Chapter A6, Problem 53E

To determine

To Evaluate: the integrals.

Answer to Problem 53E

The result is $0.787$

Explanation of Solution

Given:

  ${\displaystyle \underset{-\ln 4}{\overset{-\ln 2}{\int }}2e^{\theta }\cosh \theta d\theta }$

Calculation:

Therefore the integral can now be evaluated as,

  $\begin{array}{l}{\displaystyle \underset{-\ln 4}{\overset{-\ln 2}{\int }}2e^{\theta }\cosh \theta d\theta }={\displaystyle \underset{-\ln 4}{\overset{-\ln 2}{\int }}2e^{\theta }\left(\frac{e^{\theta }+e^{-\theta }}{2}\right)}.d\theta \\ ={\displaystyle \underset{-\ln 4}{\overset{-\ln 2}{\int }}\left(e^{2\theta }+1\right)}.d\theta \\ ={\left[\frac{e^{2\theta }}{2}+\theta \right]}_{-\ln 4}^{-\ln 2}\\ =\left(\frac{e^{-2\ln 2}}{2}-\ln 2\right)-\left(\frac{e^{-2\ln 4}}{2}-\ln 4\right)\\ =\left(\frac{1}{8}-\ln 2\right)-\left(\frac{1}{32}-\ln 4\right)\\ \approx 0.787\end{array}$

Conclusion:

The result is $0.787$