Integration of Sqrt Tangent

  • Thread starter amcavoy
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How would you compute something like:

[tex]\int \sqrt{\tan \theta}\space d\theta[/tex]

I cannot seem to be able to use any trig. identities.

Thanks.
 
1,062
6
Try [tex]u = tan\theta [/tex]

Then [tex]u = x^2[/tex]

One more hint: [tex](x^4+1)= (x^2 - \sqrt{2}x+1)(x^2 + \sqrt{2}x + 1)[/tex]
 
788
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There was a very long thread about this integral already, click here
 
663
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mattmns said:
Try [tex]u = tan\theta [/tex]

Then [tex]u = x^2[/tex]

One more hint: [tex](x^4+1)= (x^2 - \sqrt{2}x+1)(x^2 + \sqrt{2}x + 1)[/tex]
Yeah, but then the new integral has two different variables.

With [tex]u=\tan{\theta}[/tex], I come up with:

[tex]\int \sqrt{u}\cos^2{\theta}du[/tex]
 
Last edited:

dextercioby

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[tex] u^{2}=\tan\theta [/tex]

is the right one to do.

Daniel.
 
1,062
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cos^2 = 1 / sec^2 right, and sec^2 = 1 + tan^2 right

so the integral is

[tex]\int \frac{\sqrt{u}}{1 + u^2}du[/tex]

then used the hint above.

edit... looking at the other thread, you should probably do dex's substitution.
 
Last edited:
663
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mattmns said:
cos^2 = 1 / sec^2 right, and sec^2 = 1 + tan^2 right

so the integral is

[tex]\int \frac{\sqrt{u}}{1 + u^4}du[/tex]

then used the hint above.

edit... looking at the other thread, you should probably do dex's substitution.
Alright, that makes sense. The only problem I see is this:

Shouldn't the integral be [tex]\int \frac{\sqrt{u}}{1+u^2}du[/tex]?
 
1,062
6
Yes you are right, sorry.
 
663
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Alright, well I can integrate that. Thanks a lot for your help.
 

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