# How do you integrate 1 / (t^3(t^2 - 9)^(1/2))?

May 5, 2015

Well, this is a bit hard !

Let's $u = \sqrt{{t}^{2} - 9}$
$\mathrm{du} = \frac{t}{\sqrt{{t}^{2} - 9}} \mathrm{dt}$

$\mathrm{dt} = \frac{\sqrt{{t}^{2} - 9}}{t} \mathrm{du}$

Integral become :

$\implies \int \frac{\sqrt{{t}^{2} - 9}}{{t}^{4} \sqrt{{t}^{2} - 9}} \mathrm{du} = \int \frac{1}{t} ^ 4 \mathrm{du}$

$\frac{1}{t} ^ 4 = \frac{1}{{u}^{2} + 9} ^ 2$

So now we have $\implies \int \frac{1}{{u}^{2} + 9} ^ 2 \mathrm{du}$

can't do partial fraction so let's $u = 3 \tan \left(v\right)$

$\implies \mathrm{du} = \frac{3}{\cos} ^ 2 \left(v\right) \mathrm{dv}$

$\implies \frac{1}{{u}^{2} + 9} ^ 2 = \frac{1}{9 {\tan}^{2} \left(v\right) + 9} ^ 2 = {\cos}^{4} \frac{v}{81}$

So now we have

$\implies \int {\cos}^{4} \frac{v}{81} \cdot \frac{3}{\cos} ^ 2 \left(v\right) \mathrm{dv} = \frac{3}{81} \int {\cos}^{2} \left(v\right) \mathrm{dv}$

$\implies \frac{3}{162} \int 1 + \cos \left(2 v\right) \mathrm{dv} = \frac{1}{54} \int 1 + \cos \left(2 v\right) \mathrm{dv}$

$\implies \frac{1}{54} \left[v + \frac{1}{2} \sin \left(2 v\right)\right] + C$

Substitute back for $v = \arctan \left(\frac{1}{3} u\right)$

$\implies \frac{1}{54} \left[\arctan \left(\frac{1}{3} u\right) + \frac{1}{2} \sin \left(2 \arctan \left(\frac{1}{3} u\right)\right)\right]$

Substitute again for $u = \sqrt{{t}^{2} - 9}$

$\implies \frac{1}{54} \left[\arctan \left(\frac{1}{3} \sqrt{{t}^{2} - 9}\right) + \frac{1}{2} \sin \left(2 \arctan \left(\frac{1}{3} \sqrt{{t}^{2} - 9}\right)\right)\right]$