Definite Integral of Polynomial of Sine and Cosines I was wondering if there is a way to compute definite integral 
\begin{align}
I(m,n) :=\int_{0}^{\pi} \sin^m (\theta) \cdot \cos^{n}(\theta) \ \mathrm{d} \theta
\end{align}
in general for integer-valued $m$ and $n$. This problem arises when I try to compute integrals of trigonometric functions over a high dimensional sphere.

In fact I am more interested in the asymptotic order of this integral. I was wondering how $I(m,n)$ behaves when $n$ goes to infinity. For example, what is the limit of $\lim _{n\rightarrow \infty} I(2, n)$?
I'd try to expand 
$$\cos^n(x) = \frac{(e^{ix}+e^{-ix})^n}{2^n} = \frac{e^{-inx}(e^{2ix}+1)^n}{2^n} = \frac{e^{-inx}}{2^n}\sum_{k=0}^n {n\choose k} e^{2ikx}$$
$$\sin^m(x) = \frac{(e^{ix}-e^{-ix})^m}{(2i)^m} = \frac{e^{-imx}(e^{2ix}-1)^m}{(2i)^m} = (-1)^m\frac{e^{-imx}}{(2i)^m}\sum_{l=0}^m {m\choose l} (-1)^l e^{2ilx}$$

$$\cos^n(x) \sin^m(x) =\frac{(-1)^m}{2^{n+m}i^m}\sum_{l=0}^m \sum_{k=0}^n {m\choose l} {n\choose k} (-1)^l e^{\textstyle i(2l+2k-n-m)x}  $$

for $a \in \mathbb{Z}^*$ we know that $\int_0^\pi e^{2 i a x} dx = 0$ so that 

$$\int_0^\pi \sin^m(x) \cos^n(x) dx = \pi \frac{(-1)^m}{2^{n+m}i^m}\sum_{(l,k) \in E_{n,m}} {m\choose l} {n\choose k} (-1)^l $$

where $E_{n,m} = \left\{ (l,k) \ | \ 2l+2k=n+m, l \in \{0 \ldots m\}, k \in \{0 \ldots n\} \right\}$

this also tells us that the result is non-zero only when $m$ and $n$ are even (if $m$ is odd the result is imaginary and if $m+n$ is odd $E_{n,m}$ is empty)
Well, if there is a way, there is the answer
$$\int \sin^m (x) \cdot \cos^{n}(x) \ \mathrm{d} x  = -\frac{\sin ^{m+1}(x) \sin ^2(x)^{\frac{1}{2} (-m-1)} \cos ^{n+1}(x) \, _2F_1\left(\frac{1-m}{2},\frac{n+1}{2};\frac{n+3}{2};\cos ^2(x)\right)}{n+1}$$
where $_2F_1$ is the Hypergeometric Function. Hence
Well, if there is a way, there is the answer
$$\int_{0} ^{\infty} \sin^m (x) \cdot \cos^{n}(x) \ \mathrm{d} x  = -\frac{\sin ^{m+1}(x) \sin ^2(x)^{\frac{1}{2} (-m-1)} \cos ^{n+1}(x) \, _2F_1\left(\frac{1-m}{2},\frac{n+1}{2};\frac{n+3}{2};\cos ^2(x)\right)}{n+1} \Bigg | _0 ^{\pi} = -\frac{0.^{m+0.5 (-1. m-1.)+1.} (-1.)^{n+2.} \Gamma (0.5 m+0.5) \Gamma (0.5 n+1.5)}{(n+1.) \Gamma (0.5 m+0.5 n+1.)} - \left( -\frac{1. 0.^{m+0.5 (-1. m-1.)+1.} \Gamma (0.5 m+0.5) \Gamma (0.5 n+1.5)}{(n+1.) \Gamma (0.5 m+0.5 n+1.)} \right)$$