# How do you integrate int 1/sqrt(4x^2+12x-5)  using trigonometric substitution?

Jun 4, 2018

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} + 12 x - 5}} = \frac{1}{2} \ln \left\mid 2 x + 3 + \sqrt{4 {x}^{2} + 12 x - 5} \right\mid + C$

#### Explanation:

Complete the square at the denominator:

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} + 12 x - 5}} = \int \frac{\mathrm{dx}}{\sqrt{{\left(2 x + 3\right)}^{2} - 14}} = \frac{1}{\sqrt{14}} \int \frac{\mathrm{dx}}{\sqrt{{\left(\frac{2 x + 3}{\sqrt{14}}\right)}^{2} - 1}}$

Substitute for $\frac{2 x + 3}{\sqrt{14}} > 1$

$\frac{2 x + 3}{\sqrt{14}} = \sec t$

$\mathrm{dx} = \frac{\sqrt{14}}{2} \sec t \tan t \mathrm{dt}$

with $t \in \left(0 , \frac{\pi}{2}\right)$

so that:

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} + 12 x - 5}} = \frac{1}{2} \int \frac{\sec t \tan t \mathrm{dt}}{\sqrt{{\sec}^{2} t - 1}}$

Now:

$\sqrt{{\sec}^{2} t - 1} = \sqrt{\frac{1}{\cos} ^ 2 t - 1} = \sqrt{\frac{1 - {\cos}^{2} t}{\cos} ^ 2 t} = \sqrt{{\sin}^{2} \frac{t}{\cos} ^ 2 t} = \sqrt{{\tan}^{2} t} = \tan t$

as for $t \in \left(0 , \frac{\pi}{2}\right)$ the tangent is positive.

So:

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} + 12 x - 5}} = \frac{1}{2} \int \frac{\sec t \tan t \mathrm{dt}}{\tan} t$

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} + 12 x - 5}} = \frac{1}{2} \int \sec t \mathrm{dt}$

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} + 12 x - 5}} = \frac{1}{2} \ln \left\mid \sec t + \tan t \right\mid + C$

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} + 12 x - 5}} = \frac{1}{2} \ln \left\mid \sec t + \sqrt{{\sec}^{2} t - 1} \right\mid + C$

Undoing the substitution:

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} + 12 x - 5}} = \frac{1}{2} \ln \left\mid \frac{2 x + 3}{\sqrt{14}} + \sqrt{{\left(\frac{2 x + 3}{\sqrt{14}}\right)}^{2} - 1} \right\mid + C$

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} + 12 x - 5}} = \frac{1}{2} \ln \left\mid 2 x + 3 + \sqrt{4 {x}^{2} + 12 x - 5} \right\mid + C$

and by differentiating we can verify that the solution is valid also for $\frac{2 x + 3}{\sqrt{14}} < - 1$.