Integration with logarithmic expression. I'm trying to solve an integration problem from the book which is the $$\int\frac{\sqrt{9-4x^2}}{x}dx$$ using trigonometric substitution. The answer from the book is $$3\ln\left|\frac{3-\sqrt{9-4x^2}}{x}\right|+\sqrt{9-4x^2}+C.$$ 
I have almost the same solution where there's a $$3\ln|\csc\theta-\cot\theta|+3\cos\theta+C.$$ 
The problem is when the substitution comes in. I end up having 
$$3\ln\frac{|3-\sqrt{9-x^2}|}{2x}+\sqrt{9-x^2}$$ and when I tried to simplify it further, it resulted to $$3\ln\left|3-\sqrt{9-x^2}\right|-3\ln|2x|+\sqrt{9-x^2}.$$ I hope you could help me to tell where i did wrong. 
By the way, I set $a=3$ and $x=\frac{3}{2}\sin\theta$.

I tried to follow your steps. Here we are assuming that $0<x\leq 3/2$. Let $x=3/2\sin(t)$ then $$3\cos(t)=3\sqrt{1-(2x/3)^2}=\sqrt{9-4x^2}$$ and
\begin{align*}\int\frac{\sqrt{9-4x^2}}{x}dx&=\int\frac{3\cos(t)}{3/2\sin(t)}d(3/2\sin(t))\\&
=3\int\frac{\cos^2(t)}{\sin(t)}dt=3\int\frac{1-\sin^2(t)}{\sin(t)}dt\\&=3\int\frac{dt}{\sin(t)}-3\int \sin(t)dt
\\&=
3\ln\frac{1-\cos(t)}{\sin(t)}+3\cos(t)+C\\
\\&=
3\ln\frac{3-3\cos(t)}{3\sin(t)}+3\cos(t)+C\\
&=3\ln\frac{3-\sqrt{9-4x^2}}{2x}+\sqrt{9-4x^2}+C\\
&=3\ln\frac{3-\sqrt{9-4x^2}}{x}+\sqrt{9-4x^2}+C'.
\end{align*}
where $C'=C-3\ln(2)$ is an arbitrary constant.
P.S. We have that for $\sin(t)>0$,
\begin{align*}
\int\frac{dt}{\sin(t)}&=\int\frac{1+\cos(t)-\cos(t)}{\sin(t)}\,dt\\
&=\int\frac{(1+\cos(t))(1-\cos(t))}{\sin(t)(1-\cos(t)}\,dt-\int\frac{\cos(t)}{\sin(t)}\,dt\\
&=\int\frac{\sin(t)}{1-\cos(t)}\,dt-\int\frac{\cos(t)}{\sin(t)}\,dt\\
&=\ln(1-\cos(t))-\ln(\sin(t))+C=\ln\left(\frac{1-\cos(t)}{\sin(t)}\right)+C
\end{align*}
