# How do you integrate int 1/(4x^2-9)^(3/2) by trigonometric substitution?

Nov 26, 2016

$\frac{- x}{9 \sqrt{4 {x}^{2} - 9}} + C$

#### Explanation:

$\int \frac{\mathrm{dx}}{4 {x}^{2} - 9} ^ \left(\frac{3}{2}\right)$

Let $2 x = 3 \sec \theta$ such that $2 \mathrm{dx} = 3 \sec \theta \tan \theta d \theta$.

$= \frac{1}{2} \int \frac{2 \mathrm{dx}}{{\left(2 x\right)}^{2} - 9} ^ \left(\frac{3}{2}\right)$

$= \frac{1}{2} \int \frac{3 \sec \theta \tan \theta d \theta}{9 {\sec}^{2} \theta - 9} ^ \left(\frac{3}{2}\right)$

Factoring ${9}^{\frac{3}{2}} = 27$ from the denominator:

$= \frac{1}{2} \int \frac{3 \sec \theta \tan \theta}{27 {\left({\sec}^{2} \theta - 1\right)}^{\frac{3}{2}}} d \theta$

Since ${\sec}^{2} \theta - 1 = {\tan}^{2} \theta$:

$= \frac{1}{18} \int \frac{\sec \theta \tan \theta}{\tan} ^ 3 \theta d \theta$

$= \frac{1}{18} \int \sec \frac{\theta}{\tan} ^ 2 \theta d \theta$

$= \frac{1}{18} \int \cos \frac{\theta}{\sin} ^ 2 \theta d \theta$

Letting $u = \sin \theta$ so $\mathrm{du} = \cos \theta d \theta$.

$= \frac{1}{18} \int {u}^{-} 2 \mathrm{du}$

$= - \frac{1}{18 u}$

$= - \frac{1}{18} \csc \theta$

$= - \frac{1}{18} \sec \frac{\theta}{\tan} \theta$

$= - \frac{1}{54} \frac{3 \sec \theta}{\sqrt{{\sec}^{2} \theta - 1}}$

$= - \frac{3}{54} \frac{3 \sec \theta}{\sqrt{9 {\sec}^{2} \theta - 9}}$

Using $3 \sec \theta = 2 x$:

$= - \frac{1}{18} \frac{2 x}{\sqrt{4 {x}^{2} - 9}}$

$= \frac{- x}{9 \sqrt{4 {x}^{2} - 9}} + C$