# How do you integrate int 1/sqrt(4x^2-12x-7)  using trigonometric substitution?

Jun 27, 2018

(2x-a)^2=4x^2-4ax+a^2

#### Explanation:

$\int \frac{1}{\sqrt{4 {x}^{2} - 12 x - 7}} \mathrm{dx} = \int \frac{1}{\sqrt{{\left(2 x - 3\right)}^{2} - 16}} \mathrm{dx}$

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

Let $u = \frac{x}{2} - \frac{3}{4}$ therefore $2 \mathrm{du} = \mathrm{dx}$

$\frac{1}{2} \int \frac{1}{\sqrt{{u}^{2} - 1}} = \frac{i}{2} \int - \frac{1}{\sqrt{1 - {u}^{2}}}$

$\frac{i}{2} {\cos}^{-} 1 \left(\frac{2 x - 3}{4}\right)$

Jun 27, 2018

The answer is $= \frac{1}{2} \ln \left(| \frac{2 x - 3}{4} + \sqrt{\left({\left(\frac{2 x - 3}{4}\right)}^{2}\right) - 1} |\right) + C$

#### Explanation:

Rewrite

$4 {x}^{2} - 12 x - 7 = {\left(2 x - 3\right)}^{2} - 16$

$= {4}^{2} \left({\left(\frac{2 x - 3}{4}\right)}^{2} - 1\right)$

Therefore,

$\sqrt{4 {x}^{2} - 12 x - 7} = \sqrt{{4}^{2} \left({\left(\frac{2 x - 3}{4}\right)}^{2} - 1\right)}$

$= 4 \sqrt{\left({\left(\frac{2 x - 3}{4}\right)}^{2} - 1\right)}$

The integral is

$I = \int \frac{\mathrm{dx}}{\sqrt{4 {x}^{2} - 12 x - 7}} = \frac{1}{4} \int \frac{\mathrm{dx}}{\sqrt{\left({\left(\frac{2 x - 3}{4}\right)}^{2} - 1\right)}}$

Let $u = \frac{2 x - 3}{4}$, $\implies$, $\mathrm{du} = \frac{\mathrm{dx}}{2}$

Therefore,

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

Let $u = \sec v$, $\implies$, $\mathrm{du} = \sec v \tan v \mathrm{dv}$

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

The integral is

$I = \frac{1}{2} \int \frac{\sec v \tan v \mathrm{dv}}{\sqrt{{\sec}^{2} v - 1}}$

$= \frac{1}{2} \int \sec v \mathrm{dv}$

$= \frac{1}{2} \int \frac{\sec v \left(\sec v + \tan v\right) \mathrm{dv}}{\sec v + \tan v}$

Let $w = \sec v + \tan v$, $\implies$, $\mathrm{dw} = \left(\sec v \tan v + {\sec}^{2} v\right) \mathrm{dv}$

Therefore,

$I = \frac{1}{2} \int \frac{\mathrm{dw}}{w}$

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

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

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

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

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