Question

Use a computer algebra system to evaluate the integral. Compare the answer with the result of using tables. If the answer is not the same, show that they are equivalent.

\(\;\;\int {{{\sec }^4}} xdx\)

Short answer

Using any computer algebra system,\(\int {{{\sec }^4}xdx} \)= \(\frac{{{{\tan }^3}}}{3}x + \tan x + c\)

Step-by-step solution

Calculate the value of the integral using any computer software.

Using an integral calculator, we get

\(\int {{{\sec }^4}xdx} \)=\(\frac{{{{\tan }^3}}}{3}x + \tan x\)

calculating the value of the integral using the table :

I =\(\int {{{\sec }^4}xdx} \)

I= \(\int {{{\sec }^2}} x,{\sec ^2}xdx\)\(\;\;\int {{{\sec }^4}} xdx\)

I=\(\int {{{\sec }^2}} x,\left( {1 + {{\tan }^{2x}}} \right)dx\)

I=\(\int {{{\sec }^2}} x + {\sec ^2}{\tan ^2}xdx\)

I=\(\int {{{\sec }^4}xdx} \)+\(\int {{{\sec }^2}{{\tan }^2}xdx} \)

I=\(\int {{{\sec }^4}xdx} \)+\(\int {{{\left( {\tan x} \right)}^2}} ,{\sec ^2}xdx\)

I=\(\int {{{\sec }^4}xdx} \)+\(\int {{{\left( {\tan x} \right)}^2}} ,{\sec ^2}xdx\)

We know, \(\int {{{\sec }^2}xdx = \tan x} \)

Also,\({\int {f\left( x \right)} ^n}'f\left( x \right)dx = \left( {\frac{{f{{\left( x \right)}^{n + 1}}}}{{n + 1}}} \right) + c\)

:\(\int {{{\left( {\tan x} \right)}^2}} '{\sec ^2}xdx = \frac{{{{\tan }^3}x}}{3} + c\)

Hence,

\(\int {{{\sec }^4}x = \tan x + } \frac{{{{\tan }^3}x}}{3} + c\)

We see that both answers are the same