5. $\\int x^{2}\\cos xdx$

Explanation:

Step1: Apply integration by parts

Let (u = x^{2}), (dv=\cos xdx). Then (du = 2xdx), (v=\sin x). By integration by parts formula (\int u dv=uv-\int v du), we have (\int x^{2}\cos xdx=x^{2}\sin x-\int 2x\sin xdx).

Step2: Apply integration by parts again

For (\int 2x\sin xdx), let (u = 2x), (dv=\sin xdx). Then (du = 2dx), (v=-\cos x). By integration by parts formula, (\int 2x\sin xdx=- 2x\cos x+\int 2\cos xdx).

Step3: Integrate (\int 2\cos xdx)

(\int 2\cos xdx = 2\sin x + C). Putting it all together: [ \begin{align*} \int x^{2}\cos xdx&=x^{2}\sin x-(-2x\cos x + 2\sin x)+C\ &=x^{2}\sin x + 2x\cos x-2\sin x + C \end{align*} ]

Answer:

(x^{2}\sin x + 2x\cos x-2\sin x + C)