2

I'm currently learning calculus on my own, and I have started to meet integrals with arcsin(x) and arccos(x). According to the book I'm using $\frac{d}{dx}(\arccos(x))=-\frac{1}{\sqrt{1-x^2}}$. However, everywhere else, I see: $$\int -\frac{1}{\sqrt{1-x^2}} dx=-\arcsin(x) $$Why is $\int -\frac{1}{\sqrt{1-x^2}} dx≠\arccos(x) $? Help would be much appreciated.

sreysus
  • 721
Iggelo
  • 23

1 Answers1

5

You're missing the constant of integration, which solves the issue. So both are correct, since

$$\arccos(x)=-\arcsin(x)+\frac{\pi}{2}.$$

Lorago
  • 9,239
  • Thank you for your comment. While I think your comment should help me, I sadly still dont understand. Could you explain it further? Or link something that explains it. I do understand that I missed the constants of integration, but I don't see how that solves the problem. – Iggelo Sep 22 '23 at 20:26
  • @Iggelo $$\int-\frac{\mathrm{d}x}{\sqrt{1-x^2}}=\arccos(x)+C_1=\arccos(x)-\frac{\pi}{2}+C_2=-\arcsin(x)+C_2$$with $C_2=C_1+\frac{\pi}{2}$. Since the constant goes through all real numbers (it is not just one of them), we can freely change it by adding a constant. – Lorago Sep 22 '23 at 20:29
  • @Iggelo another way to look at it is that the functions $x\mapsto\arccos(x)$ and $x\mapsto-\arcsin(x)$ have the same derivative, and thus they can only differ by a constant (and the indefinite integral of their derivatives must be the same) – Lorago Sep 22 '23 at 20:32
  • I think I understand now, thank you so much for your explanation. Does this mean we can say: $\int -\frac{1}{\sqrt{1-x^2}} dx $ = $-\arcsin(x) + C_1 $ = $\arccos(x) + C_2 $ – Iggelo Sep 22 '23 at 20:54
  • Yes exactly @Iggelo – Lorago Sep 23 '23 at 00:25
  • Thank you so much Lorago! – Iggelo Sep 23 '23 at 10:44