Integral, from 0 to 1, of dx/root(1-x^2)

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Homework Statement


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Integral, from 0 to 1, of dx/root(1-x^2)

Homework Equations


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d/dx of arcsin = 1/root(1-x^2)

The Attempt at a Solution



Since d/dx of arcsin = 1/root(1-x^2), we have that the integral, from 0 to 1, of dx/root(1-x^2) equals to arcsin, from 0 to 1.

arcsin(1) - arcsin(0) = arcsin(1). I know I'm missing something here. What did I do wrong?
 
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Nothing is wrong.
 
leo255 said:
Since d/dx of arcsin = 1/root(1-x^2), we have that the integral, from 0 to 1, of dx/root(1-x^2) equals to arcsin, from 0 to 1.

arcsin(1) - arcsin(0) = arcsin(1). I know I'm missing something here. What did I do wrong?

What is arcsin(1)?
 
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arcsin(1) is pi/2. I asked someone in my class about this, and he said that I should be taking the limit, as b (or whatever other variable) approaches 1, from the left-hand side. Can you guys confirm if this is something that should be done for this problem?
 
leo255 said:
arcsin(1) is pi/2. I asked someone in my class about this, and he said that I should be taking the limit, as b (or whatever other variable) approaches 1, from the left-hand side. Can you guys confirm if this is something that should be done for this problem?
Yes, this should be done. The integrand is undefined at x = 1, so the Fund. Thm. of Calculus doesn't apply. You can get around this by evaluating this limit:
$$\lim_{b \to 1^-}\int_0^b \frac{dx}{\sqrt{1 - x^2}}$$
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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