Solutions - 0050-03

(1) Notice $x^2-4$ pattern, so we should make use of the identity ${\sec }^{2}x-1={\tan }^{2}x$.

Select $x=2\sec \theta$ and thus $dx=2\sec \theta \tan \theta d\theta$. Then:

$$x^2-4\gg \gg 4{\sec }^2\theta -4\gg \gg 4{\tan }^{2}x$$

Plug in and simplify:

$$\begin{array}{c}\int \frac{dx}{x^3\sqrt{x^2-4}}\gg \gg \int \frac{2\sec \theta \tan \theta }{8{\sec }^3\theta |2\tan \theta |}d\theta \\ \\ \gg \gg \frac{1}{8}\int {\cos }^2\theta d\theta \end{array}$$

(We must assume that $\tan \theta >0$ for the relevant values of $\theta$ here.)

---

(2) Use power-to-frequency conversion:

$$\begin{array}{c}\gg \gg \frac{1}{8}\int \frac{1}{2}(1+\cos 2\theta )d\theta \gg \gg \frac{1}{16}(\theta +\frac{1}{2}\sin 2\theta )+C\end{array}$$

---

(3) Convert back to terms of $x$:

First draw a triangle expressing $\sec \theta ={\displaystyle \frac{x}{2}}$:

Therefore:

$$\sin \theta =\frac{2}{x},\cos \theta =\frac{\sqrt{x^2-4}}{x}$$

For $\sin 2\theta$, use the double-angle identity:

$$\sin 2\theta =2\sin \theta \cos \theta \gg \gg 2\frac{\sqrt{x^2-4}}{x}\frac{2}{x}$$

Therefore:

$$\begin{array}{c}\frac{1}{16}(\theta +\frac{1}{2}\sin 2\theta )+C\\ \\ \gg \gg \frac{1}{16}({\sec }^{-1}\frac{x}{2}+\frac{2\sqrt{x^2-4}}{x^2})+C\end{array}$$