As Daniel Fischer pointed on the comment above, the integral can be computed as follows:
$\int J_o(x) \cos(x) dx=x J_0(x) \cos(x) + \int x J_1(x) \cos(x) dx + \int x J_0 \sin(x) dx +C= J_0(x) \cos(x) + \int x J_1(x) \cos(x) dx+x J_1(x) \sin(x) - \int x J_1(x) \cos(x) dx$
which produces the desired result.
Note that by dominated convegence, we have that
$$\lim_{a\to 0}\int_0^c e^{-ax}J_0(x)dx=\int_0^c J_0(x)dx.$$
Thus it suffices to show that
$$\lim_{c\to \infty}\lim_{a\to 0}\int_c^{\infty}e^{-ax}J_0(x)dx=0.$$
As you have observed, there seems to be no cheap way to do this, but it isn't that bad employing the asymptotic for $J_0(x)$ are $x\to \infty$. We can write
$$J_0(x)=\sqrt{\frac{2\pi}{x}}\cos(x-\pi/4)+\frac{1}{x^{3/2}}g(x),$$
where $g(x)$ a bounded function on $[1,\infty)$, such that say $|g(x)|\le C$.
Using this we see that
$$\int_c^{\infty}e^{-ax}J_0(x)dx=\int_c^{\infty}e^{-ax}\sqrt{\frac{2\pi}{x}}\cos(x-\pi/4)dx+\int_c^{\infty}\frac{e^{-ax}}{x^{3/2}}g(x)dx.$$
For the second integral, note that
$$|\int_c^{\infty}\frac{e^{-ax}}{x^{3/2}}g(x)dx|\le C|\int_c^{\infty}\frac{1}{x^{3/2}}dx|\le \frac{C}{2c^{1/2}},$$
so that
$$\lim_{c\to \infty}\lim_{a\to 0}\int_c^{\infty}\frac{e^{-ax}}{x^{3/2}}g(x)dx=0.$$
To bound the first integral, we can use integration by parts:
$$\int_c^{\infty}e^{-ax}\sqrt{\frac{2\pi}{x}}\cos(x-\pi/4)dx=-\int_c^{\infty}\frac{\sqrt{2\pi}}{2x^{3/2}}\int_x^{\infty}e^{-ay}\cos(y-\pi/4)dy+\sqrt{\frac{2\pi}{c}}\int_c^{\infty}e^{-ay}\cos(y-\pi/4)dy.$$
Now we can compute that
$$\int_{c}^\infty e^{-ax}cos(x-\pi/4)dx=e^{-ac}\frac{(1+a)cos(c)-(1-a)sin(c)}{\sqrt{2}(1+a^2)}.$$
We see from the right that for such that for $a<1$, $$|\int_c^{\infty}e^{-ay}\cos(y-\pi/4)dy|\le 4.$$
Thus
$$|\sqrt{\frac{2\pi}{c}}\int_c^{\infty}e^{-ay}\cos(y-\pi/4)dy|\le \sqrt{4\frac{2\pi}{c}}.$$
$$|\int_c^{\infty}\frac{\sqrt{2\pi}}{2x^{3/2}}\int_x^{\infty}e^{-ay}\cos(y-\pi/4)dy|\le-\int_c^{\infty}\frac{4\sqrt{2\pi}}{2x^{3/2}}\le \frac{4\sqrt{2\pi}}{c^{1/2}}.$$
Both of these expressions vanish taking interactive limits, so we see that
$$\lim_{c\to \infty}\lim_{a\to 0}\int_c^{\infty}e^{-ax}\sqrt{\frac{2\pi}{x}}\cos(x-\pi/4)dx=0.$$
This completes the proof.
Best Answer
Well, all we need are two formulas as below: $$\frac{d}{dx}J_0(x)=-J_1(x),$$ $$\frac{d}{dx}J_1(x)=J_0(x)-x^{-1}J_1(x).$$
Now, let's use integration by parts:$$\int xJ_0^2(x)=\frac{1}{2}x^2J_0^2(x)+\int x^2J_0(x)J_1(x).$$
So, we just need to prove: $$\int x^2J_0(x)J_1(x)=\frac{x^2}{2}J_1^2(x).$$
And, this is almost obvious since:
$$\frac{d}{dx}(\frac{x^2}{2}J_1^2(x))=xJ_1^2(x)+x^2J_1(x)(J_0(x)-x^{-1}J_1(x)).$$
Edit: The two formulas I mentioned can be found in most typical advanced engineering-mathematics books, for example by Erwin Kreyszig.