# How do you find the integral intsqrt(x^2-9)/x^3dx ?

Sep 1, 2014

By using the substitution $x = 3 \sec \theta$,
$\int \frac{\sqrt{{x}^{2} - 9}}{{x}^{3}} \mathrm{dx} = \frac{1}{6} {\sec}^{- 1} \left(\frac{x}{3}\right) - \frac{\sqrt{{x}^{2} - 9}}{2 {x}^{2}} + C$.

Let $x = 3 \sec \theta R i g h t a r r o w \mathrm{dx} = 3 \sec \theta \tan \theta d \theta$.
$\int \frac{\sqrt{{x}^{2} - 9}}{{x}^{3}} \mathrm{dx} = \int \frac{\sqrt{{\left(3 \sec \theta\right)}^{2} - 9}}{{\left(3 \sec \theta\right)}^{3}} 3 \sec \theta \tan \theta d \theta$,
which simplifies to
$= \frac{1}{3} \int \frac{{\tan}^{2} \theta}{{\sec}^{2} \theta} d \theta$
by using $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$,
$= \frac{1}{3} \int {\sin}^{2} \theta d \theta$
by using ${\sin}^{2} \theta = \frac{1 - \cos \left(2 \theta\right)}{2}$,
$\frac{1}{6} \int \left[1 - \cos \left(2 \theta\right)\right] d \theta = \frac{1}{6} \left[\theta - \sin \frac{2 \theta}{2}\right] + C$
by using $\sin \left(2 \theta\right) = 2 \sin \theta \cos \theta$,
$= \frac{1}{6} \left(\theta - \sin \theta \cos \theta\right) + C$

Now, we need to rewrite our answer above in terms of $x$.
$x = 3 \sec \theta R i g h t a r r o w \frac{x}{3} = \sec \theta R i g h t a r r o w \theta = {\sec}^{- 1} \left(\frac{x}{3}\right)$
Since $\sec \theta = \frac{x}{3}$, we can construct a right triangle with angle $\theta$ such that its hypotenuse is $x$, its adjacent is $3$, and its opposite is $\sqrt{{x}^{2} - 9}$. So, we have
$\sin \theta = \frac{\sqrt{{x}^{2} - 9}}{x}$ and $\cos \theta = \frac{3}{x}$

Hence, by rewriting $\theta$, $\sin \theta$, $\cos \theta$ in terms of $x$,
$\int \frac{\sqrt{{x}^{2} - 9}}{{x}^{3}} \mathrm{dx} = \frac{1}{6} {\sec}^{- 1} \left(\frac{x}{3}\right) - \frac{\sqrt{{x}^{2} - 9}}{2 {x}^{2}} + C$.