$\displaystyle\int_0^{\pi}|\cos x|^3dx = \int_0^{\pi/2}\cos^3x\,dx + \int_{\pi/2}^{\pi}(-\cos x)^3dx$ Wait: $|\cos x|^3$ for $x\in[\pi/2,\pi]$: $\cos x\leq0$, so $|\cos x|^3=(-\cos x)^3$? No: $|\cos x|^3=|\cos x|^2\cdot|\cos x|=\cos^2x\cdot|\cos x|$.

$=2\displaystyle\int_0^{\pi/2}\cos^3x\,dx$ (by symmetry about $\pi/2$)

$=2\displaystyle\int_0^{\pi/2}(1-\sin^2x)\cos x\,dx=2\left[\sin x-\dfrac{\sin^3x}{3}\right]_0^{\pi/2}=2\left(1-\dfrac{1}{3}\right)=\mathbf{\dfrac{4}{3}}$