Integrating $\cos^3 (x) \, dx$

Your first integral is correct.

The second has two sign errors: $$u = \cos x \implies du = -\sin x\,dx$$

So evaluating the second integral should yield $$\begin{align} \int \Big(\cos^2x\sin x-\cos^4x\sin x\Big)\,dx & = -\int \cos^2x\Big(-\sin x\,dx\Big) - (-)\int \cos^4x\Big(-\sin x\,dx\Big) \\ \\& = -\int u^2 du + \int u^4\,du \\ \\ & = -\frac{u^3}{3} + \frac{u^5}{5}+c\quad\text{or}\quad \frac{u^5}{5} - \frac{u^3}{3}+c\end{align}$$