Chapter 5.7, Problem 14E

a.

To determine

To verify the expression ${\displaystyle \int {\sec }^3\theta d\theta =\frac{1}{2}}\left(\sec \theta \tan \theta +\ln \left|\sec \theta +\tan \theta \right|\right)+C$ , Using Differentiation

Explanation of Solution

Given information :

  ${\displaystyle \int {\sec }^3\theta d\theta =\frac{1}{2}}\left(\sec \theta \tan \theta +\ln \left|\sec \theta +\tan \theta \right|\right)+C$

Formula used :

  $\frac{d}{d\theta }\left(\sec \theta \right)=\tan \theta \sec \theta$ , $\frac{d}{d\theta }\left(\sec \theta \right)=\tan \theta \sec \theta$

Verification:

Let $f\left(\theta \right)={\sec }^3\theta$ and $F\left(\theta \right)=\frac{1}{2}\left(\sec \theta \tan \theta +\ln \left|\sec \theta +\tan \theta \right|\right)+C$

Here , $C$ is constant of integration , therefore $\frac{d}{d\theta }\left(C\right)=0$

Now , $\frac{d}{d\theta }F\left(\theta \right)=\frac{d}{d\theta }\left(\frac{1}{2}\left(\sec \theta \tan \theta +\ln \left|\sec \theta +\tan \theta \right|\right)+C\right)$

  $\begin{array}{l}=\frac{1}{2}\frac{d}{d\theta }\left(\left(\sec \theta \tan \theta +\ln \left|\sec \theta +\tan \theta \right|\right)+C\right)\\ =\frac{1}{2}\left(\frac{d}{d\theta }\left(\sec \theta \tan \theta \right)+\frac{d}{d\theta }\left(\ln \left|\sec \theta +\tan \theta \right|\right)+\frac{d}{d\theta }\left(C\right)\right)\\ =\frac{1}{2}\left(\sec \theta \frac{d}{d\theta }\left(\tan \theta \right)+\tan \theta \frac{d}{d\theta }\left(\sec \theta \right)+\frac{1}{\sec \theta +\tan \theta }\frac{d}{d\theta }\left(\sec \theta +\tan \theta \right)+0\right)\\ =\frac{1}{2}\left({\sec }^3\theta +\sec \theta {\tan }^2\theta +\frac{\sec \theta }{\sec \theta +\tan \theta }\times \left(\sec \theta +\tan \theta \right)\right)\\ =\frac{1}{2}\left({\sec }^3\theta +\sec \theta \left({\sec }^2\theta -1\right)+\sec \theta \right)\\ =\frac{1}{2}\left({\sec }^3\theta +{\sec }^3\theta -\sec \theta +\sec \theta \right)\\ =\frac{1}{2}\left(2{\sec }^3\theta \right)={\sec }^3\theta =f\left(\theta \right)\end{array}$

Hence it is verified , ${\displaystyle \int {\sec }^3\theta d\theta =\frac{1}{2}}\left(\sec \theta \tan \theta +\ln \left|\sec \theta +\tan \theta \right|\right)+C$ , Using Differentiation

b.

To determine

To evaluate the integral ${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{x^2+1}dx}$

Answer to Problem 14E

The value is $\frac{1}{2}\left(\sqrt{2}+\ln \left|\sqrt{2}+1\right|\right)$

Explanation of Solution

Given information :

The integral is ${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{x^2+1}dx}$

Formula used :

  $\frac{d}{d\theta }\left(\tan \theta \right)={\sec }^2\theta$ , $\frac{d}{d\theta }\left(\sec \theta \right)=\tan \theta \sec \theta$ , ${\displaystyle \int {\sec }^2\theta d\theta =\tan \theta }$ , $1+{\tan }^2\theta ={\sec }^2\theta$

Calculation:

The integral is ${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{x^2+1}dx}$

Let $x=\tan \theta$

Differentiating both side , $dx={\sec }^2\theta d\theta$

Substituting , $x=\tan \theta$ and $dx={\sec }^2\theta d\theta$ in given integral

Lower limit : When $x=0$ then $\theta ={\tan }^{-1}0=0$

Upper limit : When $x=1$ then $\theta ={\tan }^{-1}1=\frac{\pi }{4}$

  ${\displaystyle \underset{0}{\overset{\pi /4}{\int }}\sqrt{{\tan }^2\theta +1}\times {\sec }^2\theta d\theta =}{\displaystyle \underset{0}{\overset{\pi /4}{\int }}{\sec }^3\theta d\theta =}I={\displaystyle \underset{0}{\overset{\pi /4}{\int }}\sec \theta {\sec }^2\theta d\theta }$

Using By Parts Rule of integration,

  $\text{Considering first function = sec}\theta$ , ${\text{Considering second function = sec}}^2\theta$

  $\begin{array}{l}I={\left|\sec \theta \times {\displaystyle \int {\sec }^2\theta d\theta -{\displaystyle \int \left(\frac{d}{d\theta }\left(\sec \theta \right)\times {\displaystyle \int {\sec }^2\theta d\theta }\right)d\theta }}\right|}_0^{\pi /4}\\ I={\left|\sec \theta \tan \theta -{\displaystyle \int \sec \theta \tan \theta \times \tan \theta d\theta }\right|}_0^{\pi /4}\\ I={\left|\sec \theta \tan \theta -{\displaystyle \int \sec \theta {\tan }^2\theta d\theta }\right|}_0^{\pi /4}\\ I={\left|\sec \theta \tan \theta -{\displaystyle \int \sec \theta \left({\sec }^2\theta -1\right)d\theta }\right|}_0^{\pi /4}\\ I={\left|\sec \theta \tan \theta \right|}_0^{\pi /4}-{{\displaystyle \underset{0}{\overset{\pi /4}{\int }}{\sec }^3\theta +\left|\ln \left|\sec \theta +\tan \theta \right|\right|}}_0^{\pi /4}\\ 2I=\left(\sec \frac{\pi }{4}\tan \frac{\pi }{4}-\sec 0\tan 0\right)+\left(\ln \left|\sec \frac{\pi }{4}+\tan \frac{\pi }{4}\right|-\ln \left|\sec 0+\tan 0\right|\right)\\ 2I=\left(\sqrt{2}\times 1-1\times 0\right)+\ln \left|\sqrt{2}+1\right|-\ln \left|1+0\right|\\ 2I=\sqrt{2}+\ln \left|\sqrt{2}+1\right|\\ I=\frac{1}{2}\left(\sqrt{2}+\ln \left|\sqrt{2}+1\right|\right)\end{array}$ ,

Hence the value of ${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{x^2+1}dx}$ is $\frac{1}{2}\left(\sqrt{2}+\ln \left|\sqrt{2}+1\right|\right)$