A problem on indefinite integral: $\int(\cos x)^m\sin(nx) \mathrm dx$

Question

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

Answer 1

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 $t=\cos x$.

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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.

Answer 2

A general solution

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$$\begin{align} I(m,n) &=\int(\cos x)^m \sin(nx) ~\mathrm{d}x \\ &=\int(\cos x)^m\sin((n-1)x+x) ~\mathrm{d}x \\ &=\int(\cos x)^{m+1}\sin((n-1)x) ~\mathrm{d}x+\int(\cos x)^m\cos((n-1)x)(\sin x)~\mathrm{d}x \\ &\stackrel{\color{red}{\text{IBP}}}{=}-\dfrac{\cos((n-1)x)}{n-1}(\cos x)^{m+1}-\left(\dfrac{m+1}{n-1}-1\right)\int(\cos x)^m\cos((n-1)x)(\sin x)~\mathrm{d}x \\ &=-\dfrac{\cos((n-1)x)}{n-1}(\cos x)^{m+1}-\left(\dfrac{m-n+2}{n-1}\right)\int(\cos x)^m\cos((n-1)x)(\sin x)~\mathrm{d}x \end{align}$$

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.