How to integrate $\cos^3x$ by parts I've converted $\cos^3(x)$ into $\cos^2(x)\cos(x)$ but still have not gotten the answer. 
The answer is $\dfrac{\sin(x)(3\cos^2x + 2\sin^2x)}{3}$
My answer was the same except I did not have a $3$ infront of $x$ and my $2\sin^2x$ was not squared.
Help! 

If you wish to do this using by parts,
use $\cos^2x$ as the first function (to differentiate) and integrate $\cos x$ to get:
$$\int \cos^2 x \cos x dx = \cos^2 x \sin x + 2\int \sin^2 x \cos {x}dx$$
Now use $\sin x = t$ to get $dt = \cos x dx$ and
$$\int \cos^2 x \cos x dx = \cos^2 x \sin x + \frac{2\sin^3 x}{3} + C$$
