Evaluate $\int \cos^3 x\;\sin^2 xdx$ Is this correct? I thought it would be but when I entered it into wolfram alpha, I got a different answer. 
$$\int (\cos^3x)(\sin^2x)dx  = \int(\cos x)(\cos^2x)(\sin^2x)dx

= \int (\cos x)(1-\sin^2x)(\sin^2x)dx.$$
let $u = \sin x$, $du = \cos xdx$
$$\int(1-u^2)u^2du = \int(u^2-u^4)du

= \frac{u^3}{3} - \frac{u^5}{5} +C$$
Plugging in back $u$, we get $\displaystyle\frac{\sin^3 x}{3} - \frac{\sin^5 x}{5}$ + C

Take $\sin x=t, \cos x dx=dt$
\begin{align*}
\therefore \int \cos^{3}x\sin^{2}x dx &=\int \cos^{2}x\sin^{2} x\cos x dx\
                                      &=\int (1-\sin^{2}x) \sin^{2} x \cos x dx\
                                      &= \int (1-t^{2})t^{2} dt\
                                      &=\int (t^{2}-t^{4})dt\
                                      &=\frac{t^{3}}{3}-\frac{t^{5}}{5}+c\
                                      &= \frac{\sin^{3}x}{3}-\frac{\sin^{5}x}{5}+c
\end{align*}
