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I'm trying to find formulas for hyperbolic functions, starting with this image Hyperbola

Knowing that the area between the origin, vertex and a point on hyperbola (enclosed by x-axis and hyperbola itself) is equal to $\frac{\alpha}{2}$, I wrote $$ x^2 - y^2 = 1 $$ $$ x = \sqrt{1 + y^2} $$ ... and started integrating this expression for $x$.

$$ \begin{align} \frac{\alpha}{2} & = \int^{y}_0{\sqrt{1 + y^2}\,\textrm{d}y} \\ & = \int^{\arctan y}_0{\sqrt{1 + \tan^2\theta}\sec^2\theta\,\textrm{d}\theta} && \text{let }y = \tan\theta \text{ and }\textrm{d}y = \sec^2\theta\,\textrm{d}\theta \\ & = \int^{\arctan y}_0{\sec\theta\sec^2\theta\,\textrm{d}\theta} \end{align} $$ Here I used integration by parts to find $\int{sec^3\theta\,\textrm{d}\theta}$ $$ \begin{align} \int{\sec^3\theta\,\textrm{d}\theta} & = \sec\theta\tan\theta - \int{\sec\theta\tan^2\theta \,\textrm{d}\theta} \\ & = \sec\theta\tan\theta - \int{\sec\theta(\sec^2\theta - 1) \,\textrm{d}\theta} \\ & = \sec\theta\tan\theta - \int{\sec^3\theta \,\textrm{d}\theta} + \int{\sec\theta \,\textrm{d}\theta} \\ & = \frac{1}{2}\bigg(\sec\theta\tan\theta + \int{\sec\theta \,\textrm{d}\theta}\bigg) \\ & = \frac{1}{2}\bigg(\sec\theta\tan\theta + \ln{\bigg|\sec\theta + \tan\theta \,\bigg|}\bigg) \end{align} $$

... and by substituting $\tan\theta = y$, $\sec\theta = \sqrt{1 + y^2}$ back I got $$ \alpha = y\sqrt{1 + y^2} + \ln{\bigg|\sqrt{1 + y^2} + y \,\bigg|} $$

Which doesn't make much sense, because this is not what an inverse hyperbolic sine looks like. Where have I made mistake? What should I do different?

Thanks.

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    $\begingroup$ Remember you are effectively taking a definte integral approach, so include limits and note that $\sinh^{-1}(x)=\ln(x+\sqrt{1+x^{2}})$ $\endgroup$ – Autolatry Aug 5 '15 at 9:20
  • $\begingroup$ @Autolatry The last expression is the one I got after evaluating the definite integral (maybe I did this one wrong). I noticed that I do have $\sinh^{-1}\,y$ there, but there's also that additional part which probably shouldn't be there. $\endgroup$ – user35443 Aug 5 '15 at 9:32
  • $\begingroup$ FYI: You can get a much easier integral to work with if you rotate the figure by $45^\circ$, so that the bounding hyperbola is given by $xy=1/2$. See my answer here. $\endgroup$ – Blue Aug 5 '15 at 9:39
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The integral you have calculated $\int_0^y\sqrt{1+y^2}dy$ does not represent the red area alone, but includes the area of the right angled triangle immediately above it. Therefore you need to subtract $\frac 12 xy$ and then you will get the answer you expected

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