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The limits of integration are $[0,1]$ and thus the result should be $\pi$. My book suggests an elegant way to evaluate the integral by Monte Carlo Integration but I was wondering, can we reach the same result with an easier way? Thank you.

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    $\begingroup$ Why don't you try something as x = Sin[y] ? $\endgroup$ Nov 26, 2013 at 11:57
  • $\begingroup$ @ClaudeLeibovici Brilliant! Thank you. $\endgroup$
    – JohnK
    Nov 26, 2013 at 12:06

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Hints:

$$\sin u:= x\implies du\cos u=dx\implies$$

$$\int\sqrt{1-x^2}\,dx=\int \cos^2u\,du=\frac{u+\sin u\cos u}2\;\ldots$$

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    $\begingroup$ That works! Thank you. $\endgroup$
    – JohnK
    Nov 26, 2013 at 12:02

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