# How do you integrate from -6 to-3 for sqrt((x^2-9))/2?

Apr 12, 2015

Method is this

Substitute $x = 3 \sec \theta$
For any variable substitution in the integration, Integrant, Limits and the integration operator need to be changed.

So Integrant is $\frac{\sqrt{9 {\sec}^{2} \theta - 9}}{2} = \frac{3 \tan \theta}{2}$

The limits are $\circ x = - 6 \to x = - 3$ changes as well.

Upper Limit calculation
$3 \sec \theta = - 3$
$\sec \theta = - 1$
$\theta = \pi$

Lower Limit calculation
$3 \sec \theta = - 6$
$\sec \theta = - 2$
$\theta = \frac{5 \pi}{6}$

Integral operator change
$\mathrm{dx} = \sec \theta \tan \theta d \theta$
${\int}_{-} {6}^{-} 3 \frac{\sqrt{{x}^{2} - 9}}{2} = {\int}_{\frac{5 \pi}{6}}^{\pi} \frac{3 \sec \theta {\tan}^{2} \theta d \theta}{2}$
$= \frac{3}{2} {\int}_{\frac{5 \pi}{6}}^{\pi} \left({\sec}^{3} \theta - \sec \theta\right) d \theta$

Check how to evaluate $\int {\sec}^{3} \theta d \theta$ here

$= \frac{3}{2} {\left\{\frac{1}{2} \left(\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right) - \ln | \sec \theta + \tan \theta |\right\}}_{\frac{5 \pi}{6}}^{\pi}$

$= \frac{3}{4} {\left\{\sec \theta \tan \theta - \ln | \sec \theta + \tan \theta |\right\}}_{\frac{5 \pi}{6}}^{\pi}$
$= - 0.0082$