I'm trying to solve the integral $I=\int \frac{dx}{x^3 \sqrt{x^2-4}}$ ($x<0$) which can be solved by substituting $x=2 \sec t$. One ends up with $$I=-\frac{1}{16} \left( t+\sin t \cos t \right)+C$$ (This is verified by the book I'm using.) If $x=2 \sec t$, then $t=\sec^{-1}(\frac x2)$ and $\cos t =\frac 2x$.
My problem is replacing $\sin t$ by something which is dependent on $x$. The standard method would be constructing a right-angled triangle with sides given by $\cos t=\frac 2x$, giving $\sin t$. However, since $x<0$, the triangle would have negative sides which is causing me trouble.
The solution given in the book is $\sin t = \frac{-\sqrt{x^2-4}}{x}$ but I don't see what justifies the sign. (Could someone explain it?) Another method would be using the identity $\sin^2 t+\cos^2 t = 1$ but this also only gives $\sin t = \pm \sqrt{1-\frac{4} {x^2}}$.
If $-\frac{1}{2}\pi \leq t\leq 0$, then $-1 \leq \sin t \leq 0$, and all values in this range are achieved. Therefore, since $-1 < -\frac{\sqrt{x^2-4}}x \leq 0$, there exist valid values of $t$ making this formula true, so you are allowed to make the substitution, and we could choose $t$ with $-\frac12\pi\leq t\leq 0$. There doesn't need to be an actual triangle, the function does take negative values for these arguments.