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 $x=\frac{3}{2}\sin\theta$.
2
in the denominator of your answer goes into the constant, that's all. $\endgroup$