Stosujemy całkowanie przez części: \[ \begin{split} \int x^3 \cos x\ dx&= \int x^3 \left ( \sin x \right )'\ dx=\\[6pt] &=x^3\sin x-\int3x^2\sin x\ dx=\\[6pt] &=x^3\sin x-\left ( \int3x^2\left ( -\cos x \right )'\ dx \right )=\\[6pt] &=x^3\sin x+\left ( \int3x^2\left ( \cos x \right )'\ dx \right )=\\[6pt] &=x^3\sin x+\left ( 3x^2\cos x-\int6x \cos x\ dx \right )=\\[6pt] &=x^3\sin x+3x^2\cos x-\int6x (\sin x)'\ dx=\\[6pt] &=x^3\sin x+3x^2\cos x-\left ( 6x\sin x - \int6\sin x\ dx \right )=\\[6pt] &=x^3\sin x+3x^2\cos x- 6x\sin x + 6\int\sin x\ dx =\\[6pt] &=x^3\sin x+3x^2\cos x- 6x\sin x - 6\cos x +C \end{split} \]