A problem on indefinite integral: $\int(\cos x)^m\sin(nx) \mathrm dx$ 
If
$$I(m,n)=\int(\cos x)^m\sin(nx) ~\mathrm dx,$$
How do I get $7I(4,3)-4I(3,2)$?

 A: Not the general answer, but for the specific case
$$
7I(4,3)-4I(3,2)=\int(7\cos^4x\sin 3x-4\cos^3x\sin 2x)dx
$$
using the known formulae
$$
\sin2x=2\sin x\cos x\\
\sin 3x=(4\cos^2x-1)\sin x 
$$
you get
$$
7I(4,3)-4I(3,2)=\int(28\cos^6x-15\cos^4x)\sin xdx
$$
which is easily solved setting $t=\cos x$.

Using Chebyshev polynomials we have
$$
\int\cos^mx\sin nx dx=\int\cos^mxU_{n-1}(\cos x)\sin xdx=-\left.\int t^mU_{n-1}(t)dt\right|_{t=\cos x}
$$
where $U_{n-1}$ is the $(n-1)$th Chebyshev polynomial of the second kind.
A: A general solution

$$\begin{align} 
I(m,n)
&=\int(\cos x)^m \sin(nx) ~\mathrm{d}x \\
&=\int(\cos x)^m\sin((n-1)x+x) ~\mathrm{d}x \\
&=\int(\cos x)^{m+1}\sin((n-1)x) ~\mathrm{d}x+\int(\cos x)^m\cos((n-1)x)(\sin x)~\mathrm{d}x \\ 
&\stackrel{\color{red}{\text{IBP}}}{=}-\dfrac{\cos((n-1)x)}{n-1}(\cos x)^{m+1}-\left(\dfrac{m+1}{n-1}-1\right)\int(\cos x)^m\cos((n-1)x)(\sin x)~\mathrm{d}x \\
&=-\dfrac{\cos((n-1)x)}{n-1}(\cos x)^{m+1}-\left(\dfrac{m-n+2}{n-1}\right)\int(\cos x)^m\cos((n-1)x)(\sin x)~\mathrm{d}x 
\end{align}$$
Now
$$\begin{align}
&\int(\cos x)^m\cos((n-1)x)(\sin x)~\mathrm{d}x \\ 
&\stackrel{\color{red}{\text{IBP}}}{=} -\dfrac{(\cos x)^{m+1}\cos((n-1)x)}{m+1}+\left(\dfrac{n-1}{m+1}\right)\int (\cos x)^{m+1}\sin((n-1)x) ~\mathrm{d}x \\ 
&=-\dfrac{(\cos x)^{m+1}\cos((n-1)x)}{m+1}+\left(\dfrac{n-1}{m+1}\right) I(m+1,n-1)
\end{align}$$
This is tedious but now everything can be evaluated by successively putting values into it. It would be great if someone shows a nice way of evaluating the given value by some manipulation.
