Answer
$\dfrac{3}{8}+ \ln \sqrt 2$
Work Step by Step
Given: $\int^{0}_{-\ln 2} \cosh^2(\dfrac{x}{2}) dx$
As we know, $\cosh x=\dfrac{e^{x} + e^{-x}}{2}$
Thus, $\int^{0}_{-\ln 2} \cosh^2(\dfrac{x}{2}) dx=\int^{0}_{-\ln 2} [\dfrac{e^{x/2} + e^{-x/2}}{2}]^2 dx$
This implies: $\int^{0}_{-\ln 2} [\dfrac{e^{x/2} + e^{-x/2}}{2}]^2 dx= \dfrac{1}{4}\int^{0}_{-\ln 2} (e^{x} + e^{-x}+2) dx=(\dfrac{1}{4})[ e^{x} - e^{-x}+2x]^{0}_{-\ln 2}$
and, $= -(\dfrac{1}{4})[-\dfrac{3}{2}-2 \ln 2]$
Hence, $\int^{0}_{-\ln 2} \cosh^2(\dfrac{x}{2}) dx=\dfrac{3}{8}+ \ln \sqrt 2$
