A few ways to integrate $e^{-ax}\sin(x)$:
1) Integration by parts:
$$\begin{align}\int e^{-ax}\sin(x)~dx&=-e^{-ax}\cos(x)-a\int e^{-ax}\cos(x)~dx\\&=-e^{-ax}\cos(x)-a\left(e^{-ax}\sin(x)+a^2\int e^{-ax}\sin(x)~dx\right)\end{align}$$
Let $I=\int e^{-ax}\sin(x)~dx$ to see that
$$I=-e^{-ax}\cos(x)-a\left(e^{-ax}\sin(x)+a^2I\right)$$
which is a linear equation to solve for $I$.
2) Euler's formula:
This is a more complex method (get the pun?) but pretty straight forward. One may either use
$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}\qquad or\qquad\sin(x)=\Im(e^{ix})$$
Using the second one for simplicity, we see that
$$\begin{align}I&=\Im\int e^{-ax}e^{ix}~dx\\&=\Im\int e^{(i-a)x}~dx\\&=\Im\left(\frac1{i-a}e^{(i-a)x}\right)+c\\&=\Im(u+vi)+c\\&=v+c\end{align}$$
where $v$ is the imaginary part of $\frac1{i-a}e^{(i-a)x}$.
As per the original problem, this is how I would've tackled it, using the complex method:
$$\begin{align}\int_0^1\frac{e^{-ax}\sin(x)}x\ dx&=\int_0^1e^{-ax}\sin(x)\int_0^\infty e^{-xt}\ dt\ dx\\&=\int_0^\infty\int_0^1e^{-(a+t)x}\sin(x)\ dx\ dt\\&=\int_0^\infty\Im\int_0^1e^{[i-(a+t)]x}\ dx\ dt\\&=\int_0^\infty\Im\left(\frac1{i-(a+t)}e^{[i-(a+t)]x}\bigg|_{x=0}^1\right)\ dt\\&=\int_0^\infty\frac1{1+(a+t)^2}\left(1-\frac{\cos(t)+(a+t)\sin(t)}{e^{a+t}}\right)\ dt\end{align}$$
And I think this is far as you can go this way.
I'm not sure about Feynman's trick, but there is a delightfully simple way to compute this integral using Glasser's Master Theorem. First, substitute $u=x^2$ to obtain
$$2\int_0^\infty \frac{u^2}{u^4+2u^2+5}\,du=\int_{-\infty}^\infty \frac{1}{u^2+5u^{-2}+2}\,du$$
Now, notice that $$u^2+5u^{-2}+2=\left(u-\frac{\sqrt{5}}{u}\right)^2+4\phi,$$
where $\phi=\frac{1+\sqrt{5}}{2}$.
Glasser's Master Theorem then tells us that
$$\int_{-\infty}^\infty \frac{1}{\left(u-\sqrt{5}/u\right)^2+4\phi}\,du=\int_{-\infty}^\infty \frac{1}{u^2+4\phi}\,du.$$
The last integral is just
$$\frac{\pi}{2\sqrt{\phi}}=\frac{\pi}{\sqrt{2}}\cdot\frac{1}{\sqrt{1+\sqrt{5}}}$$
Best Answer
Let
$$ I(a)=\int_0^c dx\ \frac{\sin(ax^{1/2})}{x^{1/2}}$$
We've chosen $a$ so that taking the derivative will simplify$^\dagger$ the expression.
$$ \partial_aI(a)=\int_0^c dx \ \cos(ax^{1/2})$$
Now we perform the $x$ integral
$$ \partial_aI(a)= \frac{2}{a^2} \left( a c^{1/2} \sin(ac^{1/2})+\cos(ac^{1/2})-1 \right) $$
Then the $a$ integral
$$ I(a)=\frac{2}{a} \left( 1- \cos(ac^{1/2})\right) +C$$
To find $C$ note that $I(0)=0$ from the first equation. In this case $C=0$ and we have
$$ I(a)=\frac{2}{a} \left( 1- \cos(ac^{1/2})\right)$$
Setting $a=\pi, c=441$, and multiplying by $\pi$ gives exactly $4$.
$\dagger$ In this case, it's not clear if the 'simplified' integral is actually simpler to do on paper. Still, this is the general idea behind Feynman's trick.