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$$
Plugging in back $u$, we get $\displaystyle\frac{\sin^3 x}{3} - \frac{\sin^5 x}{5}$ + C
 A: \begin{align*}
\frac{\sin^3 (x)}{3} - \frac{\sin^5 (x)}{5} &= \sin^3 (x) (\frac{1}{3} - \frac{\sin^2 (x)}{5});\\
\cos(2x) &= 1 - 2 \sin^2(x)\\
\sin^2(x) &= \frac{1- \cos(2x)}{2}.
\end{align*}
Hence, we get,
\begin{align*}
\sin^3 (x) \left(\frac{1}{3} - \frac{\sin^2 (x)}{5}\right) &= \sin^3 (x) \left(\frac{1}{3} - \frac{1 - \cos(2x)}{10}\right)\\
& = \sin^3(x) \left(\frac{10 - 3 + 3 \cos(2x)}{30}\right)\\
& = \frac{\sin^3(x)}{30} (3 \cos(2x)+7)
\end{align*}
A: Just for the heck of it, another substitution could have been $u=\sin^3 x,$ in which case, $du = 3\sin^2x\cos x \ dx.$ Now, $1-u^{2/3} = 1-\sin^2x = \cos^2x.$ Thus, 
$$
\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*}
$$
A: 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*}
