# Computing an improper integral

1. Jun 17, 2010

### stripes

1. The problem statement, all variables and given/known data

$$\int^{1}_{0}\frac{dx}{\sqrt{1-x^{2}}}$$

2. Relevant equations

None

3. The attempt at a solution

$$\int^{1}_{0}\frac{dx}{\sqrt{1-x^{2}}} = sin^{-1}x\right|^{1}_{0}$$

$$sin^{-1}x\right|^{1}_{0} = \frac{\pi}{2} - 0$$

so the final answer is just pi/2. I have no problem computing the answer, but it's in the improper integrals section of the textbook....but I don't see this as being an improper integral. There's no need to deal with infinity at all, no asymptotes, no discontinuity on the closed interval [0,1].

Am I missing something?

Last edited: Jun 17, 2010
2. Jun 17, 2010

### D H

Staff Emeritus
What happens to $$1/\sqrt{1-x^2}$$ as $x\to1\quad$?

3. Jun 17, 2010

### stripes

Good call. Fixed it and showed a few extra steps, namely:

$$\int^{1}_{0}\frac{dx}{\sqrt{1-x^{2}}} = lim_{t\rightarrow1^{-}}\int^{t}_{0}\frac{dx}{\sqrt{1-x^{2}}}$$

then just computed the integral which is arcsin x, and the limit as t approaches 1 from the left for arcsin t is just pi/2.

Yay! Thanks!