How can you find the integral of $\frac{cos(2t)}{2t^2}$ between 1 and infinity? How can you find the integral of $\frac{\cos(2t)}{2t^2}$
between 1 and infinity?
$$
I = \int\limits_1^\infty \frac{\cos(2t)}{2t^2} dt
$$ 
My problem is that I just simply do not know how to handle this. I have tried integration by parts and some trigonometric identities...

While the actual integration of $\cos(2t)/(2t^2)$ would be difficult to find explicitly, you can show that the integral converges by noting $$|I| \le \int_1^\infty \frac{|\cos(2t)|}{2t^2} dt \le \int_1^\infty \frac{1}{2t^2} dt < \infty.$$ The last integral can be integrated by standard techniques. Also we know that it converges by the $p$-test.
As for computing the value of the integral, one method that is typically used to compute improper integrals is to transform the integral to an integral over a finite interval. We can achieve this with the substitution $u=1/t$ which makes $du = -1/t^{2} dt$. Thus we have $$\int_{1}^\infty \frac{\cos(2t)}{2t^2} dt = \int_{0}^1 \frac{\cos(2/u)}{2} du.$$
This wont be any easier to find an antiderivative for, however, it will make it easier to approximate the integral. One way would be to use a Riemann sum on the interval $[0,1]$.
