Evaluate $\int{\sin^3(x)\cos^2(x)}dx$ I'm trying to solve $\int{\sin^3(x)\cos^2(x)}dx$.
I got $-\frac{1}{2}\cos(x)+C$, but the memo says $\frac{1}{5}\cos^5(x)-\frac{1}{3}\cos^3(x)+C$
This is my working:

Your help is appreciated!

I was challenged to do this question without using either substitution or even $\sin^2 x + \cos^2x = 1$,
$\sin^3 x \cdot\cos^2 x \\
= \sin x \cdot \sin^2 x \cdot \cos^2 x\\
= \sin x \cdot (\frac{1}{2}\cdot 2\sin x\cdot\cos x)^2\\
= \sin x \cdot (\frac{1}{2} \sin 2x)^2\\
= \frac{1}{4}\cdot (\sin x \cdot \sin 2x) \cdot \sin 2x\\
= \frac{1}{4} \cdot \frac{1}{2} (\cos x - \cos 3x)\cdot \sin 2x\\
= \frac{1}{8}\cdot (\sin2x\cdot \cos x - \sin2x\cdot \cos3x)\\
=\frac{1}{8}\cdot[\frac{1}{2}(\sin x + \sin 3x) - \frac{1}{2}(\sin(-x) + \sin5x) ]\\
=\frac{1}{8}\cdot\frac{1}{2} (\sin x + \sin 3x + \sin x - \sin 5x)\\
=\frac{1}{16}(2\sin x + \sin3x - \sin 5x)\\$
$$\therefore \int{(\sin^3x\cdot\cos^2x)}\cdot dx \\
= \frac{1}{16}\int{(2\sin x + \sin3x - \sin 5x)}\cdot dx\\
= \frac{(-\cos x)}{8} + \frac{(-\cos3x)}{3\times 16} - \frac{(-\cos5x)}{5\times 16} + C\\
= \frac{\cos5x}{80} - \frac{\cos3x}{48} - \frac{\cos x}{8} + C$$
$\dots$which isn't the answer in the memo but that doesn't mean it isn't right. 
Now, if there's any math teacher out there who won't accept my answer on a written examination, please speak or forever hold your peace.

Edit: This video could useful for general questions of this type
