# How do you integrate int 1/sqrt(3x-12sqrtx+53)  using trigonometric substitution?

Jul 1, 2018

$\int \frac{1}{\sqrt{3 x - 12 \sqrt{x} + 53}} \mathrm{dx}$

$= \frac{2 \sqrt{41}}{3 \sin \left({\cot}^{- 1} \left(\sqrt{\frac{3}{41}} \left(\sqrt{x} - 2\right)\right)\right)} + \frac{2}{\sqrt{3}} \ln \left(| \csc \left({\cot}^{- 1} \left(\sqrt{\frac{3}{41}} \left(\sqrt{x} - 2\right)\right)\right) + \sqrt{\frac{3}{41}} \left(\sqrt{x} - 2\right) |\right) + C$ $C \in \mathbb{R}$

See explanations below.

#### Explanation:

$\int \frac{1}{\sqrt{3 x - 12 \sqrt{x} + 53}} \mathrm{dx}$

$= \frac{1}{\sqrt{3}} \int \frac{1}{\sqrt{x - 4 \sqrt{x} + \frac{53}{3}}} \mathrm{dx}$
This form of integral doesn't seem to be know: let substitute:
Let
$u = \sqrt{x}$

u²=x

$\mathrm{dx} = 2 u \mathrm{du}$
So:
1/sqrt3int1/sqrt(x-4sqrtx+53/3)dx=1/sqrt3int(2u)/sqrt(u²-4u+53/3)du

=2/sqrt3intu/sqrt(u²-4u+53/3)du

Now we need to complete the square :

=2/sqrt3intu/sqrt((u-2)²+41/3)du

Now, let : $u - 2 = \sqrt{\frac{41}{3}} \cot \left(\theta\right)$
du=-sqrt(41/3)csc(theta)²d theta

So :

2/sqrt3int(u*du)/sqrt(u²-4u+53/3)

=2/sqrt3int((sqrt(41/3)cot(theta)+2)(-sqrt(41/3)csc(theta)²))/sqrt(41/3cot(theta)²+41/3)d theta

=-2/sqrt3int((sqrt(41/3)cot(theta)+2)(cancel(sqrt(41/3))csc(theta)²))/(cancel(sqrt(41/3))sqrt(cot(theta)²+1))d theta

Because cot(theta)²+1=csc(theta)²,

-2/sqrt3int((sqrt(41/3)cot(theta)+2)csc(theta)²)/(sqrt(cot(theta)²+1))d theta=-2/sqrt3int((sqrt(41/3)cot(theta)+2)csc(theta)^(cancel(2)))/cancel(csc(theta))d theta

$= - \frac{2}{\sqrt{3}} \int \sqrt{\frac{41}{3}} \cot \left(\theta\right) \csc \left(\theta\right) d \theta - \frac{2}{\sqrt{3}} \int \csc \left(\theta\right) d \theta$

=-2/3sqrt41intcos(theta)/(sin(theta)²)d theta-2/sqrt3intcsc(theta)d theta

Now we've got two easier integrals : I won't come back on $\int \csc \left(\theta\right) = - \ln \left(\csc \left(\theta\right) + \cot \left(\theta\right)\right)$, you can see there the demonstration.

For the second one, remarkably, ${\left(- \cos \left(x\right)\right)}^{'} = \sin \left(x\right)$
So: -intcos(theta)/(sin(theta)²)d theta=-int(f^')/(f²)=1/f=1/sin(theta)

So: -2/3sqrt41intcos(theta)/(sin(theta)²)d theta-2/sqrt3intcsc(theta)d theta

$= \frac{2 \sqrt{41}}{3 \sin \left(\theta\right)} + \frac{2}{\sqrt{3}} \ln \left(\csc \left(\theta\right) + \cot \left(\theta\right)\right)$

But $\theta = {\cot}^{- 1} \left(\sqrt{\frac{3}{41}} \left(u - 2\right)\right)$

So:

$\frac{2 \sqrt{41}}{3 \sin \left(\theta\right)} + \frac{2}{\sqrt{3}} \ln \left(\csc \left(\theta\right) + \cot \left(\theta\right)\right)$

$= \frac{2 \sqrt{41}}{3 \sin \left({\cot}^{- 1} \left(\sqrt{\frac{3}{41}} \left(u - 2\right)\right)\right)} + \frac{2}{\sqrt{3}} \ln \left(| \csc \left({\cot}^{- 1} \left(\sqrt{\frac{3}{41}} \left(u - 2\right)\right)\right) + \sqrt{\frac{3}{41}} \left(u - 2\right) |\right)$

Finally, $u = \sqrt{x}$

So:

$\int \frac{1}{\sqrt{3 x - 12 \sqrt{x} + 53}} \mathrm{dx}$

$= \frac{2 \sqrt{41}}{3 \sin \left({\cot}^{- 1} \left(\sqrt{\frac{3}{41}} \left(\sqrt{x} - 2\right)\right)\right)} + \frac{2}{\sqrt{3}} \ln \left(| \csc \left({\cot}^{- 1} \left(\sqrt{\frac{3}{41}} \left(\sqrt{x} - 2\right)\right)\right) + \sqrt{\frac{3}{41}} \left(\sqrt{x} - 2\right) |\right) + C$ $C \in \mathbb{R}$