Improper integrals, one-sided limits and trigonometric functions

Question

Evaluate $$\int \limits _{-1}^1\frac{\mathrm d x}{\sqrt{1-x^2}}$$

In my attempt I have

$$\int \limits _{-1}^0\frac{\mathrm d x}{\sqrt{1-x^2}}+\int \limits _0^1\frac{\mathrm d x}{\sqrt{1-x^2}}=\lim\limits _{a\rightarrow -1^+}-2\sin ^{-1}a +\lim\limits _{b\rightarrow 1^-}2\sin ^{-1}b=\pi +\pi$$

This solution is the same as the one in the textbook I am using.

In the first term, I have $\pi$ when I take $\lim\limits _{a\rightarrow -1^+}-2\sin ^{-1}a=-2\frac{-\pi}{2}$

I don't understand why $\lim\limits _{a\rightarrow -1^+}-2\sin ^{-1}a=-2\frac{3\pi}{2}$ is wrong since $\sin(\frac{3\pi}{2})=-1$

I think that it has something to do with the one sided limit but I really can't figure it out.

Answer 1

Note that the arcsine function is continuous on $[-\pi/2,\pi/2]$ and it is odd, i.e. $f(-x)=-f(x)$. Therefore $$\lim_{x\to -1^+}\arcsin(x)=\arcsin(-1)=-\arcsin(1)=-\frac{\pi}{2}.$$ Hence, since $D(\arcsin(x))=\frac{1}{\sqrt{1-x^2}}$ (there is an extra factor $2$ in your attempt), it follows that $$\int \limits _{-1}^1\frac{1}{\sqrt{1-x^2}}= \lim_{x\rightarrow 1^-}\arcsin(x)-\lim_{x\rightarrow -1^+}\arcsin(x)=\arcsin(1)-\arcsin(-1)=\frac{\pi}{2}+\frac{\pi}{2}=\pi.$$

Answer 2

The derivative of $\arcsin$ on the interior of its domain $[-1,1]$ is $x\mapsto \frac1{\sqrt{1-x^2}}$. From here it follows that $$\arcsin x = \int_0^x \frac{dt}{\sqrt{1-t^2}}, \quad \forall x \in \langle -1,1\rangle$$

and therefore by taking the limit it follows $$\frac{\pi}2 = \arcsin 1 = \lim_{x\to 1^-} \arcsin x = \lim_{x\to 1^-}\int_0^x \frac{dt}{\sqrt{1-t^2}} = \int_0^1 \frac{dt}{\sqrt{1-t^2}}$$ and similarly $$-\frac\pi2 = \arcsin(-1) = \int_0^{-1} \frac{dt}{\sqrt{1-t^2}}$$ Now we conclude $$\int_{-1}^1\frac{dt}{\sqrt{1-t^2}} = \int_0^1 \frac{dt}{\sqrt{1-t^2}} - \int_0^{-1} \frac{dt}{\sqrt{1-t^2}} = \frac{\pi}2 - \left(-\frac{\pi}2\right) = \pi$$