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 $\lim _{n\rightarrow \infty} I(2, n)$?

When at least one of the exponents is odd, substitute $\sin^2 x = (1 - \cos^2 x)$ and $\cos^2 x = (1 - \sin^2 x)$ as in:
$$
\begin{eqnarray}
\int \sin^3 x \cos ^4 x \, \textrm{d}x &=& \int \sin^2 x \cos^4 x(\sin x \, \textrm{d}x) \\
 &=& \int (1 - \cos^2 x) cos^4 x (\sin x \, \textrm{d}x) \\
 &=& \int \cos ^4 x \sin x \, \textrm{d}x - \int \cos^6 x \sin x \, \textrm{d}x \\
 &=& \frac{-\cos^5 x}{5} + \frac{\cos^7 x}{7} + C \\
\end{eqnarray}
$$
When both $m$ and $n$ are even, substitute $\sin^2 x = \frac{1 - \cos 2x}{2}$ and $\cos^2 x = \frac{1 + \cos 2x}{2}$ as in:
$$
\begin{eqnarray}
\int \sin^2 x \cos^4 x \, \textrm{d}x &=& \int \left(\frac{1 - \cos 2x}{2}\right) \left(\frac{1 + \cos 2x}{2}\right)^2 \, \textrm{d}x \\
 &=& \frac{1}{8} \int \, \textrm{d}x + \frac{1}{8} \int \cos 2x \, \textrm{d}x - \frac{1}{8} \int \cos^2 2x \, \textrm{d}x - \frac{1}{8} \int \cos^3 2x \, \textrm{d}x \\
 &=& \frac{x}{8} + \frac{1}{16} \sin 2x - \frac{1}{8} \int \frac{1 + \cos 4x}{2} \, \textrm{d}x - \frac{1}{8} \int (1 - \sin^2 2x) \cos 2x \, \textrm{d}x \\
 && ... \\
 &=& \frac{x}{16} + \frac{\sin^3 2x}{48} - \frac{\sin 4x}{64} + C
\end{eqnarray}
$$
This requires computing the indefinite integral on which to apply your interval, which might be painful for high values of $m$ or $n$. I hope someone has a better answer.
