Integral with inverse trig functions and u substitution I've been trying to find the integral of $\dfrac{dx}{x \sqrt{x^2-4}}$. Currently, following the example problem, I have it as $\dfrac{1}{2} S \dfrac{1\,du}{x \sqrt{4 [(\frac{x}{2})^2-1]}}$. From what I've got it looks like it's going to the formula matching $\sec^{-1}(x)$. But at this point I'm not sure what to do now about the 4 in there that will get me to the $1/2$ coefficient. I have $u = x/2$, and $du=\frac{1}{2} \, dx$. The answer given is $-\frac{1}{2}\csc^{-1}(\frac{x}{2})+C$. The opposite trig function from what I thought and with a negative somehow. 

\begin{align}
& \frac{dx}{x \sqrt{x^2-4}} = \frac{dx}{x\sqrt{4\left(\left(\frac x 2\right)^2 - 1\right)}} = \frac 1 {\sqrt 4} \cdot \frac{dx}{x\sqrt{\left( \frac x 2 \right)^2 - 1}} = \frac 1 2 \cdot \frac{dx/2}{\frac x 2 \sqrt{\left( \frac x 2 \right)^2 - 1}} \\[10pt]
 = {} & \frac 1 2 \cdot \frac{du}{u\sqrt{u^2 -1}} = \frac 1 2 \cdot \frac{\sec\theta\tan\theta \, d\theta}{\sec\theta\sqrt{\sec^2\theta - 1}} = \frac{d\theta} 2.
\end{align}
