# How do you integrate int dx/(9x^2-25)^(3/2) using trig substitutions?

Nov 24, 2016

The answer is $= - \frac{x}{25 \cdot \sqrt{9 {x}^{2} - 25}} + C$

#### Explanation:

We use ${\sec}^{2} u = 1 + {\tan}^{2} u$

Let's use the substitution

$x = \frac{5 \sec u}{3}$ $\implies$, $\mathrm{dx} = \frac{5 \sec u \tan u \mathrm{du}}{3}$

$\int \frac{\mathrm{dx}}{9 {x}^{2} - 25} ^ \left(\frac{3}{2}\right) = \int \frac{\frac{5 \sec u \tan u \mathrm{du}}{3}}{9 \cdot \frac{25}{9} {\sec}^{2} u - 25} ^ \left(\frac{3}{2}\right)$

$= \int \frac{5 \sec u \tan u \mathrm{du}}{3 \cdot 125 \left({\tan}^{3} u\right)}$

$= \frac{1}{75} \int \frac{\sec u \mathrm{du}}{\tan} ^ 2 u$

$= \frac{1}{75} \int \frac{{\cos}^{2} u \mathrm{du}}{\cos u {\sin}^{2} u}$

$= \frac{1}{75} \int \frac{\cos u \mathrm{du}}{{\sin}^{2} u}$

Let $v = \sin u$,$\implies$$\mathrm{dv} = \cos u \mathrm{du}$

Therefore,

$\frac{1}{75} \int \frac{\cos u \mathrm{du}}{{\sin}^{2} u} = \frac{1}{75} \int \frac{\mathrm{dv}}{v} ^ 2$

$= \frac{1}{75} \int {v}^{- 2} \mathrm{dv} = \frac{1}{75} {v}^{- 1} / - 1 = - \frac{1}{75 v}$

$= - \frac{1}{75 \sin u}$

$x = \frac{5 \sec u}{3}$ $\implies$, $\cos u = \frac{5}{3 x}$

${\sin}^{2} u = 1 - {\cos}^{2} u = 1 - \frac{25}{9 {x}^{2}} = \frac{9 {x}^{2} - 25}{9 {x}^{2}}$

$\frac{1}{\sin} u = \frac{3 x}{\sqrt{9 {x}^{2} - 25}}$

Therefore,

$\int \frac{\mathrm{dx}}{9 {x}^{2} - 25} ^ \left(\frac{3}{2}\right) = \left(- \frac{1}{75}\right) \cdot \frac{3 x}{\sqrt{9 {x}^{2} - 25}} + C$

$= - \frac{x}{25 \cdot \sqrt{9 {x}^{2} - 25}} + C$