# How do you evaluate int_1^sqrt(3) 4/(x^2sqrt(x^2 - 1)) dx?

Apr 24, 2017

The integral has value $4 \sqrt{\frac{2}{3}}$

#### Explanation:

Use the substitution $x = \sec \theta$. Then $\mathrm{dx} = \sec \theta \tan \theta d \theta$. You should (stylistically) change the bounds of integration, but for ease of entering on the computer, I haven't. Call the integral $I$.

$I = {\int}_{1}^{\sqrt{3}} \frac{4}{{\left(\sec \theta\right)}^{2} \sqrt{{\left(\sec \theta\right)}^{2} - 1}} \sec \theta \tan \theta d \theta$

$I = {\int}_{1}^{\sqrt{3}} \frac{4}{{\sec}^{2} \theta \sqrt{{\tan}^{2} \theta}} \sec \theta \tan \theta d \theta$

$I = {\int}_{1}^{\sqrt{3}} \frac{4}{{\sec}^{2} \theta \tan \theta} \sec \theta \tan \theta d \theta$

$I = {\int}_{1}^{\sqrt{3}} \frac{4}{\sec} \theta d \theta$

$I = {\int}_{1}^{\sqrt{3}} 4 \cos \theta d \theta$

$I = {\left[4 \sin \theta\right]}_{1}^{\sqrt{3}}$

DO NOT EVALUATE THIS INTEGRAL. SINCE WE DIDN'T CHANGE THE BOUNDS OF INTEGRATION, THIS INTEGRAL WILL BE INCORRECT. THE CORRECT PROCESS WOULD BE TO SWITCH THE VARIABLE BACK TO $X$.

From our initial substitution, we know that $\sec \theta = \frac{x}{1}$, so if we were to draw a triangle, the hypotenuse would measure $x$ and the side adjacent $\theta$ would measure $1$. By pythagoras, the side opposite would measure $\sqrt{{x}^{2} - 1}$. Therefore, $\sin \theta = \frac{\sqrt{{x}^{2} - 1}}{x}$.

$I = {\left[\frac{4 \sqrt{{x}^{2} - 1}}{x}\right]}_{1}^{\sqrt{3}}$

$I = \frac{4 \sqrt{{\sqrt{3}}^{2} - 1}}{\sqrt{3}} - \frac{4 \sqrt{{1}^{2} - 1}}{1}$

$I = \frac{4 \sqrt{2}}{\sqrt{3}} - 0$

$I = 4 \sqrt{\frac{2}{3}}$

Hopefully this helps!