Chapter 5.3, Problem 25E

To determine

To evaluate : the integral

Answer to Problem 25E

The value of ${\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\int }}\frac{1+{\cos }^2\theta }{{\cos }^2\theta }}d\theta$ is $1+\frac{\pi }{4}$

Explanation of Solution

Given information : The integral is ${\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\int }}\frac{1+{\cos }^2\theta }{{\cos }^2\theta }}d\theta$

Calculation :

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\int }}\frac{1+{\cos }^2\theta }{{\cos }^2\theta }}d\theta \\ \Rightarrow {\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\int }}\frac{1}{{\cos }^2\theta }+\frac{{\cos }^2\theta }{{\cos }^2\theta }}d\theta \\ \Rightarrow {\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\int }}\left(\frac{1}{{\cos }^2\theta }+1\right)}d\theta \\ \Rightarrow {\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\int }}\frac{1}{{\cos }^2\theta }}d\theta +{\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\int }}1}d\theta \\ \Rightarrow {\left[\tan \theta \right]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}+{\left[\theta \right]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\\ \Rightarrow \left(\tan \frac{\pi }{4}-\tan 0\right)+\left(\frac{\pi }{4}-0\right)\\ \Rightarrow \left(1-0\right)+\frac{\pi }{4}\\ \Rightarrow 1+\frac{\pi }{4}\end{array}$