# How do you integrate int 1/sqrt(4x^2+4x+-24)dx using trigonometric substitution?

Jun 28, 2016

$I = \ln | \frac{2 x + 1 + \sqrt{4 {x}^{2} + 4 x - 24}}{5} | + C ,$

OR

$I = \ln | 2 x + 1 + \sqrt{4 {x}^{2} + 4 x - 24} | + K ,$
where, $K = C - \ln 5.$

#### Explanation:

I will solve $I = \int \frac{1}{\sqrt{4 {x}^{2} + 4 x - 24}} \mathrm{dx} .$

We have, $4 {x}^{2} + 4 x - 24 = 4 {x}^{2} + 4 x + 1 - 25 = {\left(2 x + 1\right)}^{2} - {5}^{2.}$

Knowing that, ${\sec}^{2} \theta - 1 = {\tan}^{2} \theta , 2 x + 1 = 5 \sec t$ will be proper substitution.

So, let, $2 x + 1 = 5 \sec t \ldots \ldots \ldots \ldots \left(1\right) ,$ so that, $2 \mathrm{dx} = 5 \cdot \sec t \cdot \tan t \mathrm{dt} ,$ and,

$\sqrt{4 {x}^{2} + 4 x - 24} = \sqrt{{\left(2 x + 1\right)}^{2} - {5}^{2}} = \sqrt{25 {\sec}^{2} t - 25} = 5 \tan t \ldots \ldots \ldots . . \left(2\right) .$

Hence, $I = \int \frac{1}{\sqrt{4 {x}^{2} + 4 x - 24}} \mathrm{dx} = \int \frac{1}{5 \tan t} \cdot 5 \cdot \sec t \cdot \tan t \mathrm{dt} = \int \sec t \mathrm{dt} = \ln | \sec t + \tan t | .$

Hence, by (1) & (2), I=ln|(2x+1+sqrt(4x^2+4x-24))/5|+C.

One may say $I = \ln | 2 x + 1 + \sqrt{4 {x}^{2} + 4 x - 24} | + K ,$
where, $K = C - \ln 5.$