For indefinite integrals, is arccos equivalent to -arcsin?

Question

I apologize if there was any confusion from the title. I was just confused by the antiderivative: \begin{aligned} & \int-\frac{1}{\sqrt{1-x^2}} d x=\cos ^{-1}(x)+C \end{aligned} Which made me think why couldn't it just be? \begin{aligned} & \int-\frac{1}{\sqrt{1-x^2}} d x=-1 \int \frac{1}{\sqrt{1-x^2}} d x \\ & =-1\left(\sin ^{-1}(x)\right)+C \\ & =-\sin ^{-1}(x)+C \end{aligned} I'm just wondering if someone could explain why this is the way it is because I've had a hard time finding things online.

Answer 1

Let $f(x) = \arcsin(x) + \arccos(x)$ where $x\in[0,1]$.

Based on the differentiation rules and the MVT, one concludes that: \begin{align*} f'(x) = \frac{1}{\sqrt{1 - x^{2}}} - \frac{1}{\sqrt{1 - x^{2}}} = 0 \Rightarrow f(x) = k \end{align*} To find out the value of $k$, set $x = \dfrac{\sqrt{2}}{2}$ to obtain: \begin{align*} k = f\left(\frac{\sqrt{2}}{2}\right) = \arcsin\left(\frac{\sqrt{2}}{2}\right) + \arccos\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} \end{align*}

Hence one concludes that: \begin{align*} \arcsin(x) = \frac{\pi}{2} - \arccos(x) \end{align*} and we are done.

Hopefully this helps!

Answer 2

Notice that the $C$ can be different. This follows since $\sin,\cos$ are bijective from $[0, \frac{\pi}{2}]$ to $[0,1]$ and using that $\sin( \frac{\pi}{2}-x)=\cos(x)$ then letting $x=\arccos(y)$ we get $\sin(-\arccos(y) + \frac{\pi}{2})=y$ hence $\arcsin(y)=-\arccos(y) + \frac{\pi}{2}$ for $y \in [0,1]$.