Chapter 7, Problem 14RE

To determine

To evaluate the integral.

Answer to Problem 14RE

The required solution is, ${\displaystyle \int \frac{1}{{\theta }^2}\sec \frac{1}{\theta }\tan \frac{1}{\theta }d\theta }\text{ =}-\sec \frac{1}{\theta }+C$ .

Explanation of Solution

Given information: The equation is, ${\displaystyle \int \frac{1}{{\theta }^2}\sec \frac{1}{\theta }\tan \frac{1}{\theta }d\theta }$ .

Concept used: Indefinite integral.

The family of all antiderivatives of a function $f\left(x\right)$ is a indefinite integral of $f$

With respect to $x$ and is denoted by ${\displaystyle \int f\left(x\right).}$

If F is an function such that $F\text{'}\left(x\right)=f\left(x\right)$ , then ${\displaystyle \int f\left(x\right)}dx=F\left(x\right)+C,$

Where C is an arbitrary constant, called the constant of integration.

Calculation : The required solution is obtained as,

  $\begin{array}{l}{\displaystyle \int \frac{1}{{\theta }^2}\sec \frac{1}{\theta }\tan \frac{1}{\theta }d\theta }={\displaystyle \int \sec \frac{1}{\theta }\tan \frac{1}{\theta }\left(\frac{1}{{\theta }^2}\right)d\theta }\\ \text{ let }u=\frac{1}{\theta }={\theta }^{-1}\\ du=-{\theta }^{-2}\\ du=-\frac{1}{{\theta }^2}\\ \text{ =}{\displaystyle \int \sec u\tan u}\left(-du\right)\\ \text{ =}-{\displaystyle \int \sec u\tan u}\left(-du\right)\\ \text{ =}-\sec u+C\\ \text{ =}-\sec \frac{1}{\theta }+C\end{array}$

Hence, the required solution is, ${\displaystyle \int \frac{1}{{\theta }^2}\sec \frac{1}{\theta }\tan \frac{1}{\theta }d\theta }\text{ =}-\sec \frac{1}{\theta }+C$ .