Answer
$$\ln 2 + \frac{3}{8}$$
Work Step by Step
$$\eqalign{
& \int_0^{\ln 2} {2{e^{ - x}}\cosh x} dx \cr
& {\text{By definition }}\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2} \cr
& \int_0^{\ln 2} {2{e^{ - x}}\cosh x} dx = \int_0^{\ln 2} {2{e^{ - x}}\left( {\frac{{{e^x} + {e^{ - x}}}}{2}} \right)} dx \cr
& {\text{Multiplying}} \cr
& = \int_0^{\ln 2} {{e^{ - x}}\left( {{e^x} + {e^{ - x}}} \right)} dx \cr
& = \int_0^{\ln 2} {\left( {{e^{x - x}} + {e^{ - x - x}}} \right)} dx \cr
& = \int_0^{\ln 2} {\left( {1 + {e^{ - 2x}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = \left[ {x - \frac{1}{2}{e^{ - 2x}}} \right]_0^{\ln 2} \cr
& = \left[ {\ln 2 - \frac{1}{2}{e^{ - 2\left( {\ln 2} \right)}}} \right] - \left[ {0 - \frac{1}{2}{e^{ - 2\left( 0 \right)}}} \right] \cr
& {\text{ = }}\left( {\ln 2 - \frac{1}{8}} \right) - \left( { - \frac{1}{2}} \right) \cr
& = \ln 2 - \frac{1}{8} + \frac{1}{2} \cr
& = \ln 2 + \frac{3}{8} \cr} $$
