# Integration: inverse trigonometric functions

1. Homework Statement
∫(t/√(1-t4))dt

2. 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)

3. The Attempt at a Solution

Edit: I meant to write u where t2 is

Last edited:
Related Calculus and Beyond Homework Help News on Phys.org

#### Titan97

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

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

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

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 :-/

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

Staff Emeritus
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:

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

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!

Gold Member
Use LaTeX

#### stevendaryl

Staff Emeritus
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.

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

"Integration: inverse trigonometric functions"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving