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} $
 A: 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.
A: 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
$$
\begin{align}
\cos5\theta&=16\cos^5\theta-20\cos^3\theta+5\cos\theta
\end{align}
$$
A: First of all, Both answers $I=-\frac{\cos ^3 x}{3}+\frac{\cos ^5 x}{5}+C= \frac{1}{80} \cos 5 x-\frac{1}{48} \cos 3 x-\frac{1}{8} \cos x+C$  are correct, which can be confirmed by the following proof:
Let $z=\cos x+i\sin x$, then $\sin x= \frac{1}{2i}(z-\frac{1}{z})$ and $\cos x=   \frac{1}{2}(z+\frac{1}{z})$.
Expressing $\sin^3x\cos^2x$ in terms of $z$ and then sine of multiples of $x$ gives
$$
\begin{aligned}
\sin ^3 x \cos ^2 x =&-\frac{1}{32 i}\left(z-\frac{1}{z}\right)^3\left(z+\frac{1}{z}\right)^2 \\
=&-\frac{1}{32 i}\left(z^2-\frac{1}{z^2}\right)^2\left(z-\frac{1}{z}\right) \\
=&-\frac{1}{32 i}\left[\left(z^5-\frac{1}{z^5}\right)-\left(z^3-\frac{1}{z^3}\right)-2\left(z-\frac{1}{z}\right)\right] \\
=&-\frac{1}{32 i}(2 i \sin 5 x-2 i \sin 3 x-4 i \sin x) \\
=&-\frac{1}{16} \sin 5 x+\frac{1}{16} \sin 3 x+\frac{1}{8}\sin x
\end{aligned}
$$
Integrating back yields
$$
\boxed{\int \sin ^3 x \cos ^2 x d x=\frac{1}{80} \cos 5 x-\frac{1}{48} \cos 3 x-\frac{1}{8} \cos x+C}
$$
Wish it helps!
