I=int_0^oo cos(2t)/t*dt
I use Feynman's trick for evaluating integral,
I(a)=int_0^oo (e^(-at)*cos(2t))/t*dt
I note that original integral can be reached after setting b=0.
After differentiating under integral sign according to a,
I'(a)=int_0^oo -e^(-at)*cos2t*dt
=1/a*[e^(-at)*cos2t]_0^oo+2/a*int_0^oo e^(-at)*sin2t*dt
=-1/a+2/a*int_0^oo e^(-at)*sin2t*dt
=-1/a-2/a^2*[e^(-at)*sin2t]_0^oo-4/a^2*int_0^oo e^(-at)*cos2t*dt
=-1/a-2/a^2*0-4/a^2*I'(a)
(1+4/a^2)*I'(a)=-1/a
I'(a)=-a/(a^2+4)
After integrating both sides,
I(a)=-1/2*arctan(a/2)+C
Now, I choose a strategic value b = b_0 in order to make our integrand vanish so that I(b_0) =0. In this case, took b=oo, so that I(oo)=0
Hence -1/2*arctan(oo/2)+C=0, so C=pi/4
Thus I=I(0)=-1/2*arctan(0/2)+pi/4=pi/4