The easiest way of approaching this question is to use De Moivre's formula: e^(inx) = cos(nx) + isin(nx) from which it is simple to show that cos(nx) = (e^(inx) + e^(-inx)) / 2 and sin(nx) = (e^(inx))- e^(-inx)) /2i therefore, cos(4x)sin(x) = (e^(4ix) + e^(-4ix)) * ((e^(ix)) - (e^(-ix)) / 4i= [e^(5ix) - e^(-5ix) - e^(3ix) + e^(-3ix)] / 4i= sin(5x)/2 - sin(3x)/2Finally, integrating, this gives cos(3x)/6 - cos(5x)/10 + integration constantThis can also be done by using various trigonometric identities, however this method is simpler and can continue to be applied to more complex questions.
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