# How do you integrate int 1/sqrt(3x^2-12x-5)dx using trigonometric substitution?

Jan 8, 2016

Assuming you know how to do substitution :

Complete the square, you want to have ${\left(a x \pm b\right)}^{2} \pm 1$ form

$\int \frac{1}{\sqrt{3 {x}^{2} - 12 x - 5}} \mathrm{dx}$

$3 {x}^{2} - 12 x - 5$

$a = \sqrt{3} x$
$b = 2 \sqrt{3}$

${a}^{2} = 3 {x}^{2}$
${b}^{2} = 12$

$3 {x}^{2} - 12 x + 12 - 17$

${\left(\sqrt{3} x - 2 \sqrt{3}\right)}^{2} - 17$

So we have

$\int \frac{1}{\sqrt{3 {x}^{2} - 12 x - 5}} \mathrm{dx} = \int \frac{1}{\sqrt{{\left(\sqrt{3} x - 2 \sqrt{3}\right)}^{2} - 17}} \mathrm{dx}$

$= \int \frac{1}{\sqrt{17 \left(\frac{1}{17} {\left(\sqrt{3} x - 2 \sqrt{3}\right)}^{2} - 1\right)}} \mathrm{dx}$

$= \frac{1}{\sqrt{17}} \int \frac{1}{\sqrt{\frac{1}{17} {\left(\sqrt{3} x - 2 \sqrt{3}\right)}^{2} - 1}} \mathrm{dx}$

In fact, at this moment you can do $u = \frac{1}{\sqrt{17}} \left(\sqrt{3} x - 2 \sqrt{3}\right)$

$\mathrm{du} = \sqrt{\frac{3}{17}} \mathrm{dx}$
$\mathrm{dx} = \sqrt{\frac{17}{3}} \mathrm{du}$

${u}^{2} = \frac{1}{17} {\left(\sqrt{3} x - 2 \sqrt{3}\right)}^{2}$

$= \frac{1}{\sqrt{3}} \int \frac{1}{\sqrt{{u}^{2} - 1}} \mathrm{du}$

You have the perfect derivate of $\frac{1}{\sqrt{3}} \arccos h \left(u\right)$

One form of the integral of $\int \frac{1}{\sqrt{3 {x}^{2} - 12 x - 5}} \mathrm{dx}$

is $\frac{1}{\sqrt{3}} \arccos h \left(\frac{1}{\sqrt{17}} \left(\sqrt{3} x - 2 \sqrt{3}\right)\right)$

But i don't like this form, so coming back to :

$\frac{1}{\sqrt{17}} \int \frac{1}{\sqrt{\frac{1}{17} {\left(\sqrt{3} x - 2 \sqrt{3}\right)}^{2} - 1}} \mathrm{dx}$

$\sqrt{3} x - 2 \sqrt{3} = \frac{\sqrt{17}}{\cos} \left(u\right)$

note that $u = \arccos \left(\frac{\sqrt{17}}{\sqrt{3} x - 2 \sqrt{3}}\right)$

$\sqrt{3} \mathrm{dx} = \sqrt{17} \tan \frac{u}{\cos} \left(u\right) \mathrm{du}$

$\mathrm{dx} = \sqrt{\frac{17}{3}} \tan \frac{u}{\cos} \left(u\right)$

You get

$\frac{1}{\sqrt{3}} \int \frac{1}{\sqrt{\frac{1}{{\cos}^{2} \left(u\right)} - 1}} \tan \frac{u}{\cos} \left(u\right) \mathrm{dx}$

$\frac{1}{\sqrt{\frac{1}{{\cos}^{2} \left(u\right)} - 1}} = \frac{1}{|} \tan \left(u\right) |$

$\frac{1}{\sqrt{3}} \int \frac{1}{\cos} \left(u\right) \mathrm{du}$

$\frac{1}{\sqrt{3}} \int \cos \frac{u}{\cos} ^ 2 \left(u\right) \mathrm{du}$

let's $t = \sin \left(u\right)$

$\mathrm{dt} = \cos \left(u\right) \mathrm{du}$

$= \frac{1}{\sqrt{3}} \int \frac{1}{1 - {t}^{2}} \mathrm{dt}$

do partial fraction to get

1/sqrt(3)(int1/(2(t+1))dt-int1/(2(t-1))dt)

which is

$\frac{1}{2 \sqrt{3}} \left[\ln \left(| t + 1 |\right) - \ln \left(| t - 1 |\right)\right]$

Substitute back for $t = \sin \left(u\right)$

$\frac{1}{2 \sqrt{3}} \left[\ln \left(| \sin \left(u\right) + 1 |\right) - \ln \left(| \sin \left(u\right) - 1 |\right)\right]$

Substitute back for $u = \arccos \left(\frac{\sqrt{17}}{\sqrt{3} x - 2 \sqrt{3}}\right)$

$\frac{1}{2 \sqrt{3}} \left[\ln \left(| \sin \left(\arccos \left(\frac{\sqrt{17}}{\sqrt{3} x - 2 \sqrt{3}}\right)\right) + 1 |\right) - \ln \left(| \sin \left(\arccos \left(\frac{\sqrt{17}}{\sqrt{3} x - 2 \sqrt{3}}\right)\right) - 1 |\right)\right]$

assuming that
$\sin \left(\arccos \left(\frac{\sqrt{17}}{\sqrt{3} x - 2 \sqrt{3}}\right)\right) = \sqrt{1 - \frac{17}{\sqrt{3} x - 2 \sqrt{3}} ^ 2}$

You finally have

$\frac{1}{2 \sqrt{3}} \left[\ln \left(| \sqrt{1 - \frac{17}{\sqrt{3} x - 2 \sqrt{3}} ^ 2} + 1 |\right) - \ln \left(| \sqrt{1 - \frac{17}{\sqrt{3} x - 2 \sqrt{3}} ^ 2} - 1 |\right)\right]$