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It is not difficult to verify that $$ \frac{\mathrm d}{\mathrm dx} \left[ \log\Big(x+\sqrt{x^2+1}\Big) \right] = \frac{1}{\sqrt{1+x^2}} $$ for $x\geq 0$, say.

How would one calculate the indefinite integral $$ \int \frac{1}{\sqrt{1+x^2}} \ \!\mathrm dx$$ without knowing this? I have tried many of the usual tricks, without success.

The title of the question is chosen because Mathematica outputs $\text{Arcsinh}(x)+C$ as the answer.

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You can do a substitution for $x=\tan\theta$ and $\mathrm dx=\sec^2\theta\,\mathrm d\theta$ to get

$$ \int\frac{\sec^2\theta}{\sqrt{1+\tan^2\theta}}\,\mathrm d\theta. $$

Then use that $1+\tan^2\theta=\sec^2\theta$ to get the integral

$$ \int\sec\theta\,\mathrm d\theta. $$

EDIT: In response to the comment by Sasha: This is one of the standard integrals in a calculus class. I usually just derive it in class and have students memorize it along with the other trigonometric functions. Just multiply top and bottom by $\sec\theta + \tan\theta$ to get

$$ \int \frac{\sec^2\theta + \tan\theta\sec\theta}{\sec\theta+\tan\theta}\mathrm d\theta. $$

Then $u=\sec\theta + \tan\theta$, $\mathrm du=\sec\theta\tan\theta + \sec^2\theta\;\mathrm d\theta.$ The answer is then $\ln|\sec\theta+\tan\theta|+C$. Putting this in terms of $x$ one gets $\ln|\sqrt{1+x^2}+x|+C$.

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One would use Euler's substitution. Also on PlanetMath.

Alternatively, one could use $x = \sinh(t)$, because $1+x^2 = 1+\sinh^2(t) = \cosh^2(t)$, and because $\mathrm{d}x = \sinh^\prime(t) \mathrm{d}t = \cosh(t) \mathrm{d}t$. Therefore, using that $\cosh(t) >0 $ for real $t$:

$$ \int \frac{\mathrm{d} x}{\sqrt{1+x^2}} = \int \frac{\cosh(t)}{\sqrt{\cosh^2(t)}} \mathrm{d} t = \int \mathrm{d} t = t + C $$ Substituting back $\int \frac{\mathrm{d} x}{\sqrt{1+x^2}} = \operatorname{arcsinh}(x) + C$

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  • $\begingroup$ Your answer gives arcsinh($x$), not the function involving the logarithm. I guess after you'd have to further solve $e^z-e^{-z} = 2w$ which ends up being a quadratic in $e^z$ to get the answer I want. $\endgroup$ – Jim Oct 7 '11 at 21:07
  • $\begingroup$ @Jum The answer you want is provided by Euler's substitution. I was explaining how Mathematica arrived at arcsine hyperbolic. $\endgroup$ – Sasha Oct 7 '11 at 21:35

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