# How do you integrate int 1/sqrt(4x+8sqrtx+12)  using trigonometric substitution?

Sep 7, 2016

$\sqrt{x + 2 \sqrt{x} + 3} - \ln \left(\left\mid \sqrt{x} + \sqrt{x + 2 \sqrt{x} + 3} + 1 \right\mid\right) + C$

#### Explanation:

We have:

$\int \frac{\mathrm{dx}}{\sqrt{4 x + 8 \sqrt{x} + 12}}$

Factor $\sqrt{4} = 2$ from the denominator:

$= \frac{1}{2} \int \frac{\mathrm{dx}}{\sqrt{x + 2 \sqrt{x} + 3}}$

Complete the square in the denominator. Think about it in terms of ${x}^{2} + 2 x + 3$, and then switch $x$ terms to $\sqrt{x}$ and ${x}^{2}$ to $x$:

$= \frac{1}{2} \int \frac{\mathrm{dx}}{\sqrt{{\left(\sqrt{x} + 1\right)}^{2} + 2}}$

Now, let $\sqrt{x} + 1 = \sqrt{2} \tan \theta$. Note that this implies that $\frac{\mathrm{dx}}{2 \sqrt{x}} = \sqrt{2} {\sec}^{2} \theta d \theta$.

Rearranging some:

$= \int \frac{\sqrt{x} \mathrm{dx}}{2 \sqrt{x} \sqrt{{\left(\sqrt{x} + 1\right)}^{2} + 2}}$

Now, performing the substitutions. We have $\frac{\mathrm{dx}}{2 \sqrt{x}} = \sqrt{2} {\sec}^{2} \theta d \theta$ present, as well as $\sqrt{x} = \sqrt{2} \tan \theta - 1$ in the numerator. Don't forget the switch in the radical in the denominator either:

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

Factoring $\sqrt{2}$ from the denominator and using $\sqrt{{\tan}^{2} \theta + 1} = \sec \theta$:

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

$= \int \left(\sqrt{2} \tan \theta - 1\right) \sec \theta d \theta$

$= \sqrt{2} \int \tan \theta \sec \theta d \theta - \int \sec \theta d \theta$

Both of which are common integrals:

$= \sqrt{2} \sec \theta - \ln \left(\left\mid \sec \theta + \tan \theta \right\mid\right)$

Write $\sec \theta$ in terms of $\tan \theta$:

$= \sqrt{2} \sqrt{{\tan}^{2} \theta + 1} - \ln \left(\left\mid \sqrt{{\tan}^{2} \theta + 1} + \tan \theta \right\mid\right)$

Using $\tan \theta = \frac{\sqrt{x} + 1}{\sqrt{2}}$:

$= \sqrt{2} \sqrt{{\left(\sqrt{x} + 1\right)}^{2} / 2 + 1} - \ln \left(\left\mid \sqrt{{\left(\sqrt{x} + 1\right)}^{2} / 2 + 1} + \frac{\sqrt{x} + 1}{\sqrt{2}} \right\mid\right)$

$= \sqrt{2} \sqrt{\frac{{\left(\sqrt{x} + 1\right)}^{2} + 2}{2}} - \ln \left(\left\mid \sqrt{\frac{{\left(\sqrt{x} + 1\right)}^{2} + 2}{2}} + \frac{\sqrt{x} + 1}{\sqrt{2}} \right\mid\right)$

$= \sqrt{{\left(\sqrt{x} + 1\right)}^{2} + 2} - \ln \left(\left\mid \frac{\sqrt{{\left(\sqrt{x} + 1\right)}^{2} + 2} + \sqrt{x} + 1}{\sqrt{2}} \right\mid\right)$

Note that the $\frac{1}{\sqrt{2}}$ can actually be taken from the denominator of the logarithm as $- \ln \sqrt{2}$, which will be absorbed into the constant of integration in the next step:

$= \sqrt{x + 2 \sqrt{x} + 3} - \ln \left(\left\mid \sqrt{x} + \sqrt{x + 2 \sqrt{x} + 3} + 1 \right\mid\right) + C$