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#### find_the_fun

##### Active member

- Feb 1, 2012

- 166

rewriting the equation gives \(\displaystyle \frac{1}{\sqrt{1-x^2}}dx = \frac{1}{\sqrt{1-y^2}}dy\)

Isn't this the integral for \(\displaystyle \sin^{-1}(x)\) & \(\displaystyle \sin^{-1}(y)\)? The back of book has \(\displaystyle y=1/2x+\frac{\sqrt{3}}{2}\sqrt{1-x^2}\)