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

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

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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.