Recurrent integral formula $$ I_{m,n}= \int {\sin(x)^m}{\cos(x)^n} dx $$
Determine the recurrence formula for the given integral,where "$n$" and "$m$" are $2$ natural parameters.
EDIT: Can it be solved by using only definite integrals formulas or the integration by parts method?
 A: Use integration by parts on $I_{m,n}$ with:
$$\begin{array}{ll}
u=\sin^{m-1}x                     & dv=\sin x \cos^n x\,dx  \\
du=(m-1)\sin^{m-2}x \cos x\,dx    & v =\frac{-1}{n+1}\cos^{n+1}x
\end{array}$$
Then we have:
$$
I_{m,n} = \frac{-1}{n+1}\sin^{m-1}x\cos^{n+1}x + \frac{m-1}{n+1}\color{blue}{\int{\sin^{m-2}x\,\cos^{n+2}x\,dx}} \tag{1}$$
where the integral in blue is
$$\begin{align}
\int{\sin^{m-2}x\cos^{n+2}x\,dx}&=\int{\sin^{m-2}x(1-\sin^2x)\cos^{n}x\,dx} \\ 
                                &=\int{\sin^{m-2}x\,\cos^{n}x\,dx}-\int{\sin^{m}x\cos^{n}x\,dx} \\[1em]
                                &=I_{m-2,n}-I_{m,n}
\end{align}$$
The "trick" lies in replacing $\cos^2x$ by $1-\sin^2x$ in the integrand, so as to get the recursive relation.
So (1) reduces to
$$\begin{align}
I_{m,n} &= \frac{-1}{n+1}\sin^{m-1}x\,\cos^{n+1}x + \frac{m-1}{n+1}(I_{m-2,n}-I_{m,n}) \implies \\[1em]
\frac{m+n}{n+1}I_{m,n} &= \frac{-1}{n+1}\sin^{m-1}x\,\cos^{n+1}x + \frac{m-1}{n+1}I_{m-2,n} \implies \\[1em]
I_{m,n} &= \frac{-1}{m+n}\sin^{m-1}x\,\cos^{n+1}x + \frac{m-1}{m+n}I_{m-2,n} \tag{*}
\end{align}$$
