How do I use the hypergeometric function? I was given a long list of integrals involving sines and cosines as homework. Being the slothful person that I am, I tried to find a general formula for these integrals as a function of the sine's and cosine's exponents. I turned to WA after no success. There I found that the general integral needed a Hypergeometric Function, a thing that I never even knew existed. I went to check what it was and, in summary, I found that:

Given a power series $$\sum_{i \ge 0} \beta_i x^i$$
Define two polynomials:
$$P(n) = \prod _{i=1}^p (a_i+n), \qquad p=0 \to P(n)=1$$
$$Q(n) = \prod _{i=1}^q (b_i+n), \qquad q=0 \to Q(n)=1+n$$
Such that:
$$\frac {\beta_{r+1}}{\beta_r} = \frac {P(r)}{Q(r)}$$
Denote this function by:
$${}_pF_q(a_1,...,a_p;b_1,...,b_q;x)$$
So of course I tried to put this new knowledge to use by trying my hand at a few Hypergeometric Functions. Given that WA gave me ${}_2F_1(3,-3/2;4;x)$ I tried finding the series that this HF would describe but I quickly ran into the problem of determining the coefficients. Given that the coefficients are only specified by a ratio, how can I actually determine them so that I can use them in the power series? Is it possible to reconstruct the power series from a hypergeometric function? Could you walk me through an example using ${}_2F_1(3,-3/2;4;x)$? Is my basic knowledge wrong on some fundamental level or am I on the right track?
 A: You are on the right track. The hypergeometric function $_p F_q$ is specified by the series
$$_p F_q\left(\mathbf a;\mathbf b;x\right)=\sum_{k=0}^{\infty}
\frac{\left(a_1\right)_n\ldots \left(a_p\right)_n}{
\left(b_1\right)_n\ldots \left(b_q\right)_n}\frac{x^n}{n!},$$
where $\left(a\right)_n=a\left(a+1\right)\ldots \left(a+n-1\right)$ is called the Pochhammer's symbol. Thus for example
$$_2 F_1\left(3,-\frac32;4;x\right)=\sum_{k=0}^{\infty}
\frac{\left(3\right)_n \left(-\frac32\right)_n}{
\left(4\right)_n}\frac{x^n}{n!}=9\sum_{n=0}^{\infty}
\frac{\left(2n-5\right)!!\,x^n}{\left(n+3\right)2^nn!},$$
where to get the second equality we use $\displaystyle\frac{\left(a\right)_n}{\left(a+1\right)_n}=\frac{a}{a+n}$ and $\left(-\frac32\right)_n=3\times\frac{\left(2n-5\right)!!}{2^n}$. Actually due to special parameter values this particular case of hypergeometric function admits more explicit evaluation as
$$_2 F_1\left(3,-\frac32;4;x\right)=\frac{16}{105x^3}-\frac{70x^4-100x^3+6x^2+8x+16}{105x^3}\sqrt{1-x}.$$
