I am facing trouble to prove: $$\displaystyle\int\frac{\mathrm d x}{\sqrt{a+bx+cx^2}}= \frac{1}{\sqrt{-c}}\arccos\left(-\frac{b+2cx}{\sqrt{-\Delta}} \right) \tag{1}\\ \text{when } c<0, \; \Delta=4ac-b^2<0 $$ It's used here: https://youtu.be/O4SIw6cYSow?t=577
But, so far, I have come up with this form: $$\displaystyle\int\frac{\mathrm d x}{\sqrt{a+bx+cx^2}}= \frac{-1}{\sqrt{-c}}\arcsin\left(\frac{b+2cx}{\sqrt{-\Delta}} \right)= \frac{1}{\sqrt{-c}}\arccos\left(\frac{b+2cx}{\sqrt{-\Delta}} \right) \tag{2}$$ As you can see, the necessary minus sign isn't appearing inside the inverse cosine function!
By the way, the form with arcsine function is also mentioned in section 2.261 of the book "Table of Integrals, Series, and Products". (https://www.amazon.com/Table-Integrals-Products-Daniel-Zwillinger/dp/0123849330)
TIA