Question

How do you integrate $\int \dfrac{{\sqrt {{x^2} - 25} }}{x}dx$ using trigonometric substitution?

Answer 1

Hint: To solve this question, first we will assume any of the set of variables or constant be another variable to get the expression easier. And conclude until the non-operational state is not achieved. And finally substitute the assumed value. Here we substitute x= secu by looking at the nature of expression given.

Complete step by step solution:
Let us suppose that, $I = \int \dfrac{{\sqrt {{x^2} - 25} }}{x}dx$ , and
let $x = 5\sec u$ .
Differentiate $x = 5\sec u$ :
$ \Rightarrow dx = 5\sec u\tan u du$
Now,
$
  \therefore I = \int \dfrac{{\sqrt {25{{\sec }^2}u - 25} }}{{5\sec u}}.5\sec u\tan udu \\
   = \int 5\tan u.\tan udu \\
   = 5\int {\tan ^2}udu \\
   = 5\int ({\sec ^2}u - 1)du \\
   = 5(\tan u - u) \;
 $
Here,
$5\tan u = \sqrt {{x^2} - 25} $ and
$
  x = 5\sec u \\
   \Rightarrow u = arc\,\sec (\dfrac{x}{5}) \;
 $
Hence,
$ \Rightarrow I = \sqrt {{x^2} - 25} - 5arc\,\sec (\dfrac{x}{5}) + C$ .
So, the correct answer is “$ I = \sqrt {{x^2} - 25} - 5arc\,\sec (\dfrac{x}{5}) + C$ ”.

Note: Usually the method of integration by substitution is extremely useful when we make a substitution for a function whose derivative is also present in the integrand. Doing so, the function simplifies and then the basic formulas of integration can be used to integrate the function.