# Are $\arcsin x$ and $\arccos x$ equal up to a constant?

Are the functions $\arcsin x$ and $\arccos x$ equal up to a constant?

When I was solving the indefinite integral $\int\frac{\mathrm dx}{\sqrt{1-x^2}}$ I got two different results depending on the kind of the trigonometric substitution I make:

$\displaystyle \int\frac{\mathrm dx}{\sqrt{1-x^2}}=-\arcsin x$, if $x=\sin \theta$

$\displaystyle \int\frac{\mathrm dx}{\sqrt{1-x^2}}=\arccos x$, if $x=\cos\theta$

Where's my mistake?

• You mean like $\arcsin x + \arccos x = \pi/2$ ?
– user312097
Sep 10, 2017 at 16:47
• @A---B I didn't understand, if $-\arcsin x=\arccos x$, then $\arcsin x+ \arccos x=0$, no? Sep 10, 2017 at 16:52
• $-\arcsin x \ne \arccos x$ but $\arccos x =\pi/2 - \arcsin x$. So, $\arccos x + C=C_2 - \arcsin x$, where $C, C_2 \in \Bbb R$ and $C_2 = C + \pi/2$.
– user312097
Sep 10, 2017 at 16:53
• @user42912 No! What you can deduce for your computations is that $\int\frac{\mathrm dx}{\sqrt{1+x^2}}=\arcsin x+C_1$ and that $\int\frac{\mathrm dx}{\sqrt{1+x^2}}=-\arccos x+C_2$, where $C_1$ and $C_2$ are constants. Note, by the way, that there is an error in your signs. Sep 10, 2017 at 16:54
• @A---B ha ok, I got it. Thank you! Sep 10, 2017 at 16:58

No mistake. It turns out that$$\left(\forall x\in\left[-1,1\right]\right):\arcsin(x)+\arccos(x)=\frac\pi2.$$It is easy to justify this geometrically.
• I didn't understand, if $-\arcsin x=\arccos x$, then $\arcsin x+ \arccos x=0$, no? Sep 10, 2017 at 16:53