# How do we prove that $\int_{0}^{\pi/2}\sin(t)^{2n+3}dt=\frac{4^n(2n+2)}{(2n+3)(2n+1){2n\choose n}}$?

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 $$\frac{\Gamma(1/2)\Gamma(n)}{\Gamma(n+1/2)}=\frac{4^n}{n{2n\choose n}}$$

• Just set $t=\arcsin\sqrt{u}$ to convert $S(n)$ into a value of the Beta function. It is worth to recall that $$\beta(1/2,n)= \frac{4^n}{n\binom{2n}{n}}.$$ Oct 24, 2018 at 16:36
• I guess the paper is the one born from math.stackexchange.com/questions/2123298/… Oct 24, 2018 at 16:47
• @JackD'Aurizio I know how to get from the original integral to $\frac{2n+2}{(2n+3)(2n+1)}n\beta(1/2,n)$, but my point of confusion is on the derivation of $$\beta(1/2,n)=\frac{4^n}{n{2n\choose n}}$$ Oct 24, 2018 at 16:50
• The integrals $\int_{0}^{\pi/2}\sin(\theta)^n\,d\theta$ can be simply computed by integration by parts or the lemma $\int_{0}^{2\pi} e^{mi\theta}\,d\theta = 2\pi\delta(m)$. They are typically involved in the derivation of Wallis' product. Oct 24, 2018 at 16:52
• @JackD'Aurizio and that's the Dirac delta? Oct 24, 2018 at 16:54

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.
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 $$\frac{(2n+2)!!}{(2n+3)!!} = \frac{4^n (n!)^2(2n+2)}{(2n+1)(2n+3)(2n)!}$$
• So instead of using the gamma and beta functions, how do I get to $$\frac{4^n(2n+2)}{(2n+3)(2n+1){2n\choose n}}$$ Oct 24, 2018 at 17:44
• When you integrate the polynomial, how do you get the ${2n\choose j}$ term? When I integrate it, I get $\sum_{j=0}^{n+1}\frac{(-1)^j{n+1\choose j}}{2j+1}$ because I integrate the binomial series of $(1-x^2)^{n+1}$ Oct 24, 2018 at 22:57
• I think the ${2n \choose j}$ term might be an error Oct 24, 2018 at 23:17