1

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.

Attached images:

enter image description here enter image description here enter image description here

Curious
  • 541

1 Answers1

1

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 $$ -\sin^{-1}\bigg(\frac{1}{cr^2}\bigg)=\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)-\frac{\pi}{2}=\tan^{-1}\bigg(\sqrt{c^2r^4-1}\bigg)-\frac{\pi}{2}. $$

xpaul
  • 44,000
  • There is only one thing I don't understand in your solution. It consists of the last line of your final equation, that is, why do you get that $\cos^{-1}(\frac{1}{cr^2}) = \tan^{-1}(\sqrt{c^2 r^4 -1})$? I need an explanation please. – Curious Nov 30 '20 at 01:13
  • Taking the inverse tangent, you will get it. See the update. – xpaul Nov 30 '20 at 01:23
  • $\frac{1}{y}=cos(x)$, so $\frac{\sqrt{y^2-1}}{y}=sin(x)$ and $\sqrt{y^2-1}=tan(x)$.. – herb steinberg Nov 30 '20 at 01:37