# How do you integrate int 1/sqrt(9x-12sqrtx-1)  using trigonometric substitution?

Jul 18, 2018

$I = \frac{2}{9} \sqrt{9 x - 12 \sqrt{x} - 1} + \frac{4}{9} \ln \left(3 \sqrt{x} - 2 + \sqrt{9 x - 12 \sqrt{x} - 1}\right) + c$

#### Explanation:

Let
$I = \int \frac{\mathrm{dx}}{\sqrt{9 x - 12 \sqrt{x} - 1}}$

We want to get rid of the square roots in the square roots, so let $u = \sqrt{x}$. Therefore,
$I = \int \frac{2 u \mathrm{du}}{\sqrt{9 {u}^{2} - 12 u - 1}} = \int \setminus \frac{2 u \mathrm{du}}{\sqrt{{\left(3 u - 2\right)}^{2} - 5}}$
To simplify, let $v = 3 u - 2$. Therefore,
$I = \frac{2}{9} \int \setminus \setminus \frac{\left(v + 2\right) \mathrm{dv}}{\sqrt{{v}^{2} - 5}} = \frac{2}{9} \int \frac{v \mathrm{dv}}{\sqrt{{v}^{2} - 5}} + \frac{4}{9} \int \frac{\mathrm{dv}}{\sqrt{{v}^{2} - 5}}$

Now we have some tools for things like this, but that 5 is pesky, so let's let $y = \frac{v}{\sqrt{5}}$, such that
$I = \frac{2}{9} \int \frac{5 y \mathrm{dy}}{\sqrt{5 \left({y}^{2} - 1\right)}} + \frac{4}{9} \int \frac{\mathrm{dy} \sqrt{5}}{\sqrt{5 \left({y}^{2} - 1\right)}}$
$= \frac{2 \sqrt{5}}{9} \int \frac{y \mathrm{dy}}{\sqrt{{y}^{2} - 1}} + \frac{4}{9} \int \frac{\mathrm{dy}}{\sqrt{{y}^{2} - 1}}$

These two sides now will have different methods. The left side is a simple u-subsitution, with $w = {y}^{2}$, and the right with a more complicated trig substitution with $y = \sec z$. Therefore,

$= \frac{\sqrt{5}}{9} \int \frac{\mathrm{dw}}{\sqrt{w - 1}} + \frac{4}{9} \int \setminus \setminus \sec z \setminus \mathrm{dz}$

Finally, using $g = w - 1$ and $\setminus \Omega = \sec z + \tan z$ (this final substitution is famous, I can't really give any logical jump to easily deduce it):
$\frac{2 \sqrt{5}}{9} \cdot \int \setminus \frac{1}{2} {g}^{- \frac{1}{2}} \mathrm{dg} + \frac{4}{9} \int \setminus \frac{\mathrm{dO} m e g a}{\Omega}$

So
$I = \frac{2 \sqrt{5}}{9} \sqrt{g} + \frac{4}{9} \ln \Omega + C$
for some constant $C$, and where
$\sqrt{g} = \sqrt{w - 1} = \sqrt{{y}^{2} - 1} = \sqrt{{v}^{2} / 5 - 1} = {5}^{- \frac{1}{2}} \sqrt{{v}^{2} - 5}$
$= \frac{1}{\sqrt{5}} \sqrt{{\left(3 u - 2\right)}^{2} - 5} = \frac{1}{\sqrt{5}} \cdot \sqrt{9 x - 12 \sqrt{x} - 1}$

and
$\Omega = \sec z + \tan z = \sec z + \sqrt{{\sec}^{2} z - 1} = y + \sqrt{{y}^{2} - 1}$
$= \frac{1}{\sqrt{5}} \left(v + \sqrt{{v}^{2} - 5}\right) = \frac{1}{\sqrt{5}} \left(3 \sqrt{x} - 2 + \sqrt{9 x - 12 \sqrt{x} - 1}\right)$