Evaluate $\int \cos^3 x\;\sin^2 xdx$

\begin{align*} \frac{\sin^3 (x)}{3} - \frac{\sin^5 (x)}{5} &= \sin^3 (x) (\frac{1}{3} - \frac{\sin^2 (x)}{5});\\ \cos(2x) &= 1 - 2 \sin^2(x)\\ \sin^2(x) &= \frac{1- \cos(2x)}{2}. \end{align*}

Hence, we get,

\begin{align*} \sin^3 (x) \left(\frac{1}{3} - \frac{\sin^2 (x)}{5}\right) &= \sin^3 (x) \left(\frac{1}{3} - \frac{1 - \cos(2x)}{10}\right)\\ & = \sin^3(x) \left(\frac{10 - 3 + 3 \cos(2x)}{30}\right)\\ & = \frac{\sin^3(x)}{30} (3 \cos(2x)+7) \end{align*}

Just for the heck of it, another substitution could have been $u=\sin^3 x,$ in which case, $du = 3\sin^2x\cos x \ dx.$ Now, $1-u^{2/3} = 1-\sin^2x = \cos^2x.$ Thus, $$ \begin{align*} \int \cos^3x\sin^2x \ dx &= \frac{1}{3}\int \cos^2x(3\sin^2x\cos x) dx \\ &= \frac{1}{3} \int(1-u^{2/3})du \\ &= \frac{u}{3} - \frac{u^{5/3}}{5} + C \\ &= \frac{\sin^3x}{3} - \frac{\sin^5x}{5} + C \ . \end{align*} $$