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} $

To check the equivalence of different trigonometric forms, the simplest way is to use the following type of relation
$$
\cos5\theta=\Re (e^{5i\theta})
$$
and the binomial theorem to derive relations between multiples of angles. For example,
$$
\begin{align}
\left({e^{i\theta}}\right)^5&=(\cos\theta+i\sin\theta)^5\\
&=\cos^5\theta+5i\cos^4\theta \sin\theta-10\cos^3\theta \sin^2\theta\\
&-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^5\theta\\
\end{align}
$$
Discarding the complex part and writing $\sin^2=1-\cos^2$ gives us
$$
\begin{align}
\cos5\theta&=16\cos^5\theta-20\cos^3\theta+5\cos\theta
\end{align}
$$
