arccos and arcsin integral contradiction: I am shown:
$$f(x) = \arcsin x \implies f'(x) = \frac{1}{\sqrt{1-x^2}}$$
$$f(x) = \arccos x \implies f'(x) = -\frac{1}{\sqrt{1-x^2}}$$
These two derivatives can be very readily derived by a bit of implicit differentiation. However... I want to undo my differentiation for both expressions.
I set up the integrals and now note that the only difference between the two derivatives is a negative sign. Suppose that for the arccos(x) derivative, I factor out the negative sign before integrating the expression, as $-1$ is a constant and the constnat factor rule states that:
$$\int k \frac{dy}{dx} dx = k \int \frac{dy}{dx} dx$$
However, due to this, I end up with the false equality:
$$-\arccos(x)=\arcsin(x)$$
Which doesn't hold for any $x$ at all!
What have I done wrong, why cant one simply factor out the negative sign for the arccos derivative expression?

Two functions defined over an interval that have the same derivative  differ by a constant.
Since the derivatives of $\arcsin x$ and $-\arccos x$ are both equal to $\dfrac{1}{\sqrt{1-x^2}}$ on the interval $(-1,1)$, you can say that there exists $c$ such that
$$
\arcsin x=c-\arccos x
$$
If you evaluate at $0$, you get
$$
0=\arcsin 0=c-\arccos 0=c-\frac{\pi}{2}
$$
Therefore, for all $x\in(-1,1)$,
$$
\arcsin x=\frac{\pi}{2}-\arccos x
$$
and this holds by continuity also for $x=-1$ and $x=1$.

You get a similar “contradiction” if you do
$$
\log x=\int\frac{1}{x}\,dx=\int\frac{2}{2x}\,dx=\int\frac{1}{2x}\,d(2x)=\log(2x)
$$
without taking into account the “constant of integration”; indeed
$$
\log x = \log(2x)-\log 2
$$
