Problem: Prove that $\int\limits_{0}^{\pi/2} \sin^n x\, dx=\int\limits_{0}^{\pi/2} \cos^n x\, dx=\begin{cases}
\frac{n-1}{n}.\frac{n-3}{n-2}.\frac{n-5}{n-4}…\frac{2}{3}, & \mbox{ when $n$ is odd } \\
\frac{n-1}{n}.\frac{n-3}{n-2}.\frac{n-5}{n-4}…\frac{1}{2}\frac{\pi}{2}, & \mbox{ when $n$ is even}.
\end{cases}$
Let $I_n = \int \sin^n{x} \ dx $
$$
\begin{align*}
I_n &= -\cos x \hspace{3pt} \sin^{n-1}x + \int \cos^2 x \hspace{4pt} (n-1) \sin^{n-2} x \hspace{4pt} dx \\
&= -\cos x \hspace{3pt} \sin^{n-1}x + (n-1) \int \sin^{n-2} x \hspace{4pt} dx – (n-1) \int \sin^{n} dx\\
&= -\cos x \hspace{3pt} \sin^{n-1}x + (n-1) \int \sin^{n-2} x \hspace{4pt} dx – (n-1) I_n
\end{align*}
$$
$$ \Rightarrow (1+n-1)I_n = -\cos x \hspace{3pt} \sin^{n-1}x + (n-1) \int \sin^{n-2} x dx $$
$$ \Rightarrow I_n = \frac{-\cos x \hspace{3pt} \sin^{n-1}x}{n} + \frac{(n-1)}{n}I_{n-2}$$
Best Answer
Using reduction formula for $\sin(x)$ we have:
$$\int_{0}^{\frac{\pi}{2}}\sin^{n}\left(x\right)dx=\frac{n-1}{n}\int_{0}^{\frac{\pi}{2}}\sin^{\left(n-2\right)}\left(x\right)dx-\frac{\sin^{\left(n-1\right)}\left(x\right)\cos\left(x\right)}{n} \Bigg|_0^\frac{\large\pi}{2} $$
Now a simple comupation shows that:
$$\frac{\sin^{\left(k-1\right)}\left(x\right)\cos\left(x\right)}{k}\Bigg|_0^\frac{\large\pi}{2}=\left(\frac{\sin^{\left(k-1\right)}\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi}{2}\right)}{k}\right)-\left(\frac{\sin^{\left(k-1\right)}\left(0\right)\cos\left(0\right)}{k}\right)=0$$
for $$k\ge2$$
Finally the least even power which the reduction is valid for is $k=2$ ,setting $k=2↦n$ and substituting for $n-1$,$n-2$,...,$2$ respectively we get the same result, so I will prevent to write it again and again.
Continuing this way we have:$$\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\frac{n-5}{n-4}...\left[\int_{0}^{\frac{\pi}{2}}\sin^{\left(2\right)}\left(x\right)dx\right]$$$$=\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\frac{n-5}{n-4}...\left[\frac{2-1}{2}\int_{0}^{\frac{\pi}{2}}\sin^{0}\left(x\right)dx\right]$$$$=\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\frac{n-5}{n-4}...\frac{2-1}{2}\left[x\right]\Bigg|_0^\frac{\large\pi}{2}=$$$$\boxed {\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\frac{n-5}{n-4}\cdot...\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}}$$ or $$\frac{\pi}{2}\prod_{k=1}^{\frac{n}{2}}\frac{2k-1}{2k}$$
So all we have to do is just a simple substitution. for the case $n$ odd just let $k=3↦n$ , (clearly the least odd power is $3$) ,then we have:
$$\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\frac{n-5}{n- 4}\cdot\cdot\cdot\left[\int_{0}^{\frac{\pi}{2}}\sin^{5}\left(x\right)dx\right]$$$$=\frac{n- 1}{n}\cdot\frac{n-3}{n-2}\cdot\frac{n-5}{n-4}\cdot\cdot\cdot\left[\frac{5- 1}{5}\int_{0}^{\frac{\pi}{2}}\sin^{3}\left(x\right)dx\right]$$$$=\frac{n-1}{n}\cdot\frac{n-3}{n- 2}\cdot\frac{n-5}{n-4}\cdot\cdot\cdot\frac{5-1}{5}\cdot\left[\frac{3- 1}{3}\int_{0}^{\frac{\pi}{2}}\sin^{1}\left(x\right)dx\right]=$$$$\boxed{\frac{n-1}{n}\cdot\frac{n-3}{n- 2}\cdot\frac{n-5}{n-4}\cdot...\cdot\frac{4}{5}\cdot\frac{2}{3}\cdot1}$$ or $$\prod_{k=1}^{\frac{n-1}{2}}\frac{2k}{2k+1}$$
and the result follows.