Integration of $ \int \frac{dx}{x^2\sqrt{x^2 + 9}} $ using trigonometric substitution

Question

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.

Answer 1

$$x = 3\tan \theta \iff \tan\theta = \frac x3 \implies \sin\theta = \frac x{\sqrt{x^2 + 9}} \implies \csc \theta = \frac{\sqrt{x^2 + 9}}{x}$$

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 $$-\frac{\csc\theta}{9} = -\frac{\sqrt{x^2 + 9}}{9x}$$

Answer 2

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 $

Answer 3

$$x=3\sinh{y}\Rightarrow dx=3\cosh{y}$$ $$\int\frac{dx}{x^{2}\sqrt{x^{2}+9}}=\int\frac{(3\cosh{y})dy}{(\sinh^{2}{y})(3\cosh{y})}=\int\frac{dy}{\sinh^{2}{y}}$$ $$\coth{y}=\frac{\cosh{y}}{\sinh{y}}\Rightarrow d(\coth{y})=\frac{\sinh^{2}{y}-\cosh^{2}{y}}{\sinh^{2}{y}}dy=-\frac{dy}{\sinh^{2}{y}}$$

$$\int\frac{dx}{x^{2}\sqrt{x^{2}+9}}=-\int d(\coth{y})=-\coth{y}+c=-\frac{\cosh{y}}{\sinh{y}}+c$$ $$=-\frac{\sqrt{1+(\frac{x}{3})^{2}}}{\frac{x}{3}}+c=-\frac{\sqrt{x^{2}+9}}{x}+c$$