Answer
$\dfrac{3}{8}+ \ln \sqrt 2$
Work Step by Step
Given: $\int^{0}_{-\ln 2} \cosh^2(\dfrac{x}{2}) dx$
Since, $\cosh x=\dfrac{e^{x} + e^{-x}}{2}$
Re-write as:$\int^{0}_{-\ln 2} [\dfrac{e^{x/2} + e^{-x/2}}{2}]^2 dx$
Then, $\int^{0}_{-\ln 2} [\dfrac{e^{x/2} + e^{-x/2}}{2}]^2 dx= \dfrac{1}{4}\int^{0}_{-\ln 2} (e^{x} + e^{-x}+2) dx$
or, $=\dfrac{1}{4}[ e^{x} - e^{-x}+2x]^{0}_{-\ln 2}$
or, $= -\dfrac{1}{4}[-\dfrac{3}{2}-2 \ln 2]$
or, $=\dfrac{3}{8}+ \ln \sqrt 2$
