Chapter 5.5, Problem 38E

To determine

To evaluate: the indefinite integral .illustrate and check that your answer is reasonable by graphing both the function and its anti derivative (take $\text{C=0}$ )

Answer to Problem 38E

Evaluation of the integral is $\frac{1}{3}{\tan }^3\left(\theta \right)+\text{c}$

Explanation of Solution

Given information: the indefinite integral is ${\displaystyle \int {\text{tan}}^{\text{2}}}{\text{(θ)sec}}^{\text{2}}\left(\text{θ}\right)\text{dθ}$

we have, ${\displaystyle \int {\text{tan}}^{\text{2}}}{\text{(θ)sec}}^{\text{2}}\left(\text{θ}\right)\text{dθ }\mathrm{......}\text{(1)}$

Let, $\text{u=tan}\left(\text{θ}\right)$

Now, we have $\text{u=tan}\left(\text{θ}\right)$

Derivative both side by $\left(\text{θ}\right)$ , we have

  $\frac{\text{du}}{\text{d}\theta }\text{=}\frac{\text{d}}{\text{d}\theta }\text{tan}\left(\text{θ}\right)$

  $\text{du}={\sec }^2\theta \text{ d}\theta$

Now , put the value of $\tan \theta$ in (1), than we have

  $={\displaystyle \int {\text{u}}^2}\text{du}$

Now apply the power rule : ${\displaystyle \int {\text{x}}^{\text{a}}}\text{dx=}\frac{{\text{x}}^{\text{a+1}}}{\text{a+1}}\text{,a¹-1}$

Then, we have $\frac{{\text{u}}^{\text{2+1}}}{\text{2+1}}+\text{c}$ [ where c is constant]

Now put the vale of u , we have $\frac{{\text{tan}}^{\text{2+1}}\left(\text{θ}\right)}{\text{2+1}}+\text{c}$

We are Simplify, $\frac{{\tan }^{2+1}\left(\theta \right)}{2+1}+\text{c}=\frac{1}{3}{\tan }^3\left(\theta \right)+\text{c}$

Evaluation of the integral is $\frac{1}{3}{\tan }^3\left(\theta \right)+\text{c}$

Graph of the evaluate integral is given below