Key to "Use trigonometric substitution to integrate (Don’t forget to..."

mronjo7n

Answered question

2021-11-17

Use trigonometric substitution to integrate (Don’t forget to write your answer in terms of the original variable xand do not leave trigonometric functions evaluated at inverse trigonometric functions)

\int \frac{\sqrt{x^{2} - 1}}{x^{4}} d x

Answer & Explanation

Ralph Lester

Beginner2021-11-18Added 16 answers

Step 1
Given:
$\int \frac{\sqrt{x^{2} - 1}}{x^{4}} d x$
To find:
Use trigonometric substitution to integrate above function.
Step 2
Apply trigonometric substitution: $x = \sec (u)$
$\int \frac{\sqrt{x^{2} - 1}}{x^{4}} d x = \int \frac{\sqrt{\sec^{2} (u) - 1}}{\sec^{4} (u)} d u$
$= \int \frac{\tan^{2} (u)}{\sec^{3} (u)} d u$
Rewrite using trigonometric identities,
$\tan (u) = \frac{\sin (u)}{\cos (u)}$ and $\sec (u) = \frac{1}{\cos (u)}$
$\int \frac{\sqrt{x^{2} - 1}}{x^{4}} d x = \int \sin^{2} (u) \cos (u) d u$
Substitute $v = \sin (u)$ , and $d v = \cos (u) d u$
$\int \frac{\sqrt{x^{2} - 1}}{x^{4}} d x = \int v^{2} d v$
using the formula, $\int x^{n} d x = \frac{x^{n + 1}}{n + 1}$
$\int \frac{\sqrt{x^{2} - 1}}{x^{4}} d x = \frac{v^{2 + 1}}{2 + 1}$
On back substitution,
$\int \frac{\sqrt{x^{2} - 1}}{x^{4}} d x = \frac{\sin^{3} (u)}{3}$
$\int \frac{\sqrt{x^{2} - 1}}{x^{4}} d x = \frac{\sin^{3} (\sec^{- 1} (x))}{3}$
On simplifying above function,
$\int \frac{\sqrt{x^{2} - 1}}{x^{4}} d x = \frac{{(x^{2} - 1)}^{\frac{3}{2}}}{3 x^{3}}$

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