<s>[INST] Why $\int{\frac{dr}{r\sqrt{c^{2}r^4 -1}}} = -\frac{1}{2} \arcsin\left(\frac{1}{cr^2}\right)$? I must solve the following integral, where $c$ is a constant $$\int{\frac{dr}{r\sqrt{c^{2}r^4 -1}}}$$ When I compute and do the calculations, I obtain that $$\int{\frac{dr}{r\sqrt{c^{2}r^4 -1}}} = \frac{1}{2} \arctan \left(\sqrt{c^{2}r^4 -1} \right)$$ I have rectified my calculations many times and I always get the same result, but in the book I'm studying I have that $$\int{\frac{dr}{r\sqrt{c^{2}r^4 -1}}} = -\frac{1}{2} \arcsin\left(\frac{1}{cr^2}\right)$$  Why? Is it some mistake in the book or is it my own mistake? The book I mention is: The calculus of variations by Van Brunt page 47, there the equation arises through an example on the invariance of the Eule-Lagrange equation.
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 [/INST] In fact
$$ \tan\bigg(\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\bigg)=\frac{\sin\bigg(\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\bigg)}{\cos\bigg(\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\bigg)}=\frac{\sqrt{1-\frac1{c^2r^4}}}{\frac{1}{cr^2}}=\sqrt{c^2r^4-1}. $$
Since $\frac{1}{cr^2}>0$, $\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\in(0,\pi/2)$ and hence
$$ \cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\bigg)=\tan^{-1}\sqrt{c^2r^4-1}. $$
Using
$$ \sin^{-1} x+\cos^{-1} x=\frac{\pi}{2} $$
and hence
$$ -\sin^{-1}x=\cos^{-1}x-\frac{\pi}{2}. $$
one has
$$ -\sin^{-1}\bigg(\frac{1}{cr^2}\bigg)=\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)-\frac{\pi}{2}=\tan^{-1}\bigg(\sqrt{c^2r^4-1}\bigg)-\frac{\pi}{2}. $$
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