Chapter 7.2, Problem 84E

(a)

To determine

To show ${\displaystyle \underset{0}{\overset{\sqrt{3}}{\int }}\frac{dx}{\sqrt{1+x^2}}}={\displaystyle \underset{0}{\overset{\pi /3}{\int }}\sec udu}$ by substitution.

Explanation of Solution

Given information: $u={\tan }^{-1}x$

Calculation :

Put $x=\tan u$ so that $dx={\sec }^{2}udu$

Also, when $x=0;u=0$ and when $x=\sqrt{3};u=\pi /3$

The given integral becomes as:

  ${\displaystyle \underset{0}{\overset{\sqrt{3}}{\int }}\frac{dx}{\sqrt{1+x^2}}}={\displaystyle \underset{0}{\overset{\pi /3}{\int }}\frac{{\sec }^{2}udu}{\sqrt{1+{\tan }^{2}u}}}={\displaystyle \underset{0}{\overset{\pi /3}{\int }}\frac{{\sec }^{2}udu}{\sec u}}={\displaystyle \underset{0}{\overset{\pi /3}{\int }}\sec udu}$

This is the required proof

(b)

To determine

To evaluate ${\displaystyle \underset{0}{\overset{\sqrt{3}}{\int }}\frac{dx}{\sqrt{1+x^2}}}$

Explanation of Solution

Given information: $u={\tan }^{-1}x$

Calculation :

  ${\displaystyle \underset{0}{\overset{\sqrt{3}}{\int }}\frac{dx}{\sqrt{1+x^2}}}={\displaystyle \underset{0}{\overset{\pi /3}{\int }}\sec udu}={\left[\ln (\sec u+\tan u)\right]}_0^{\pi /3}$

  $=\left[\ln (\sec (\pi /3)+\tan (\pi /3)-0\right]=\left[\ln (2+\sqrt{3})\right]$

The value of the integral is $\left[\ln (2+\sqrt{3})\right]$