How to show the following cool equality: I am looking for a proof of the following relationship:
$\newcommand{\ds}[1]{\displaystyle{#1}}$
$$
\frac{\ds{\int_{0}^{\pi}\sin^{n-2}\left(t\right)\,{\rm d}t}}
     {\ds{\int_{0}^{\pi}\sin^{n-3}\left(t\right)\,{\rm d}t}}
     =
     {\ds{\Gamma^{\,2}\left(\left[n - 1\right]/2\right)}
      \over
      \ds{\Gamma\left(n/2\right) \Gamma\left(\left[n - 2\right]/2\right)}},
$$
where $\Gamma$ is the Gamma-function.
But: The proof must only contain things that a person could prove that has just a knowledge of $1$-d calculus (you do not need to show everything, but the things you use should be provable with standard techniques). Especially no Complex Analysis and Functional Analysis.
My problem is, that I do not see any kind of relationship between the left and right-hand side.
 A: Let $I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^n(x) dx$. We then have
$I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{n-1}(x) d(-\cos(x)) = -\sin^{n-1}(x) \cos(x) |_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} (n-1) \sin^{n-2}(x) \cos^2(x) dx$
The first expression on the right hand side is zero since $\sin(0) = 0$ and $\cos(\frac{\pi}{2}) = 0$.
Now rewrite $\cos^2(x) = 1 - \sin^2(x)$ to get
$I_n = (n-1) \left(\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{n-2}(x) dx - \int_{0}^{\frac{\pi}{2}} \sin^{n}(x) dx \right) = (n-1) I_{n-2} - (n-1) I_n$.
Rearranging we get  $n I_n = (n-1) I_{n-2}$, $I_n = \frac{n-1}{n}I_{n-2}$.
Using this recurrence we get
$$I_{2k+1} = \frac{2k}{2k+1}\frac{2k-2}{2k-1} \cdots \frac{2}{3} I_1$$
$$I_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} I_0$$
$I_1$ and $I_0$ can be directly evaluated to be $1$ and $\frac{\pi}{2}$ respectively and hence,
$$I_{2k+1} = \frac{2k}{2k+1}\frac{2k-2}{2k-1} \cdots \frac{2}{3} = 4^k \dfrac{(k!)^2}{(2k+1)!}$$
$$I_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} \frac{\pi}{2} = \dfrac{(2k)!}{4^k (k!)^2} \dfrac{\pi}2$$
Use this to get what you want.
A: They are called Wallis' integrals, since they were studied by the great British mathematician John Wallis more than $3$ centuries ago. (The first link contains the answer to your question, and more).
A: Recalling the $\beta$ function$

$$\beta(x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,d\theta, \qquad \mathrm{Re}(x)>0,\ \mathrm{Re}(y)>0, $$

we can relate your integrals to it. Splitting the integral in the numerator as 

$$ I=\int_{0}^{\pi}\sin^{n-2}\left(t\right)\,{\rm d}t= \int_{0}^{\pi/2}\sin^{n-2}\left(t\right)\,{\rm d}t + \int_{\pi/2}^{\pi}\sin^{n-2}\left(t\right)\,{\rm d}t\,= I_1+I_2. $$

Now, $I_1$ can readily related to the $\beta$ function while $I_2$ can be related to the $\beta $ function by using the substitution $x=t+\pi/2$ which results in 

$$I_2 =\int_{0}^{\pi/2}\cos^{n-2}\left(t\right)\,{\rm d}t. $$

You can do the same with the other integral. I let you to finish the problem. 
