secant_sub2

$\displaystyle \int_{-2\sqrt{3}}^{-\sqrt{3}} \frac{\sqrt{x^2-3}}{x} \, dx$ .

$x=\sqrt{3} \sec \theta$ :

x=\sqrt{3}\sec \theta, \quad \mbox{where} \quad 0 \le \theta < \frac{\pi}{2} \mbox{ or } \frac{\pi}{2} < \theta \le \pi,\\[0.25 in] dx = \sqrt{3} \sec \theta \tan \theta \, d\theta, \\[0.25 in] x=-2\sqrt{3} \quad \Longrightarrow \quad \theta = \frac{2\pi}{3},\\[0.25 in] x=-\sqrt{3} \quad \Longrightarrow \quad \theta = \pi

With the substitution, we have

\int_{-2\sqrt{3}}^{-\sqrt{3}} \frac{\sqrt{x^2-3}}{x} \, dx = \int_{2\pi/3}^{\pi} \frac{\sqrt{3\sec^2 \theta -3} \, \sqrt{3} \sec \theta \tan \theta \, d\theta}{\sqrt{3} \sec \theta},

and this reduces to

\sqrt{3} \, \int_{2\pi/3}^{\pi} |\tan \theta| \tan \theta \, d\theta.

$\tan \theta < 0$ $[2\pi/3,\pi]$ $|\tan \theta| = -\tan \theta$ . Therefore the integral is

-\sqrt{3} \int_{2\pi/3}^{\pi} \tan^2 \theta \, d\theta.

$\tan^2 \theta = \sec^2 \theta -1$ , and we have

-\sqrt{3} \int_{2\pi/3}^{\pi} (\sec^2 \theta - 1) \, d\theta = -\sqrt{3}(\tan \theta - \theta) \Big|_{2\pi/3}^{\pi} = \frac{\sqrt{3} \pi}{3}-3.