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I would like to evaluate

$\int_{-1/2}^{1/2}$ $1 \over \sqrt{1 - x^2}$ $dx$

Approach

We substitute $x = \sin u$ $\Rightarrow$ $dx \over du$ $=$ $\cos u \Rightarrow dx$ $=$ $\cos u$ $du.$

This leads to

$\int_{\sin(-1/2)}^{\sin(1/2)} 1$ $du$ = $[u]_{\sin(-1/2)}^{\sin(1/2)}$ = $[\arcsin(x)]_{-1/2}^{1/2}$

Now, my calculator tells me that the result would be $1$, but this doesn't work with my solution that I got so far. Where am I mistaken?

Community
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Julian
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    ...and if further evidence is needed: https://www.wolframalpha.com/input/?i=integrate+1%2F(%5Csqrt(1-x%5E2))+between+-1%2F2+and+1%2F2... – user1729 Jul 05 '16 at 10:07
  • $\pi$ is indeed away from $3$ by $0.141592654...$ but $\frac{\pi}{3}$ is away from $1$ by $0.04...$. Does your calulator understand something about $fixing$ digits shown? – Math-fun Jul 05 '16 at 12:29

2 Answers2

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Your calculation is correct. The final result should be :

$$\arcsin(1/2) - \arcsin(-1/2) = \frac{\pi}{6} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{3} = 60^{\circ}$$

Zain Patel
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Zubzub
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  • Well, then my calculator was wrong. It looks like one can't rely one them when the integral becomes a little bit harder. :D Thank you! – Julian Jul 05 '16 at 10:00
  • \arcsin looks like $\arcsin$ in math form :-) – Math-fun Jul 05 '16 at 10:01
  • Although I agree with your calculation, I must ask why you have 60${}^\circ$ in the final step. What does the integral have to do with that angle? – mickep Jul 05 '16 at 11:09
  • @mickep Yes you're right. I added the value in degree because I suspected the OP had some problems with the angle unit on his calculator. – Zubzub Jul 05 '16 at 11:36
  • No, I simply chose the wrong integration limits, it was my mistake. Now the calculator gives me indeed $\pi \over 3$. :-) – Julian Jul 05 '16 at 12:31
  • Indeed the result of the integral is a real number, not an angle, hence the answer $\frac\pi3$ is correct while the answer $60°$ is not. – Did Jul 05 '16 at 12:38
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What is the purpose of the calculator here? $$ \int_{-1/2}^{1/2}\frac{dx}{\sqrt{1-x^2}}=2\int_{0}^{1/2}\frac{dx}{\sqrt{1-x^2}}=2\arcsin\frac{1}{2} = 2\cdot\frac{\pi}{6} = \frac{\pi}{3}.$$ Anyway, I doubt you calculator uses the approximation $\pi\approx 3$!

Jack D'Aurizio
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