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I've stumble into this integral on a paper (page 5, right column)

$\displaystyle\int_{\tfrac {\cos(k)}{\sqrt{1+\cos^2(k)}} \leq~y}~ \dfrac{\mathrm{d}k}{4\pi}$

with solution

$\displaystyle{1-\frac{1}{\pi}\arccos \left(\frac{k}{\sqrt{1-k^2}}\right)}$.

What type of integral is this? How would you solve it? How do the authors of this paper solve this?

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    @Marcos: You are more than welcome.2011-06-30

1 Answers 1

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This is as easy an integral as could be imagined, since no integration is necessary (the integrand is constant). All we need to do is to solve the inequality $\frac{\cos k}{\sqrt{1+\cos^2 k}} \le y$ for $k$ in terms of $y$.

The result is, however, unfortunately marred by a bad typo. In the answer, the $k$'s should be $y$'s.

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    thank you very much, now is crystal clear!2011-06-30