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Why does this approach to integration not work? If there is an integral $1/\sqrt{a^2-x^2}$, the answer is $\arcsin(x/a)$. But if the integral is $1/\sqrt{x^2-a^2}$ then it is $\log(x+\sqrt{x^2-a^2})$.

My question is, why can't we take $i=\sqrt{-1}$ and integrate as in the first case to get the answer as $-i \arcsin(x/a)$?

Nathaniel Bubis
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Rohan Rao
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  • We can do it. There are some caveats - you need to know arcsin$(x/a)$ for $x>a$; one should really understand arcsin as a multivalued function of a complex variable to fully understand what's going on. But otherwise it's perfectly OK. – user8268 Aug 27 '13 at 06:55
  • Related : http://math.stackexchange.com/questions/268591/how-to-derive-inverse-hyperbolic-trigonometric-functions – lab bhattacharjee Aug 27 '13 at 06:56
  • Thanks @labbhattacharjee, this helps me a lot! But it still doesn't exactly answer my question...could someone please elaborate? And in case it helps you answer, I have completed upto 12th standard level of Calculus in India, so the answer needn't be over explained. Thanks! – Rohan Rao Aug 27 '13 at 07:16
  • @RohanRao, added a solution here, too – lab bhattacharjee Aug 27 '13 at 07:17

1 Answers1

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Let $x=a\cosh A,$

$\implies \cos (iA)=\frac xa\implies iA=\arccos\frac xa=\frac\pi2-\arcsin\frac xa$

$\implies\ln(x+\sqrt{x^2-a^2})$

$=\ln|a|+\ln(\cosh A+\sinh A)$

$=\ln|a|+\ln(e^A)$

$=\ln|a|+A$

$=\ln|a|+i(\frac\pi2-\arcsin\frac xa)$

$=-i\arcsin \frac xa+K$ where $K=\ln|a|+i\frac\pi2$ which is contant

lab bhattacharjee
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