Chapter A6, Problem 55E

To determine

To Evaluate: the integrals.

Answer to Problem 55E

The result is $2.350$

Explanation of Solution

Given:

  ${\displaystyle \underset{-\pi /4}{\overset{\pi /4}{\int }}\cosh \left(\tan \theta \right)}{\sec }^2\theta d\theta$

Calculation:

assume a variable $u$ such that

  $u=\tan \theta ,du={\sec }^2\theta d\theta$

The limit now changes as follows:

Lower Limit $\tanh \left(-\frac{\pi }{4}\right)=-1$

Upper Limit $\tanh \left(\frac{\pi }{4}\right)=1$

Therefore the integral can now be evaluated as,

  $\begin{array}{l}{\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\cosh \left(\tan \theta \right){\sec }^2\theta .d\theta ={\displaystyle \underset{-1}{\overset{1}{\int }}\cosh u.du}}\\ ={\left[\sinh u\right]}_{-1}^1\\ =\sinh \left(1\right)-\sinh \left(-1\right)\\ =\left(\frac{e^1-e^{-1}}{2}\right)-\left(\frac{e^{-1}-e^1}{2}\right)\\ =e-e^{-1}\\ \approx 2.350\end{array}$

Conclusion:

The result is $2.350$