Chapter 5.8, Problem 25E

To determine

To evaluate and compare the value of integral.

Answer to Problem 25E

The value of the integral using CAS is ${\displaystyle \int {\sec }^{4}xdx}=\frac{1}{3}\left[\cos \left(2x\right)+2\right]\tan x{\sec }^{2}x+C$ and using table of integral is $\frac{{\tan }^{3}x}{3}+\tan x+C$ . Here both the values are not same but they are equivalent.

Explanation of Solution

Given information:

The integral is ${\displaystyle \int {\sec }^{4}xdx}$ .

Calculation:

Using Computer algebra system to evaluate ${\displaystyle \int {\sec }^{4}xdx}$ ,

Derive and Mathematica reports the answer,

  ${\displaystyle \int {\sec }^{4}xdx}=\frac{1}{3}\left[\cos \left(2x\right)+2\right]\tan x{\sec }^{2}x+C$

Now, use table of integral to evaluate the same function,

  ${\displaystyle \int {\sec }^{4}xdx}$

Prepare the substitution

  $\begin{array}{l}{\displaystyle \int {\sec }^2}x\left({\tan }^{2}x+1\right)dx\\ ={\displaystyle \int \left(u^2+1\right)du}\left[\text{substitute }u=\tan x\to \frac{du}{dx}={\sec }^{2}x\text{}\text{}\text{}\text{}\text{}\text{}\to dx=\frac{1}{{\sec }^{2}x}du\right]\\ \text{=}{\displaystyle \int u^{2}du+}{\displaystyle \int 1du}\\ =\frac{u^3}{3}+u\left[{\displaystyle \int u^{n}du=\frac{u^{n+1}}{n+1}\text{ with }n=2}\text{ and }{\displaystyle \int 1du=u}\right]\\ =\frac{{\tan }^{3}x}{3}+\tan x+C\left[\text{undo substition }u=\tan x\right]\end{array}$

On comparing both the values of the integral, it is concluded that both the answer is not same.

Now, simplify the value of the integral to check whether it is equivalent or not.

  $\begin{array}{l}\frac{{\tan }^{3}x}{3}+\tan x+C=\tan x\left(\frac{{\tan }^{2}x}{3}+1\right)+C\\ \frac{{\tan }^{3}x}{3}+\tan x+C=\tan x\left(\frac{{\tan }^{2}x+3}{3}\right)+C\\ \frac{{\tan }^{3}x}{3}+\tan x+C=\tan x\left[\frac{\left({\sec }^{2}x-1\right)+3}{3}\right]+C\\ \frac{{\tan }^{3}x}{3}+\tan x+C=\tan x\left[\frac{{\sec }^{2}x+2}{3}\right]+C\\ \frac{{\tan }^{3}x}{3}+\tan x+C=\tan x{\sec }^{2}x\left[\frac{1+2{\cos }^{2}x}{3}\right]+C\\ \frac{{\tan }^{3}x}{3}+\tan x+C=\tan x{\sec }^{2}x\left[\frac{1+\left(1+\cos 2x\right)}{3}\right]+C\\ \frac{{\tan }^{3}x}{3}+\tan x+C=\tan x{\sec }^{2}x\left[\frac{\cos 2x+2}{3}\right]+C\end{array}$

Hence,

Both the values are not same but they are equivalent.