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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.

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  • $\begingroup$ Have you been instructed to solve this by trig. substitutions? It can be done just by $u$ substitution. Let $u=\frac{\sqrt{x^2+9}}{x}$ $\endgroup$ – TheMobiusLoops Mar 23 '14 at 17:38
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$$x = 3\tan \theta \iff \tan\theta = \frac x3 \implies \sin\theta = \frac x{\sqrt{x^2 + 9}} \implies \csc \theta = \frac{\sqrt{x^2 + 9}}{x}$$

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

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    $\begingroup$ Thank you very much, I just didn't think of drawing a triangle. Now I have a better understanding of what to do for these trig substitution problems. $\endgroup$ – user137420 Mar 23 '14 at 18:14
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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 $

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