Is $\int \frac{\sqrt{1-x^2}}{x}\;dx = \ln\lvert x\rvert + C$? I have the integral
$$
\int \frac{\sqrt{1-x^2}}{x}\;dx 
$$
I do a trig. substitution $x = \sin(\theta)$. Then I get $\sqrt{1-x^2} = \cos(\theta)$. So
$$
\int \frac{\sqrt{1-x^2}}{x}\;dx = \int \frac{\cos(\theta)}{\sin(\theta)}\; d\theta = \ln\lvert \sin(\theta)\rvert + C= \ln\lvert x\rvert + C.
$$
But I am thinking that I made a mistake somewhere because the derivative of $\ln\lvert x\rvert$ is not $\frac{\sqrt{1-x^2}}{x}$. 
Where is my mistake?

$$x=\sin\theta\implies dx=\cos(\theta)\,d\theta$$
$$\begin{align}
\int{ \sqrt{1-x^2} \over x}\,dx
&= \int{ \sqrt{1-\sin^2\theta} \over \sin\theta} \cos(\theta)\,d\theta \\
&= \int{ \sqrt{\cos^2\theta} \over \sin\theta } \cos(\theta)\,d\theta \\
&= \int{ \cos^2\theta \over \sin\theta}\,d\theta \\
&= \int{ -\sin^2(\theta) + 1 \over \sin\theta}\,d\theta \\
&= \int\bigl[ -\sin(\theta) + \csc(\theta)\bigr]\,d\theta
\end{align}$$
You got it from there.
