I saw this integral in a paper on hypergeometric functions:
$$S(n)=\int_{0}^{\pi/2}\sin(t)^{2n+3}dt=\frac{4^n(2n+2)}{(2n+3)(2n+1){2n\choose n}}\;\;\;\;\;\;\;\;\;\;\;(1)$$
I tried to prove it and got this far.
Given $$\int_{0}^{\pi/2}\sin^{\alpha}tdt=\frac1{2}\beta(\frac{\alpha+1}{2},\frac1{2})$$
We arrive at
$$S(n)=\frac1{2}\beta(n+2,\frac1{2})$$
$$S(n)=\frac{\sqrt{\pi}\,\Gamma(n+2)}{2\Gamma(n+5/2)}$$
$$S(n)=\frac{\sqrt{\pi}}{2}\frac{n(n+1)}{(n+3/2)(n+1/2)}\frac{\Gamma(n)}{\Gamma(n+1/2)}$$
$$S(n)=\frac{2n+2}{(2n+3)(2n+1)}n\beta(n,\frac{1}{2})$$
And assuming that $(1)$ is true, we get
$$\beta(n,1/2)=\beta(1/2,n)=\frac{4^n}{n{2n\choose n}}\;\;\;\;\;\;\;(2)$$
But I do not know how to prove $(2)$.
I tried to prove it by starting with $$\beta(1/2,n)=\frac{\Gamma(1/2)\Gamma(n)}{\Gamma(n+1/2)}$$
But I do not know how to prove $$\frac{\Gamma(1/2)\Gamma(n)}{\Gamma(n+1/2)}=\frac{4^n}{n{2n\choose n}}$$
Best Answer
A nice method to integrate $\sin t$ to an odd power is to keep one factor of sine, and convert all the rest to cosine (using $\sin^2 t = 1-\cos^2 t$), then substitute $x = \cos t$, $dx = -\sin t\;dt$. $$ \int_0^{\pi/2} (\sin t)^{2n+3} dt = \int_0^{\pi/2} (1-\cos^2 t)^{n+1} \sin t\;dt = \int_0^1 (1-x^2)^{n+1} \;dx $$ the integral of a polynomial, which you already know how to do.
This way, there is no need to know about gamma or beta functions.
Integrating the polynomial, we get $$ \sum_{j=0}^{n+1}\frac{(-1)^j\binom{n+1}{j}}{2j+1} $$ So we have to do some combinatorics to reach your answer $$ \frac{(2n+2)!!}{(2n+3)!!} = \frac{4^n (n!)^2(2n+2)}{(2n+1)(2n+3)(2n)!} $$