# How do you integrate int 1/(x^2sqrt(16x^2-9)) by trigonometric substitution?

Mar 3, 2018

$\frac{\sqrt{16 {x}^{2} - 9}}{9 x} + C$.

#### Explanation:

Suppose that, $I = \int \frac{1}{{x}^{2} \sqrt{16 {x}^{2} - 9}} \mathrm{dx} = \int \frac{1}{{x}^{2} \sqrt{{\left(4 x\right)}^{2} - {3}^{2}}} \mathrm{dx}$.

We subst. $4 x = 3 \sec y , \text{ so that, } 4 \mathrm{dx} = 3 \sec y \tan y \mathrm{dy}$.

$\therefore I = \int \frac{1}{{\left(\frac{3}{4} \cdot \sec y\right)}^{2} \sqrt{9 {\sec}^{2} y - 9}} \cdot \left(\frac{3}{4} \cdot \sec y \tan y\right) \mathrm{dy}$,

$= \frac{16}{9} \cdot \frac{1}{4} \int \frac{1}{{\sec}^{2} y \tan y} \cdot \sec y \tan y \mathrm{dy}$,

$= \frac{4}{9} \int \frac{1}{\sec} y \mathrm{dy}$,

$= \frac{4}{9} \int \cos y \mathrm{dy}$,

$= \frac{4}{9} \sin y$,

$= \frac{4}{9} \sqrt{1 - {\cos}^{2} y}$,

$= \frac{4}{9} \sqrt{1 - \frac{1}{\sec} ^ 2 y}$,

$= \frac{4}{9} \sqrt{1 - \frac{1}{\frac{4}{3} \cdot x} ^ 2}$,

$= \frac{4}{9} \sqrt{1 - \frac{9}{16 {x}^{2}}}$.

$\Rightarrow I = \frac{\sqrt{16 {x}^{2} - 9}}{9 x} + C$.