# Improper integrals, one-sided limits and trigonometric functions

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.

• $\arcsin$ is the inverse function of $\sin$ on the interval $[-\pi/2,\pi/2]$. So $\arcsin(-1)=-\pi/2$ and not $3\pi/2$. – TheSilverDoe Mar 8 '19 at 10:16
• Thank you TheSilverDoe – mamotebang Mar 8 '19 at 11:39

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 $$\int \limits _{-1}^1\frac{1}{\sqrt{1-x^2}}= \lim_{x\rightarrow 1^-}\arcsin(x)-\lim_{x\rightarrow -1^+}\arcsin(x)=\arcsin(1)-\arcsin(-1)=\frac{\pi}{2}+\frac{\pi}{2}=\pi.$$
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 $$\int_{-1}^1\frac{dt}{\sqrt{1-t^2}} = \int_0^1 \frac{dt}{\sqrt{1-t^2}} - \int_0^{-1} \frac{dt}{\sqrt{1-t^2}} = \frac{\pi}2 - \left(-\frac{\pi}2\right) = \pi$$