Integrating $\cos^3 (x) \, dx$ I am wondering whether I integrated the following correctly.


*

*$\int \cos^3 x \, dx$
I did 
\begin{align}
\int \cos^3 x \, dx &= \int \cos(x)(1-\sin^{2}x) \, dx \\
&= \int \cos(x)-\sin^{2}x \cos x \, dx \\
&= \sin(x)-\frac{u^{3}}{3} + c, \quad(u=\sin(x)) \\
&= \sin(x)-\dfrac{\sin^{3}x}{3}+c
\end{align}

2.$\int \sin^{3}x \cos^{2}x\,dx$
\begin{align}
\int \sin^{3}x \cos^{2}x\,dx &= \int(1-\cos^2x)(\cos^2x)\sin(x)\,dx \\
&= \int \cos^2x\sin(x)-\cos^4x\sin(x)\,dx, \quad u=\cos(x) \\
&= \dfrac{u^3}{3}-\dfrac{u^5}{5}+c
\end{align}
And plug in my u.

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