Help figuring out all the alternative solutions to the integrals of sine and cosine I always worry a lot when doing integrals with trigonometric functions because there's always many ways to write the final answer.  I am trying to figure out the general pattern for the various different solutions.
The integral
$$ \int \sin^3\left(x\right)\cos^2\left(x\right)\, \mathrm{d}x $$
My answer (which never seems to be the same with wolfram alpha's):
$$\frac{\cos^5\left(x\right)}{5} + \frac{-\cos^3\left(x\right)}{3} + C$$
wolfram alpha lists many more:
$$\cos^3\left(x\right)\left(\frac{1}{10}\cos\left(2x\right)-\frac{7}{30}\right) + C$$
$$\frac{1}{240}\left(-30\cos\left(x\right) - 5\cos\left(3x\right)+3\cos\left(5x\right)\right)+C$$
$$\frac{\cos^5\left(x\right)}{80} -\frac{\cos^3\left(x\right)}{48}-\frac{\cos\left(x\right)}{8}-\frac{1}{8}\sin^2\left(x\right)\cos^3\left(x\right) + \frac{1}{16}\sin^4\left(x\right)\cos\left(x\right)+\frac{1}{16}\sin^2\left(x\right)\cos\left(x\right) + C$$ 
Could anyone help me understand what integration procedure would result in those answers?

Since
$$\sin^2x\cos^2x=\frac{1}{4}\left(2\sin x\cos x\right)^2=\frac{1}{4}\sin 2x$$
we get
$$\int\sin^2x\cos^2x=\frac{1}{4}\int\sin 2x\,dx=-\frac{1}{8}\cos 2x\;\;\;(**)$$
and then, integrating by parts:
$$u=\sin x\;\;,\;\;u'=\cos x\\v'=\sin^2x\cos^2 x\;;\,\;\;v=-\frac{1}{8}\cos 2x\,\Longrightarrow$$
$$\int \sin^3x\cos^2 x\,dx=\int\sin x\left(\sin^2x\cos^2 x\right)dx=$$
$$-\frac{1}{8}\cos 2x\sin x+\frac{1}{8}\int\cos x\cos 2x\,dx=-\frac{1}{8}\cos 2x\sin x+\frac{1}{8}\int (1-\sin^2x)\cos x\,dx=$$
$$=-\frac{1}{8}\left(\cos 2x\sin x+\sin x-\frac{1}{3}\sin^3x\right)+C$$
