Chapter A6, Problem 59E

To determine

To Evaluate: the integrals.

Answer to Problem 59E

The result is $0.722$

Explanation of Solution

Given:

  ${\displaystyle \underset{-\ln 2}{\overset{0}{\int }}{\cosh }^2\left(\frac{x}{2}\right)}dx$

Calculation:

Therefore the integral can now be evaluated as,

  $\begin{array}{l}{\displaystyle \underset{-\ln 2}{\overset{0}{\int }}{\cosh }^2\left(\frac{x}{2}\right)}.dx={\displaystyle \underset{-\ln 2}{\overset{0}{\int }}\frac{\cosh x+1}{2}.dx}\\ =\frac{1}{2}{\displaystyle \underset{-\ln 2}{\overset{0}{\int }}\left(\cosh x+1\right)dx}\\ =\frac{1}{2}{\left[\sinh x+x\right]}_{-\ln 2}^0\end{array}$

Continuation of the above step

  $\begin{array}{l}=\frac{1}{2}\left[\left(\left(\sinh 0+0\right)-\left(\sinh \left(-\ln 2\right)\right)-\ln 2\right)\right]\\ =\frac{1}{2}\left[\left(0+0\right)-\left(\frac{e^{-\ln 2}-e^{\ln 2}}{2}-\ln 2\right)\right]\\ =\frac{1}{2}\left[-\frac{\left(1/2\right)-2}{2}+\ln 2\right]\\ =\frac{1}{2}\left(1-\frac{1}{4}+\ln 2\right)\\ =\frac{3}{8}+\ln \sqrt{2}\\ \approx 0.722\end{array}$

Conclusion:

The result is $0.722$