Do you remember the Probability Integral ?

$J =\int_{-\infty}^{\infty} \cos(2x){e^\frac{-x^2}{2}}dx = 2\int_0^{\infty} \cos(2x){e^\frac{-x^2}{2}}dx$

Let $I(a) = \int_0^{\infty} \cos(ax){e^\frac{-x^2}{2}}dx$

Differentiating both sides wrt to a using Leibnitz's Rule , $I'(a) = \int_0^{\infty} -x\sin(ax){e^\frac{-x^2}{2}}dx$

Now by Integration by parts $I'(a) = \sin(ax){e^\frac{-x^2}{2}} |_{x=0}^{x=\infty} -\int_0^{\infty} a\cos(ax){e^\frac{-x^2}{2}}dx$ $\therefore I'(a) = -aI(a)$

$\implies \int\frac{I'(a)}{I(a)}da = - \int ada$ $\ln{(I(a))} = -\frac{a^2}{2} + C$

$I(0) = \int_0^{\infty}e^\frac{-x^2}{2}dx$ $= \sqrt{\frac{\pi}{2}}$ ...( Probability Integral ) $\therefore C = \ln{\sqrt{\frac{\pi}{2}}}$

$\therefore \boxed{I(a) = e^{\frac{-a^2}{2} + \ln{\sqrt{\frac{\pi}{2}}}} = \sqrt{\frac{\pi}{2}}{e^{\frac{-a^2}{2}}}}$

$\implies J = 2I(2) = \sqrt{2\pi}{e^{-2}} \approx 0.33923$

$\therefore 100×J = \boxed{33.923}$