integration about $\arcsin $ and $\arccos $ To integrate $\frac{1}{\sqrt{a^2-x^2}}$, we get $\arcsin(x/a)+C $

To integrate $\frac{-1}{\sqrt{a^2-x^2}}$, we get $\arccos(x/a) + C$

but what if I take out the $-1$? 

So 

$\displaystyle\int \frac{-1}{\sqrt{a^2-x^2}}\,dx$ =  -$\arcsin(x/a)+C $?

Because $-1$ is a constant and we can just take it out of the integral? 

I am confused.
HINT:

$$\arcsin(x)+\arccos(x)=\pi/2 \tag 1$$



SPOILER ALERT:  Scroll over the highlighted area to reveal the solution


  Note that we have $\arcsin(x)=\int_0^x \frac{1}{\sqrt{1-t^2}}\,dt$ and therefore in terms of anti-derivatives, we can write $$\arcsin(x)=\int \frac{1}{\sqrt{1-x^2}}\,dx+C$$Using $(1)$, the arccosine function can be written as the anti-derivative $$\begin{align}\arccos(x)&amp;=\pi/2-\arcsin(x)\\\\&amp;=\pi/2 -\left(\int \frac{1}{\sqrt{1-x^2}}\,dx+C\right)\\\\&amp;=-\int \frac{1}{\sqrt{1-x^2}}\,dx +C'  \end{align}$$where $C'=\pi/2-C$
arcsinx + arccosx = pi/2 (some constant).
They hold a linear relationship.
So yeah, it's somehow correct.
Try solving more differential equations of first order and first degree.