\( \int _ { 0 } ^ { \ln 2 } 2 e ^ { - x } \sinh ( x ) d x = \)

To solve the integral \( \int _ { 0 } ^ { \ln 2 } 2 e ^ { - x } \sinh ( x ) d x \), we start by recalling the definition of the hyperbolic sine function: \[ \sinh(x) = \frac{e^x - e^{-x}}{2}. \] Substituting this into the integral, we get: \[ \int _ { 0 } ^ { \ln 2 } 2 e ^ { - x } \sinh ( x ) d x = \int _ { 0 } ^ { \ln 2 } 2 e^{-x} \left( \frac{e^x - e^{-x}}{2} \right) d x. \] This simplifies to: \[ \int _ { 0 } ^ { \ln 2 } (e^{-x} e^x - e^{-x} e^{-x}) d x = \int _ { 0 } ^ { \ln 2 } (1 - e^{-2x}) d x. \] Calculating the integral: \[ = \int _ { 0 } ^ { \ln 2 } 1 d x - \int _ { 0 } ^ { \ln 2 } e^{-2x} d x = \left[ x \right]_{0}^{\ln 2} - \left[ -\frac{1}{2} e^{-2x} \right]_{0}^{\ln 2}. \] Evaluating each part involves: 1. \(\left[ x \right]_{0}^{\ln 2} = \ln 2 - 0 = \ln 2\). 2. \(\left[ -\frac{1}{2} e^{-2x} \right]_{0}^{\ln 2} = -\frac{1}{2} e^{-2 \ln 2} - \left(-\frac{1}{2} e^{0}\right) = -\frac{1}{2} \cdot \frac{1}{4} + \frac{1}{2} = -\frac{1}{8} + \frac{1}{2} = \frac{3}{8}\). Putting it all together: \[ \int _ { 0 } ^ { \ln 2 } 2 e ^ { - x } \sinh ( x ) d x = \ln 2 - \frac{3}{8}. \] So the final result is: \[ \ln 2 - \frac{3}{8}. \]