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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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up vote 3 down vote accepted

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.

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