# How do you integrate int dx/sqrt(9x^2+4) using trig substitutions?

Apr 16, 2017

$\int \setminus \frac{1}{\sqrt{9 {x}^{2} + 4}} \setminus \mathrm{dx} = \frac{1}{3} \setminus \ln \left(\sqrt{1 + \frac{9 {x}^{2}}{4}} + \frac{3 x}{2}\right) + c$

#### Explanation:

We want to find:

$I = \int \setminus \frac{1}{\sqrt{9 {x}^{2} + 4}} \setminus \mathrm{dx}$

For the integrand $\frac{1}{\sqrt{9 {x}^{2} + 4}}$ a suitable substitution should be:

$\tan u = \frac{3 x}{2} \implies {\tan}^{2} u = \frac{9 {x}^{2}}{4}$
$\text{ } \implies 9 {x}^{2} = 4 {\tan}^{2} u$

And differentiating wrt $x$ we get:

${\sec}^{2} u \frac{\mathrm{du}}{\mathrm{dx}} = \frac{3}{2} \implies \frac{2}{3} {\sec}^{2} u \frac{\mathrm{du}}{\mathrm{dx}} = 1$

Applying the substitution along with $1 + {\tan}^{2} A \equiv {\sec}^{2} A$ we get;

$I = \int \setminus \frac{\frac{2}{3} {\sec}^{2} u}{\sqrt{4 {\tan}^{2} u + 4}} \setminus \mathrm{du}$
$\setminus \setminus = \frac{2}{3} \setminus \int \setminus \frac{{\sec}^{2} u}{\sqrt{4 \left({\tan}^{2} u + 1\right)}} \setminus \mathrm{du}$
$\setminus \setminus = \frac{2}{3} \setminus \int \setminus \frac{{\sec}^{2} u}{\sqrt{4 {\sec}^{2} u}} \setminus \mathrm{du}$
$\setminus \setminus = \frac{2}{3} \setminus \int \setminus \frac{{\sec}^{2} u}{2 \sec u} \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{3} \setminus \int \setminus \sec u \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{3} \setminus \int \setminus \sec u \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{3} \setminus \ln \left(\sec u + \tan u\right) + c$

And again using $1 + {\tan}^{2} A \equiv {\sec}^{2} A$ we write

${\sec}^{2} u = 1 + {\tan}^{2} u$
$\text{ } = 1 + \frac{9 {x}^{2}}{4}$
$\implies \sec u = \sqrt{1 + \frac{9 {x}^{2}}{4}}$

And restoring the substituting we get:

$I = \frac{1}{3} \setminus \ln \left(\sqrt{1 + \frac{9 {x}^{2}}{4}} + \frac{3 x}{2}\right) + c$

Apr 16, 2017

$\int \frac{\mathrm{dx}}{\sqrt{9 {x}^{2} + 4}}$

$= \frac{1}{2} \int \frac{\mathrm{dx}}{\sqrt{\frac{9}{4} {x}^{2} + 1}}$

We then say that: ${y}^{2} = \frac{9}{4} {x}^{2} \implies y = \frac{3}{2} x$ so $\mathrm{dy} = \frac{3}{2} \mathrm{dx}$.

$\implies \frac{1}{3} \int \frac{\mathrm{dy}}{\sqrt{{y}^{2} + 1}}$

Now we note that: ${\cosh}^{2} z - {\sinh}^{2} z = 1$ so we can sub $y = \sinh z$ which means that: $\mathrm{dy} = \cosh z \setminus \mathrm{dz}$

$\implies \frac{1}{3} \int \frac{\cosh z \setminus \mathrm{dz}}{\sqrt{{\sinh}^{2} z + 1}}$

$= \frac{1}{3} \int \mathrm{dz}$

$= \frac{1}{3} \left(z + C\right)$

Now to reverse the sub's:

$= \frac{1}{3} {\sinh}^{- 1} y + C$, where C is arbitrary.

$= \frac{1}{3} {\sinh}^{- 1} \left(\frac{3}{2} x\right) + C$

Now: ${\sinh}^{- 1} \alpha = \ln \left(\alpha + \sqrt{{\alpha}^{2} + 1}\right)$

$\implies \frac{1}{3} \left(\ln \left(\frac{3}{2} x + \sqrt{{\left(\frac{3}{2} x\right)}^{2} + 1}\right)\right) + C$