Answer
$\dfrac{3}{4}$
Work Step by Step
Given: $\int^{2}_{1} \dfrac{\cosh (\ln t)}{t} dt$
Plug in: $\ln t= a$ and $da=\dfrac{dt}{t}$
Thus, the limits of integration will also get changed.
Then, $\int^{2}_{1} \dfrac{\cosh (\ln t)}{t} dt=\int^{\ln 2}_{0} \cosh a da$
or, $=[ \sinh h a]^{\ln 2}_{0}$
or, $= \sinh (\ln 2) -0$
Since, $\sinh \theta=\dfrac{e^{\theta} -e^{-\theta}}{2}$
Thus, we have, $= \dfrac{e^{\ln 2} -e^{-\ln 2}}{2}$
or, $=\dfrac{3}{4}$
