Finding anti-derivative of $f(x)= \sin^3 x \cos^2 x $ 
Finding anti-derivative of $f(x)= \sin^3 x \cos^2 x $

so, integrate $\int \sin^3 x \cos^2 x dx = \int \sin x (1- \cos^2 x) \cos^2 (x)  dx $
let $u = \cos x$
$\int -u^2 (1-u^2) du = \int -u^2 dx + \int u^4 dx = \frac{-u^3}{3} + \frac{u^5}{5} + C $
Therefore $ \frac{\cos^3 x}{3}+\frac{\cos^5 x}{5} + C$
Why am I wrong? My teacher said the answer is $ \frac{-\cos x}{8} - \frac{\cos (3x)}{48} + \frac{\cos (5x)}{80} $

Note that
\begin{align}
\cos^3 x &= \frac34\cos x +\frac14\cos 3x\\
\cos^5 x & = \frac58\cos x +\frac5{16}\cos 3x+\frac1{16}\cos 5x
\end{align}
and
\begin{align}
&-\frac13\cos^3x+\frac15\cos^5x\\
=& -\frac13\left(\frac34\cos x +\frac14\cos 3x\right)+\frac15\left(\frac58\cos x +\frac5{16}\cos 3x+\frac1{16}\cos 5x\right)\\
=&-\frac18\cos x-\frac1{48}\cos 3x +\frac1{80}\cos5x
\end{align}
Thus, the two anti-derivatives are the same.
