From 0 to $\pi$

Express $x^2$ $cos^2 x$ as $x^2$ ( $\frac{1+cos 2x}{2}$ ) first. Upon integration of $\frac{1}{2}$ $x^2$ , we have $\frac{1}{2} (\frac{x^3}{3})$ = $\frac{x^3}{6}$ which gives us $\frac{\pi^3}{6}$ upon substitution of 0 and $\pi$ .

Now to intergrate $\frac{1}{2}$ $x^2 cos 2x$ , we let $\frac{1}{2}$ $x^2$ be $u$ and $\frac{dv}{dx}$ be $cos 2x$ . As such, $\frac{du}{dx}=x$ while $v=\frac{sin 2x}{2}$ and we have $\frac{1}{2}$ $x^2 (\frac{sin 2x}{2})$ - the antiderivative of $x(\frac{sin 2x}{2})$ .

It can be seen that we need to perform integration by parts a second time, but thank goodness $\frac{1}{2}$ $x^2 (\frac{sin 2x}{2})$ will give us 0 when we substitute 0 and $\pi$ into it.

To integrate $-x(\frac{sin 2x}{2})$ , we let $u = x$ and $\frac{dv}{dx}$ be $\frac{-sin 2x}{2}$ such that $\frac{du}{dx} =1$ while $v=\frac{1}{2}$ $\frac{cos 2x}{2}$ = $\frac{cos 2x}{4}$ and we have $x(\frac{cos 2x}{4})$ - the antiderivative of $(1)(\frac{cos 2x}{4})$ = $x(\frac{cos 2x}{4}) - \frac{1}{4} (\frac{sin 2x}{2})$ which simplifies to $\frac{\pi}{4}$ upon substitution.

So we have $\frac{\pi^3}{6}$ + $\frac{\pi}{4}$ where $a=3$ , $b=6$ and $c=4$ . So $abc = \boxed{72}$ .