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

Mar 21, 2018

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} - 12 x - 16}} = \frac{1}{2} \ln \left\mid \left(2 x - 3\right) + \sqrt{4 {x}^{2} - 12 x - 16} \right\mid + C$

#### Explanation:

Complete the square under the root at the denominator:

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} - 12 x - 16}} = \int \frac{\mathrm{dx}}{\sqrt{{\left(2 x - 3\right)}^{2} - 25}}$

Substitute now:

$2 x - 3 = 5 \sec t$

$\mathrm{dx} = \frac{5}{2} \sec t \tan t \mathrm{dt}$

then:

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} - 12 x - 16}} = \frac{5}{2} \int \frac{\sec t \tan t \mathrm{dt}}{\sqrt{25 {\sec}^{2} t - 25}}$

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} - 12 x - 16}} = \frac{1}{2} \int \frac{\sec t \tan t \mathrm{dt}}{\sqrt{{\sec}^{2} t - 1}}$

Use now the trigonometric identity:

${\sec}^{2} t - 1 = {\tan}^{2} t$

Note now that the function is defined for:

$\left\mid 2 x - 3 \right\mid > 5$

If we restrict to the interval:

$2 x - 3 > 5$

then $\sec t$ is positive and:

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

so that:

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} - 12 x - 16}} = \frac{1}{2} \int \frac{\sec t \tan t \mathrm{dt}}{\tan} t = \frac{1}{2} \int \sec t \mathrm{dt}$

To solve this last integral there is a well known trick:

$\int \sec t \mathrm{dt} = \int \sec t \frac{\sec t + \tan t}{\sec t + \tan t} \mathrm{dt}$

$\int \sec t \mathrm{dt} = \int \frac{{\sec}^{2} t + \sec t \tan t}{\sec t + \tan t} \mathrm{dt}$

$\int \sec t \mathrm{dt} = \int \frac{d \left(\sec t + \tan t\right)}{\sec t + \tan t} \mathrm{dt}$

$\int \sec t \mathrm{dt} = \ln \left\mid \sec t + \tan t \right\mid + C$

so:

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} - 12 x - 16}} = \frac{1}{2} \ln \left\mid \sec t + \tan t \right\mid + C$

and undoing the substitution:

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} - 12 x - 16}} = \frac{1}{2} \ln \left\mid \frac{2 x - 3}{5} + \sqrt{{\left(\frac{2 x - 3}{5}\right)}^{2} - 1} \right\mid + C$

and simplifying:

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} - 12 x - 16}} = \frac{1}{2} \ln \left\mid \left(2 x - 3\right) + \sqrt{4 {x}^{2} - 12 x - 16} \right\mid + C$

Mar 21, 2018

$I = \frac{1}{2} \ln | \left(2 x - 3\right) + \sqrt{4 {x}^{2} - 12 x - 16} | + C$

#### Explanation:

Here,
$4 {x}^{2} - 12 x - 16 = 4 {x}^{2} - 12 x + 9 - 25 = {\left(2 x - 3\right)}^{2} - {\left(5\right)}^{2}$
So,

$I = \int \frac{1}{\sqrt{4 {x}^{2} - 12 x - 16}} \mathrm{dx}$
$= \int \frac{1}{\sqrt{{\left(2 x - 3\right)}^{2} - {\left(5\right)}^{2}}} \mathrm{dx}$

Take ,$\textcolor{red}{2 x - 3 = 5 \sec u} \implies \sec u = \frac{2 x - 3}{5} \mathmr{and}$
$\implies 2 \mathrm{dx} = 5 \sec u \tan u \mathrm{du}$
$\implies \mathrm{dx} = \frac{5}{2} \sec u \tan u \mathrm{du}$
$I = \int \frac{1}{\sqrt{25 {\sec}^{2} u - 25}} \left(\frac{5}{2} \sec u \tan u\right) \mathrm{du}$
$I = \frac{5}{2} \int \frac{\sec u \tan u}{5 \sqrt{{\sec}^{2} u - 1}} \mathrm{du}$
$= \frac{1}{2} \int \frac{\sec u \tan u}{\tan u} \mathrm{du}$
$= \frac{1}{2} \int \sec u \mathrm{du}$
$= \frac{1}{2} \ln | \sec u + \tan u | + c$
$= \frac{1}{2} \ln | \sec u + \sqrt{{\sec}^{2} u - 1} | + c$
$= \frac{1}{2} \ln | \frac{2 x - 3}{5} + \sqrt{{\left(\frac{2 x - 3}{5}\right)}^{2}} - 1 | + c$
$= \frac{1}{2} \ln | \frac{2 x - 3}{5} + \frac{\sqrt{{\left(2 x - 3\right)}^{2} - 25}}{5} | + c$
$= \frac{1}{2} \left[\ln | \left(2 x - 3\right) + \sqrt{4 {x}^{2} - 12 x - 16} | - \ln 5\right] + c$
$I = \frac{1}{2} \ln | \left(2 x - 3\right) + \sqrt{4 {x}^{2} - 12 x - 16} | + C ,$
where, $C = c - \frac{1}{2} \ln 5$
It is better to use $\int \frac{1}{\sqrt{{t}^{2} - {A}^{2}}} \mathrm{dt} = \ln | t + \sqrt{{t}^{2} - {A}^{2}} | + c$
for ,$2 x - 3 = t \implies \mathrm{dx} = \frac{1}{2} \mathrm{dt} \implies I = \int \frac{1}{\sqrt{{t}^{2} - {5}^{2}}} \left(\frac{1}{2} \mathrm{dt}\right) = \frac{1}{2} \ln | t + \sqrt{{t}^{2} - 25} | + c = \frac{1}{2} \ln | \left(2 x - 3\right) + \sqrt{4 {x}^{2} - 12 x - 16} | + c$

Mar 21, 2018

The answer is $= \frac{1}{2} \ln \left(| \sqrt{{\left(2 x - 3\right)}^{2} / 25 - 1} + \frac{2 x - 3}{5} |\right) + C$

#### Explanation:

The denominator is

$\sqrt{4 {x}^{2} - 12 x - 16} = 2 \sqrt{{x}^{2} - 3 x - 4}$

$= 2 \sqrt{{x}^{2} - 3 x + \frac{9}{4} - 4 - \frac{9}{4}}$

$= 2 \sqrt{{\left(x - \frac{3}{2}\right)}^{2} - \frac{25}{4}}$

$= 5 \sqrt{{\left(\frac{2 x - 3}{5}\right)}^{2} - 1}$

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} - 12 x - 16}} = \frac{1}{2} \int \frac{\mathrm{dx}}{\sqrt{{\left(x - \frac{3}{2}\right)}^{2} - \frac{25}{4}}}$

Let $u = \frac{2 x - 3}{5}$, $\implies$, $\mathrm{du} = \frac{2}{5} \mathrm{dx}$

Therefore,

$\int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} - 12 x - 16}} = \frac{1}{2} \int \frac{\mathrm{du}}{\sqrt{{u}^{2} - 1}}$

Let $u = \sec \theta$, $\implies$, $\mathrm{du} = \sec \theta \tan \theta d \theta$

$\frac{1}{2} \int \frac{\mathrm{du}}{\sqrt{{u}^{2} - 1}} = \frac{1}{2} \int \frac{\sec \theta \tan \theta d \theta}{\sqrt{{\sec}^{2} \theta - 1}}$

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

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

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

Let $v = \sec \theta + \tan \theta$

$\implies$, $\mathrm{dv} = \left({\sec}^{2} \theta + \sec \theta \tan \theta\right) d \theta$

And finally,

$\int \sec \theta d \theta = \int \frac{\mathrm{dv}}{v}$

$= \frac{1}{2} \ln \left(v\right)$

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

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