Why isn't $\int \frac{-dx}{\sqrt{1-x^2}}$ equal to $\arccos(x)$?

Question

The reason I'm asking this is that because $d/dx(\arccos(x))=-\frac{1}{\sqrt{1-x^2}}$, why to textbooks opt to write $\int -\frac{1}{\sqrt{1-x^2}} dx=-\arcsin(x) $ instead?

Answer 1

They are both correct, since the two answers differ by a constant. This is because $$\arccos(x)+\arcsin(x)=\pi/2.$$

Answer 2

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

Answer 3

Indefinite integrals $I=\int f(x) dx$ are always uncertain by an additive constant. This constant is called constant of integration. Consequently, If an integral $I(x)$ gives two expressions by two different methods as $I_1(x)$ and $I_2(x)$, then $I_1(x)-I_2(x)=$ constant (independent of $x$). For example $$I(x)=\int \sin x \cos x dx = \int \frac{\sin 2x}{2} dx= -\frac{\cos 2x}{4}+C_1=I_1(x).$$ Also by parts $$I(x)=\sin x \sin x-\int \cos x \sin x \Rightarrow I(x)=\frac{sin^2 x}{2}+C_2 = I_2(x).$$ Next by substitution $$I(x)=\int- d(\cos x) \cos x =-~\frac{\cos^2 x}{2} =I_3(x)+C_3.$$

Finally, $$I_1(x)-I_2(x)=-1/4+C_1-C_2 ~ =\mbox{constant}.$$ Here, in the question, $I_1(x)=-\arccos(x)$ and $I_2(x)=\arcsin(x)$ So $$I_1(x)-I_2(x)=-\arccos(x)-\arcsin(x)=-\pi/2 = \mbox{constant}.$$

Answer 4

The remedy is to add a constant to either arcsine or arccosine and note that both answers are the same. The integrand as you have it is negative so arccosine is my choice.