1. ## trig sub

$\int \frac {48 dx} {x^2 \sqrt{x^2+36}}$

I finally get down to;

$= 48/216 \int \frac {1} {tan^2 t}$

Could you please tell me if that's correct? I doubt it is because there's no indication of a integral of that on my table sheet ...

2. $I = \int \frac{ dx}{ x^2 \sqrt{x^2+1} }
$

For this integral, it is good to applt trig. sub. but remind that

$I = \int \frac{ \sec^2(t) dt }{ \tan^2(t) \sqrt{ \sec^2(t) } }$

$= \int \frac{ \sec^2(t) dt }{ \tan^2(t) \sec(t) }$

$= \int \frac{ \sec(t) dt }{ \tan^2(t) }$

$= \int \frac{dt}{ \tan(t) \sin(t) }$

$= \int \csc(t) \cot(t)~dt$

$= - csc(t) + C$

$= - \frac{1}{ \sin( \tan^{-1}(x))} + C$

$= - \frac{1}{ \frac{x}{\sqrt{x^2+1}} } + C$

$= -\frac{ \sqrt{x^2+1} }{x} + C$

Of course , it is not the answer to your integral , you need to change a little .

I would also prefer the substitution $x = \frac{1}{t}$

$dx = - \frac{dt}{t^2}$

the integral becomes

$- \int \frac{dt/t^2}{ 1/t^2 (\sqrt{1+t^2}/t)}$

$= -\int \frac{tdt}{\sqrt{1+t^2}}$

$= - \sqrt{1+t^2} + C$

$= - \frac{ \sqrt{1+x^2}}{x} + C$

3. Thank you very much for your response, but is there another way to solve it without the I = and the non-trig sub?

4. Originally Posted by Archduke01
$\int \frac {48 dx} {x^2 \sqrt{x^2+36}}$

I finally get down to;

$= 48/216 \int \frac {1} {tan^2 t}$

Could you please tell me if that's correct? I doubt it is because there's no indication of a integral of that on my table sheet ...
Did you use $x=6tan(t)$?

Draw a right triangle. The leg opposite t is x and the adjacent leg is 6.

Then the hypotenuse is $\sqrt{x^2+36}$

So $sec(t)=\frac{\sqrt{x^2+36}}{6}$

$\sqrt{x^2+36}=6sec(t)$

$x^2=36tan^2(t)$

$dx=6sec^2(t)\,dt$

Then $\int \frac {48 dx} {x^2 \sqrt{x^2+36}}=48\int\frac{6sec^2(t)\,dt}{(36tan^2 (t))(6sec(t))}$

Simplify and integrate