Computing the Integral of Square Root Tangent without Trig Identities

[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

 

[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.