Not hard, just long

Let us name the main integral as $V$ .

$V = \large \int_0^{\infty} x^2 e^{-2x} \sin \: 4x \: \cos \: 2x \: dx$

Now, for us to be able to simplify the integral, let us make use of these two identities:

$\sin (A+B) = \sin A \cos B + \cos A \sin B$ $\sin (A-B) = \sin A \cos B - \cos A \sin B$

Adding these two equations, we can derive such equation below:

$\sin A \cos B = \frac{ \sin (A+B) + \sin (A-B) }{2}$

This changes $V$ into

$V = \large \frac{1}{2} \int_0^{\infty} x^2 e^{-2x} (\sin 6x + \sin 2x) \: dx$

Later on we can split this into $V_1$ and $V_2$ so that the computation becomes less tedious, such that $V = \frac{1}{2} (V_1 + V_2)$ .

So now we have an integral of the form

$\large \int_0^{\infty} x^2 e^{-ax} \sin bx dx$

Let us name this integral $Q$ , with $V_1$ and $V_2$ having the same form as $Q$ , differing only on $b$ .

We can do integration by parts immediately here, but to avoid such long computation, let us do this via Feynman's differentiation under the integral sign.

Let us assign (well, again) an integral $I$ .

$I = \int_0^{\infty} e^{-ax} \sin bx dx$

Differentiating this under $a$ two times will yield the following:

$I' = \frac{d}{da} \int_0^{\infty} e^{-ax} \sin bx dx = \int_0^{\infty} -xe^{-ax} \sin bx dx$

$I'' = \int_0^{\infty} x^2e^{-ax} \sin bx dx$

Now here, we can clearly see that $Q = I''$ with respect to $a$ .

So now, what is $I$ ?

Although this integral is a lot easier to integrate than the original integral $V$ , it will still involve a considerable amount of work. Here, we will use integration by parts twice.

$I = \int_0^{\infty} e^{-ax} \sin bx dx$

let $u = e^{-ax}$ , and $dv = \sin bx dx$

$I = e^{-ax}(-\frac{1}{b} \cos bx) - \int_0^{\infty} \frac{a}{b} e^{-ax} \cos bx dx$

setting again another $u =e^{-ax}$ and $dv = \cos bx dx$

$I = e^{-ax}(-\frac{1}{b}\cos x) - \frac{a}{b}[ e^{-ax}(\frac{1}{b}\sin bx) + \int_0^{\infty}\frac{1}{a} e^{-ax} \sin x dx]$

$I = e^{-ax}(-\frac{1}{b}\cos x) - \frac{a}{b}[ e^{-ax}(\frac{1}{b}\sin bx) + I]$

Further manipulation will lead us to

$I = - \frac {e^{-ax} (a \sin bx + b \cos bx)}{a^2 + b^2} \big|_0^{\infty}$

This simplifies to

$I = \frac{b}{a^2 + b^2}$

Now we have an $I$ solved in terms of $a$ .

Differentiating this twice, and we will get

$I'' = \frac {2b(3a^2 -b^2)}{b^2 + a^2}$

Now, this will be the equivalent of the integral $Q$ !

Since $V$ takes the form of $Q$ , we can now use the above equation to find for its value. Substituting for $V_1$ , with $a= 2$ , and $b= 6$ , we will have

$V_1 = - \frac {9}{2000}$

while $V_2$ on the other hand, for $a=2$ and $b=2$ , we have

$V_2 = \frac {1}{16}$

So now we have $V$ , and that is

$\frac{1}{2}(V_1 + V_2) = \frac {29}{1000}$

That now gives us $29 + 1000 = \boxed {1029}$