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**Prove It** I should have read the question properly.

As an alternative to Integration by parts, there's the "Find the derivative and create a new integral equation" method.

$\displaystyle \frac{d}{dx}(x^2\cos{2x}) = 2x^2\sin{2x} + 2x\cos{2x}$.

So $\displaystyle \int{2x^2\sin{2x}\,dx} + \int{2x\cos{2x}\,dx} = x^2\cos{2x}$

Now to find these integrals, differentiate each of them...

$\displaystyle \frac{d}{dx}(2x^2\sin{2x}) = 4x^2\cos{2x} + 4x\sin{2x}$

So $\displaystyle \int{4x^2\cos{2x}\,dx} + \int{4x\sin{2x}\,dx} = 2x^2\sin{2x}$

Therefore $\displaystyle 4\int{x^2\cos{2x}\,dx} = 2x^2\sin{2x} - 4\int{x\sin{2x}\,dx}$

$\displaystyle \int{x^2\cos{2x}\,dx} = \frac{1}{2}x^2\sin{2x} - \int{x\sin{2x}\,dx}$.

Now we need to try to find $\displaystyle \int{x\sin{2x}\,dx}$.

We should differentiate the other term in our original equation:

$\displaystyle \frac{d}{dx}(2x\cos{2x}) = 2\cos{2x} - 4x\sin{2x}$.

So $\displaystyle \int{2\cos{2x}\,dx} - \int{4x\sin{2x}\,dx} = 2x\cos{2x}$

$\displaystyle 4\int{x\sin{2x}\,dx} = 2\int{\cos{2x}\,dx} - 2x\cos{2x}$

$\displaystyle \int{x\sin{2x}\,dx} = \frac{1}{2}\int{\cos{2x}\,dx} - \frac{1}{2}x\cos{2x}$

$\displaystyle \int{x\sin{2x}\,dx} = \frac{1}{4}\sin{2x} - \frac{1}{2}x\cos{2x}$.

So... going back to what we were originally trying to find

$\displaystyle \int{x^2\cos{2x}\,dx} = \frac{1}{2}x^2\sin{2x} - \int{x\sin{2x}\,dx}$

$\displaystyle = \frac{1}{2}x^2\sin{2x} + \frac{1}{2}x\cos{2x} - \frac{1}{4}\sin{2x} + C$.