Integration of $ \int \frac{dx}{x^2\sqrt{x^2 + 9}} $ using trigonometric substitution I have been stuck trying to figure out an integration problem involving trigonometric substitution. 
$$ \int \frac{dx}{x^2\sqrt{x^2 + 9}} $$
So I substituted 
$$ x = 3\tan\theta $$
$$ dx = 3\sec^2\theta \,d\theta $$
and I plug in everything and simplify until I get 
$$ \frac{1}{9} \int \frac{ \cos \theta \,d \theta}{\sin^2 \theta} $$
I substitute $$ u = \sin \theta $$ and $$ du = \cos \theta \,d \theta $$
and I integrate and plug back in, in terms of theta and my answer is 
$$ \frac{-\csc( \theta )}{9} + C $$
However, I have to substitute back in my first substitution, which was the trigonometric substitution, to get the final answer, but I don't really know how.
I do know we need :
$$ \theta = \arctan(\frac{x}{3}) $$ but I'm not sure what to do afterwards.
Sorry if this is a little messy, this is my first time working with latex and my first post.
 A: $$x = 3\tan \theta \iff \tan\theta = \frac x3 \implies \sin\theta = \frac x{\sqrt{x^2 + 9}} \implies \csc \theta = \frac{\sqrt{x^2 + 9}}{x}$$
This follows if we consider a right-triangle with legs $3, x,$ and hypotenuse $\sqrt{x^2 + 9}$, where the $\theta $ is the angle formed by the intersection of the leg of length $3$ with the hypotenuse.
This means $$-\frac{\csc\theta}{9} = -\frac{\sqrt{x^2 + 9}}{9x}$$
A: now, using your first expression $x = 3\tanθ$, $\tanθ = x/3 = $opp/adj you get all 3 sides of your triangle and substitute back 
$= - \sqrt{\frac{1+\cot^2(t)}{9}}
= - \sqrt{\frac{ 1+ \frac{1}{\tan^2(t)}}{9}}
= -\sqrt{ 1+ \frac{1}{(\frac{x}{3})^2}} 
= -\sqrt{ \frac{1+\frac{9}{x^2}}{ 9}}
= \frac{- \sqrt{9+x^2}} { 9x} + C $
A: $$x=3\sinh{y}\Rightarrow dx=3\cosh{y}$$
$$\int\frac{dx}{x^{2}\sqrt{x^{2}+9}}=\int\frac{(3\cosh{y})dy}{(\sinh^{2}{y})(3\cosh{y})}=\int\frac{dy}{\sinh^{2}{y}}$$
$$\coth{y}=\frac{\cosh{y}}{\sinh{y}}\Rightarrow  d(\coth{y})=\frac{\sinh^{2}{y}-\cosh^{2}{y}}{\sinh^{2}{y}}dy=-\frac{dy}{\sinh^{2}{y}}$$
$$\int\frac{dx}{x^{2}\sqrt{x^{2}+9}}=-\int d(\coth{y})=-\coth{y}+c=-\frac{\cosh{y}}{\sinh{y}}+c$$
$$=-\frac{\sqrt{1+(\frac{x}{3})^{2}}}{\frac{x}{3}}+c=-\frac{\sqrt{x^{2}+9}}{x}+c$$
