2
$\begingroup$

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.

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

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 $$\int \limits _{-1}^1\frac{1}{\sqrt{1-x^2}}= \lim_{x\rightarrow 1^-}\arcsin(x)-\lim_{x\rightarrow -1^+}\arcsin(x)=\arcsin(1)-\arcsin(-1)=\frac{\pi}{2}+\frac{\pi}{2}=\pi.$$

$\endgroup$
1
$\begingroup$

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 $$\int_{-1}^1\frac{dt}{\sqrt{1-t^2}} = \int_0^1 \frac{dt}{\sqrt{1-t^2}} - \int_0^{-1} \frac{dt}{\sqrt{1-t^2}} = \frac{\pi}2 - \left(-\frac{\pi}2\right) = \pi$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.