- #1

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

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

Thanks.

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- Thread starter amcavoy
- Start date

- #1

- 665

- 0

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

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

Thanks.

- #2

- 1,085

- 6

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]

- #3

- 789

- 4

There was a very long thread about this integral already, click here

- #4

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mattmns said:

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:

- #5

- 13,070

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

is the right one to do.

Daniel.

is the right one to do.

Daniel.

- #6

- 1,085

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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.

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:

- #7

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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]?

- #8

- 1,085

- 6

Yes you are right, sorry.

- #9

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

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