# Integration: inverse trigonometric functions

1. Sep 13, 2015

1. The problem statement, all variables and given/known data
∫(t/√(1-t4))dt

2. Relevant 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)

3. The attempt at a solution

Edit: I meant to write u where t2 is

Last edited: Sep 13, 2015
2. Sep 13, 2015

### 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 $$\frac{1}{2}\int\frac{1+t^2+1-t^2}{\sqrt{(1+t^2)(1-t^2)}}dt$$

3. Sep 13, 2015

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

4. Sep 13, 2015

### stevendaryl

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

5. Sep 13, 2015

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 :-/

6. Sep 13, 2015

What can I do next?

7. Sep 13, 2015

### stevendaryl

Staff Emeritus
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:

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

8. Sep 13, 2015

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!

9. Sep 13, 2015

### Titan97

Use LaTeX

10. Sep 13, 2015

### stevendaryl

Staff Emeritus
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.

11. Sep 13, 2015

Think I got it, thanks

12. Sep 13, 2015