# How do you integrate? int_sqrt(2)^2dx/(x^5*sqrt(x^2-1))

## ${\int}_{\sqrt{2}}^{2} \frac{\mathrm{dx}}{{x}^{5} \cdot \sqrt{{x}^{2} - 1}}$

Apr 14, 2018

${\int}_{\sqrt{2}}^{2} \frac{\mathrm{dx}}{{x}^{5} \sqrt{{x}^{2} - 1}} = {\int}_{\frac{\pi}{4}}^{\frac{\pi}{3}} {\cos}^{4} \theta d \theta$, which I leave to you to solve.

#### Explanation:

Try letting $x = \sec \theta$. Note this implies that $\mathrm{dx} = \sec \theta \tan \theta d \theta$.

Also, the bounds will change. Observe that $x = \sqrt{2} \implies \theta = \frac{\pi}{4}$ and $x = 2 \implies x = \frac{\pi}{3}$.

Then:

${\int}_{\sqrt{2}}^{2} \frac{\mathrm{dx}}{{x}^{5} \sqrt{{x}^{2} - 1}} = {\int}_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sec \theta \tan \theta d \theta}{{\sec}^{5} \theta \sqrt{{\sec}^{2} \theta - 1}}$

Recall the trigonometric identity ${\sec}^{2} \theta - 1 = {\tan}^{2} \theta$.

$= {\int}_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sec \theta \tan \theta d \theta}{{\sec}^{5} \theta \tan \theta}$

$= {\int}_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{d \theta}{\sec} ^ 4 \theta$

$= {\int}_{\frac{\pi}{4}}^{\frac{\pi}{3}} {\cos}^{4} \theta d \theta$

From here, I would use the identity ${\cos}^{2} \alpha = \frac{1}{2} \left(1 + \cos 2 \alpha\right)$, or, in this case, ${\cos}^{4} \alpha = \frac{1}{4} {\left(1 + \cos 2 \alpha\right)}^{2}$ to further simplify the integral. It will definitely be a little messy.