Integration of Sqrt Tangent

[amcavoy]

How would you compute something like:

\int \sqrt{\tan \theta}\space d\theta

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

Thanks.

[mattmns]

Try u = tan\theta

Then u = x^2

One more hint: (x^4+1)= (x^2 - \sqrt{2}x+1)(x^2 + \sqrt{2}x + 1)

[Jameson]

There was a very long thread about this integral already, click here (http://www.physicsforums.com/showthread.php?t=77439)

[amcavoy]

Try u = tan\theta

Then u = x^2

One more hint: (x^4+1)= (x^2 - \sqrt{2}x+1)(x^2 + \sqrt{2}x + 1)

Yeah, but then the new integral has two different variables.

With u=\tan{\theta}, I come up with:

\int \sqrt{u}\cos^2{\theta}du

[dextercioby]

u^{2}=\tan\theta

is the right one to do.

Daniel.

[mattmns]

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

so the integral is

\int \frac{\sqrt{u}}{1 + u^2}du

then used the hint above.

edit... looking at the other thread, you should probably do dex's substitution.

[amcavoy]

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

so the integral is

\int \frac{\sqrt{u}}{1 + u^4}du

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 \int \frac{\sqrt{u}}{1+u^2}du?

[mattmns]

Yes you are right, sorry.

[amcavoy]

Alright, well I can integrate that. Thanks a lot for your help.