-(n-1)\int\sin^{n-2}x.(-\cos x)dx\\
=-\cos x.\sin^{n-1}x-(n-1)\int\sin^{n-2}x.\cos x.(-\cos x)dx\\
=-\cos x.\sin^{n-1}x+(n-1)\int\sin^{n-2}x(1-\sin^2x)dx\\
=-\cos x.\sin^{n-1}x+(n-1)\int\sin^{n-2}xdx-(n-1)\int\sin^{n}xdx\\
I_n=-\cos x.\sin^{n-1}x+(n-1)I_{n-2}-(n-1)I_n\\
nI_n=-\cos x.\sin^{n-1}x+(n-1)I_{n-2}\\
\color{red}{\boxed{I_n=\dfrac{-\cos x.\sin^{n-1}x}{n}+\dfrac{n-1}{n}I_{n-2}}}
$$
So far so good I am able to obtain the recurrence formula, but from here how do I obtain the above result ?

Note: I am actually seeking to get a proof other that using mathematical induction.

Answer 1

Hint:

$$I_n = -\frac{\cos x\cdot\sin^{n - 1}x}n\Bigg|_{\large0}^{\Large\frac{\pi}2} + \frac{n - 1}nI_{n - 2} = 0 + \frac{n - 1}nI_{n - 2} = \frac{n - 1}nI_{n - 2}$$

From here,

• for $n = 2p$, $I_n = \dfrac{(2p - 1)(2p - 3)\cdots1}{2p(2p-2)\cdots2}I_0$.
• for $n = 2p + 1$, $I_n = \dfrac{2p(2p - 2)\cdots2}{(2p + 1)(2p - 1)\cdots3}I_1$.