Evaluate: ∫ x² cos x dx

Integrate by parts, taking $u = x^2$ and $dv = \cos x\,dx$: $\int x^2\cos x\,dx = x^2\sin x - \int 2x\sin x\,dx$ For $\int x\sin x\,dx$ take $u = x$, $dv = \sin x\,dx$: $= -x\cos x + \int\cos x\,dx = -x\cos x + \sin x$ $\int x^2\cos x\,dx = x^2\sin x - 2\left(-x\cos x + \sin x\right) + C$ Hence $\int x^2\cos x\,dx = x^2\sin x + 2x\cos x - 2\sin x + C$.