Is $\int\frac{-1}{\sqrt{-x^2+1}}$ equal to $\arccos$ or $\arcsin$?

Question

Is this correct?

$$\int\frac{-1}{\sqrt{-x^2+1}}dx = \arccos(x) +c = -\arcsin(x) +c $$

Or is it just $$\int\frac{-1}{\sqrt{-x^2+1}}dx = -\arcsin(x) +c $$

because $$\frac{d}{dx}(\arccos(x))=\frac{-1}{\sqrt{-x^2+1}}$$

Answer 1

We know that $$ \int\frac{-1}{\sqrt{-x^2+1}}dx = \arccos x +C_1. $$ The constant $C_1$ can be manipulated as $1+C_2$, $1000-C_2$, or what ever we want. In this case, we will choose $C_1=-\dfrac\pi2+C_2$, then $$ \arccos x +C_1=\arccos x -\dfrac\pi2+C_2. $$ Using basic trigonometry identity, we know that $\arcsin x+\arccos x=\dfrac\pi2$. Try to draw the right triangle, you will easily prove it the identity. Hence $$ \int\frac{-1}{\sqrt{-x^2+1}}dx $$ can also be written as $$ \arccos x +C_1\quad\text{or}\quad-\arcsin x +C_2. $$ Both are interchangeably.