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

May 30, 2018

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

#### Explanation:

Here,

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

$= \int \frac{1}{\sqrt{4 {x}^{2} + 4 x + 1 - 25}} \mathrm{dx}$

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

Substituting,

$2 x + 1 = 5 \sec u \implies 2 \mathrm{dx} = 5 \sec u \tan u \mathrm{du}$

=>dx=5/2secutanudu and color(blue)(secu=(2x+1)/5

So,

$I = \int \frac{1}{\sqrt{{5}^{2} {\sec}^{2} u - {5}^{2}}} \times \frac{5}{2} \sec u \tan u \mathrm{du}$

$= \frac{5}{2} \int \frac{\sec u \tan u}{5 \tan u} \mathrm{du}$

$= \frac{1}{2} \int \sec u \mathrm{du}$

$= \frac{1}{2} \ln | \sec u + \tan u | + c$

=1/2ln|secu+sqrt(sec^2u-1)|+c,where, color(blue)(secu=(2x+1)/5

$= \frac{1}{2} \ln | \frac{2 x + 1}{5} + \sqrt{{\left(\left(\frac{2 x + 1}{5}\right)\right)}^{2} - 1} | + c$

$= \frac{1}{2} \ln | \frac{2 x + 1}{5} + \frac{\sqrt{4 {x}^{2} + 4 x + 1 - 25}}{5} | + c$

$= \frac{1}{2} \ln | \frac{\left(2 x + 1\right) + \sqrt{4 {x}^{2} + 4 x - 24}}{5} | + c$

$= \frac{1}{2} \ln | 2 x + 1 + \sqrt{4 {x}^{2} + 4 x - 24} | - \frac{1}{2} \ln \left(5\right) + c$

$= \frac{1}{2} \ln | 2 x + 1 + \sqrt{4 {x}^{2} + 4 x - 24} | + C$ , Put , $C = c - \frac{1}{2} \ln \left(5\right)$