Using trig substitution to solve for integration? So I used a trig sub for this problem:
$$\int \frac{1}{x^2\sqrt{9-x^2}}dx.$$
${x=3\sin\theta}$
${dx=3\cos\theta\ d\theta}$
${\sqrt{9-x^2}= 3\cos\theta}$
I ended up with 
$$\frac19 \int \frac{ d\theta}{\sin^2\theta}$$
From there how do I solve that? 
 A: Hint: $\dfrac{d}{d\theta}\cot\theta = -\csc^2\theta$
And $d\theta$ in your integral should not be in the denominator!
A: $${\int{1\over x^2\sqrt{9-x^2}}}dx$$
Your substitutions are fine:
$$x = 3\sin \theta \implies dx = 3\cos \theta \,d\theta$$
But be careful about $$\sqrt{9 - (3\sin \theta)^2}=\sqrt{9\cos^2 \theta} = 3|\cos\theta|$$
So we have $$\int \frac{3\cos \theta d\theta}{9\sin^2 \theta\cdot 3|\cos \theta|} \;=  \;\frac{1}9\int \frac{\pm d\theta}{\sin^2 \theta}\; = \;\frac 19 \int \pm \csc^2\theta \,d\theta \;\;= \;\; \pm \frac 19 \int -\frac{d}{dx}\left(\cot \theta\right)\,dx= \;\;\mp \frac 19 \cot\theta + C$$


*

*Let me explain the use of $\pm, \mp$. When $\pi/2 \lt \theta\lt
   3\pi/2, $ then $\cos \theta < 0$ and so $|\cos \theta| = -\cos
   \theta$.


Now you need only back-substitute by expressing $\theta$ in terms of $x$.
A: The simpliest way is guessing: note that $(-\cot x)'=\frac{1}{\sin^2 x}$, so:
$$\int \frac{1}{\sin^2 \theta}dx=-\cot \theta+C$$
Next $\theta=\arcsin (x/3)$,so:
$$\int \frac{1}{\sin^2 \theta}dx=-\cot \theta+C=-\cot(\arcsin (x/3))+C$$
A: Alternatively, substitute $t=\frac3x$
$$\int \frac{1}{x^2\sqrt{9-x^2}}dx=-\frac19\int \frac t{\sqrt{t^2-1}}dt =-\frac19\sqrt{t^2-1}+C
$$
