Integration: inverse trigonometric functions

[Maddie1609]

Homework Statement

∫(t/√(1-t⁴))dt

Homework Equations

∫(du/√(a² - u²)) = arcsin (u/a) + C

∫(du/(a² + u²) = (1/a) arctan (u/a)

∫(du/(u√(u² - a²))) = (1/a) arcsec (|u|/a)

The Attempt at a Solution

Edit: I meant to write u where t² is[/B]

 

[Titan97]

That is not a proper way to write an integral. You should express t in terms of u.
$dt=\frac{dt}{2t}=\frac{dt}{2\sqrt{u}}$.
But this does not make the integral easy.
Instead, have you tried expressing it as $$\frac{1}{2}\int\frac{1+t^2+1-t^2}{\sqrt{(1+t^2)(1-t^2)}}dt$$

 

[Maddie1609]

That is not a proper way to write an integral. You should express t in terms of u.
$dt=\frac{dt}{2t}=\frac{dt}{2\sqrt{u}}$.
But this does not make the integral easy.
Instead, have you tried expressing it as $$\frac{1}{2}\int\frac{1+t^2+1-t^2}{\sqrt{(1+t^2)(1-t^2)}}dt$$

I edited under that I meant to replace t² with u :) No, but I wouldn't know what to do with it anyway?

 

[stevendaryl]

My guess is that that integral does not have a solution in terms of the usual functions.

 

[Maddie1609]

My guess is that that integral does not have a solution in terms of the usual functions.

In the back of my book it says the answer is (1/2) arcsin (t²) + C, but I don't know how to get there :-/

 

[Maddie1609]

That is not a proper way to write an integral. You should express t in terms of u.
$dt=\frac{dt}{2t}=\frac{dt}{2\sqrt{u}}$.
But this does not make the integral easy.
Instead, have you tried expressing it as $$\frac{1}{2}\int\frac{1+t^2+1-t^2}{\sqrt{(1+t^2)(1-t^2)}}dt$$

What can I do next?

 

[stevendaryl]

In the back of my book it says the answer is (1/2) arcsin (t²) + C, but I don't know how to get there :-/

Hmm. Is it possible that there is a typo in your original integral? I get that answer if the original integral was as shown on the left-hand side below:

$\int \dfrac{t dt}{\sqrt{1-t^4}} = \frac{1}{2} arcsin(t^2)$

 

[Maddie1609]

Hmm. Is it possible that there is a typo in your original integral? I get that answer if the original integral was as shown on the left-hand side below:

$\int \dfrac{t dt}{\sqrt{1-t^4}} = \frac{1}{2} arcsin(t^2)$

Yes that is the integral, did I not write it correctly in my original post? How do you get your equations to look like that?
Edit: I just saw I wrote 1 instead of t, so sorry!

 

[Titan97]

Use LaTeX

 

[stevendaryl]

Yes that is the integral, did I not write it correctly in my original post? How do you get your equations to look like that?
Edit: I just saw I wrote 1 instead of t, so sorry!

To prove that, you make the substitution:

$t = \sqrt{u}$

then the integral in terms of $u$ should be one of the ones you listed in your first post.

 

[Maddie1609]

To prove that, you make the substitution:

$t = \sqrt{u}$

then the integral in terms of $u$ should be one of the ones you listed in your first post.

Think I got it, thanks

 

[Mark44]

How do you get your equations to look like that?

Use LaTeX

We have a tutorial on LaTeX: https://www.physicsforums.com/help/latexhelp/