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sirinivas · Sep 28, 2010, 02:46 AM

∫dy/(1+y^2)^1/3

galactus · Sep 28, 2010, 03:36 AM

Are you sure that's not:

$\int\frac{1}{\sqrt{1+y^{2}}}dy=sinh^{-1}(y)$ ?

I ask because $\int\frac{1}{\sqrt[3]{1+x^{2}}}dy$ can not be integrated using the usual means.

ebaines · Sep 28, 2010, 08:38 AM

Assuming that Galactus is correct, and what you want to prove is
$ \int \frac {dy} {\sqrt{1+y^2}} = \sinh ^{-1} y $

Here's a general overview of the proof - you can fill in the details:

Start with the equation:
$y = \sinh(x) $

Take it's derivative, and then use the identity
$ \cosh^2x -sinh^2x =1 $
to convert the derivative to a function of $\sinh x$ . Replace the $\sinh x$ terms with y. Rearrange a couple of terms to get dx on one side and the y terms on the other, and integrate. This gives you x is equal to an integral of y. But $x = \sinh ^{-1} y$ ; substitute that, and you're done.

galactus · Sep 28, 2010, 02:29 PM

If you like, here is a long way to go about it.

$\int\frac{1}{\sqrt{1+x^{2}}}dx$

Let $x=tan(t), \;\ dx=sec^{2}(t)dt$

Make the subs and it becomes:

$\int \frac{sec^{2}(t)}{\sqrt{\underbrace{1+tan^{2}(t)}_ {\text{sec^2(t)}}}}dt$

$\int sec(t)dt$

This can be evaluated, but it is

$ln(sec(t)+tan(t))$

Resub $t=tan^{-1}(x)$

Then, it becomes:

$ln(x+\sqrt{1+x^{2}})$

This can be shown to be equal to $sinh^{-1}(x)$

$y=sinh^{-1}(x)$

$y=\frac{e^{y}-e^{-y}}{2}$

$e^{y}-2x-e^{-y}=0$

Multiply by $e^{y}$

$e^{2y}-2xe^{y}-1=0$

Using the quadratic formula gives:

$e^{y}=x\pm\sqrt{x^{2}+1}$

Since $e^{y}>0$ , disregrard the negative solution.

Take logs and we get:

$y=ln(x+\sqrt{x^{2}+1})=sinh^{-1}(x)$

ebaines · Sep 28, 2010, 03:00 PM

Originally Posted by galactus

If you like, here is a long way to go about it.
...

Wow - that is a different way of doing it. But I still like the short way:

$ y = \sinh(x) \\ \frac {dy} {dx} \ =\ \cosh(x)\ = \ \sqrt{1+\sinh^2x} \ = \ \sqrt{1+y^2} \\ \frac {dy} {\sqrt{1+y^2}} = dx \\ \int \frac {dy} {\sqrt{1+y^2}} = \int dx \ = \ x \ = \sinh^{-1} y $

galactus · Sep 28, 2010, 04:22 PM

Yes, me too.:)

galactus · Sep 29, 2010, 02:35 AM

Why don't you find it helpful? I just thought I would show you another way. It may not be the simplest, but it shows there is always more than one way to tackle a problem. Do you know what I find not helpful? Someone who can't post the correct problem, so we have to interpret. Don't worry, you will not find me helping you anymore.