# How do you integrate int 1/sqrt(4x^2+12x-16)  using trigonometric substitution?

Jan 14, 2017

$\frac{1}{2} \ln | \frac{\sqrt{4 {x}^{2} + 12 x - 16} + 2 x + 3}{5} | + C$

#### Explanation:

Start by factoring the denominator.

$\int \frac{1}{\sqrt{4 \left({x}^{2} + 3 x - 4\right)}} \mathrm{dx}$

$\int \frac{1}{2 \sqrt{{x}^{2} + 3 x - 4}} \mathrm{dx}$

$\frac{1}{2} \int \frac{1}{\sqrt{{x}^{2} + 3 x - 4}} \mathrm{dx}$

Now complete the square in the denominator.

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

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

$\frac{1}{2} \int \frac{1}{\sqrt{{\left(x + \frac{3}{2}\right)}^{2} - \frac{25}{4}}} \mathrm{dx}$

Now let $u = x + \frac{3}{2}$: Then $\mathrm{du} = \mathrm{dx}$.

$\frac{1}{2} \int \frac{1}{\sqrt{{u}^{2} - \frac{25}{4}}} \mathrm{du}$

Apply the substitution $u = \frac{5}{2} \sec \theta$. Then $\mathrm{du} = \frac{5}{2} \sec \theta \tan \theta d \theta$:

$\frac{1}{2} \int \frac{1}{\sqrt{{\left(\frac{5}{2} \sec \theta\right)}^{2} - \frac{25}{4}}} \cdot \frac{5}{2} \sec \theta \tan \theta d \theta$

$\frac{1}{2} \int \frac{1}{\sqrt{\frac{25}{4} {\sec}^{2} \theta - \frac{25}{4}}} \cdot \frac{5}{2} \sec \theta \tan \theta d \theta$

$\frac{1}{2} \int \frac{1}{\sqrt{\frac{25}{4} \left({\sec}^{2} \theta - 1\right)}} \cdot \frac{5}{2} \sec \theta \tan \theta d \theta$

Apply ${\sec}^{2} \theta - 1 = {\tan}^{2} \theta$:

$\frac{1}{2} \int \frac{1}{\sqrt{\frac{25}{4} {\tan}^{2} \theta}} \cdot \frac{5}{2} \sec \theta \tan \theta d \theta$

$\frac{1}{2} \int \frac{1}{\frac{5}{2} \tan \theta} \cdot \frac{5}{2} \sec \theta \tan \theta d \theta$

$\frac{1}{2} \int \sec \theta d \theta$

This is a known integral: $\int \sec x \mathrm{dx} = \ln | \sec x + \tan x |$. For more details on how to integrate this, go [here].(http://math2.org/math/integrals/more/sec.htm)

We must determine an expression for $\sec \theta$ and $\tan \theta$. Draw a triangle:

As you can see, $\tan \theta = \frac{\sqrt{4 {u}^{2} - 25}}{5}$ and $\sec \theta = \frac{2 u}{5}$

$\frac{1}{2} \ln | \frac{\sqrt{4 {u}^{2} - 25}}{5} + \frac{2 u}{5} | + C$

$\frac{1}{2} \ln | \frac{\sqrt{4 {u}^{2} - 25} + 2 u}{5} | + C$

$\frac{1}{2} \ln | \frac{\sqrt{4 {\left(x + \frac{3}{2}\right)}^{2} - 25} + 2 \left(x + \frac{3}{2}\right)}{5} | + C$

$\frac{1}{2} \ln | \frac{\sqrt{4 {x}^{2} + 12 x - 16} + 2 x + 3}{5} | + C$

Hopefully this helps!