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Integration: inverse trigonometric functions

  • Thread starter Maddie1609
  • Start date
81
11
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

Screenshot_2015-09-13-14-47-36.png


Edit: I meant to write u where t2 is
 
Last edited:

Titan97

Gold Member
451
18
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 $$\frac{1}{2}\int\frac{1+t^2+1-t^2}{\sqrt{(1+t^2)(1-t^2)}}dt$$
 
81
11
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

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My guess is that that integral does not have a solution in terms of the usual functions.
 
81
11
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 :-/
 
81
11
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 $$\frac{1}{2}\int\frac{1+t^2+1-t^2}{\sqrt{(1+t^2)(1-t^2)}}dt$$
Screenshot_2015-09-13-15-19-06.png

What can I do next?
 

stevendaryl

Staff Emeritus
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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:

[itex]\int \dfrac{t dt}{\sqrt{1-t^4}} = \frac{1}{2} arcsin(t^2)[/itex]
 
81
11
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:

[itex]\int \dfrac{t dt}{\sqrt{1-t^4}} = \frac{1}{2} arcsin(t^2)[/itex]
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?o0)
Edit: I just saw I wrote 1 instead of t, so sorry!
 

Titan97

Gold Member
451
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Use LaTeX
 

stevendaryl

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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?o0)
Edit: I just saw I wrote 1 instead of t, so sorry!
To prove that, you make the substitution:

[itex]t = \sqrt{u}[/itex]

then the integral in terms of [itex]u[/itex] should be one of the ones you listed in your first post.
 
81
11
To prove that, you make the substitution:

[itex]t = \sqrt{u}[/itex]

then the integral in terms of [itex]u[/itex] should be one of the ones you listed in your first post.
Think I got it, thanks :smile:

Screenshot_2015-09-13-15-36-37.png
 

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