# 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.

• Have you been instructed to solve this by trig. substitutions? It can be done just by $u$ substitution. Let $u=\frac{\sqrt{x^2+9}}{x}$ Mar 23, 2014 at 17:38

$$x = 3\tan \theta \iff \tan\theta = \frac x3 \implies \sin\theta = \frac x{\sqrt{x^2 + 9}} \implies \csc \theta = \frac{\sqrt{x^2 + 9}}{x}$$
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 $$-\frac{\csc\theta}{9} = -\frac{\sqrt{x^2 + 9}}{9x}$$
now, using your first expression $$x = 3\tanθ$$, $$\tanθ = x/3 =$$opp/adj you get all 3 sides of your triangle and substitute back
$$= - \sqrt{\frac{1+\cot^2(t)}{9}} = - \sqrt{\frac{ 1+ \frac{1}{\tan^2(t)}}{9}} = -\sqrt{ 1+ \frac{1}{(\frac{x}{3})^2}} = -\sqrt{ \frac{1+\frac{9}{x^2}}{ 9}} = \frac{- \sqrt{9+x^2}} { 9x} + C$$