Chapter 5.7, Problem 12E

To determine

To evaluate the integral ${\displaystyle \int \frac{\sqrt{x^2-1}}{x^4}}dx$

Answer to Problem 12E

The value is $\frac{{\left(1-x^2\right)}^{3/2}}{3}+C$

Explanation of Solution

Given information:

The integral is ${\displaystyle \int \frac{\sqrt{x^2-1}}{x^4}}dx$

Formula used :

  $\frac{d}{d\theta }\left(\sec \theta \right)=\sec \theta \tan \theta$ , $\frac{d}{d\theta }\left(\sin \theta \right)=\cos \theta \mathrm{t}$ , ${\displaystyle \int x^{n}dx=\frac{x^{n+1}}{n+1}}+C$ , $1+{\tan }^2\theta ={\sec }^2\theta$

Calculation:

The integral is ${\displaystyle \int \frac{\sqrt{x^2-1}}{x^4}}dx$

Given $x=\sec \theta$ , therefore $\sin \theta =\sqrt{1-x^2}$

Differentiating both side , $dx=\sec \theta \tan \theta d\theta$

Substituting $x=\sec \theta$ and $dx=\sec \theta \tan \theta d\theta$ in the given integral

  ${\displaystyle \int \frac{\sqrt{{\sec }^2\theta -1}}{{\sec }^4\theta }}\times \sec \theta \tan \theta d\theta ={\displaystyle \int \frac{{\tan }^2\theta }{{\sec }^3\theta }}d\theta ={\displaystyle \int \frac{{\sin }^2\theta }{{\cos }^2\theta }}\times {\cos }^3\theta d\theta ={\displaystyle \int {\sin }^2\theta }\cos \theta d\theta$

Let $\sin \theta =t$

Differentiating both side , $\cos \theta d\theta =dt$

  $={\displaystyle \int t^2}dt=\frac{t^3}{3}+C=\frac{{\sin }^3\theta }{3}+C=\frac{{\left(1-x^2\right)}^{3/2}}{3}+C$

Hence the value of ${\displaystyle \int \frac{\sqrt{x^2-1}}{x^4}}dx$ is $\frac{{\left(1-x^2\right)}^{3/2}}{3}+C$