# Integral of (tan x)^(1/2)

1. Sep 20, 2007

### zachnorious

Can please anyone show me how to solve the

integral of (tan x)^(1/2) ? (integral of the square root of tangent)

Panos

I can't do anything else than the obvious tan=sin/cos. I really can't figure out which differentiation is giving you the sin^(1/2) * cos^(-1/2) :(

Last edited: Sep 21, 2007
2. Sep 20, 2007

### Dick

3. Sep 21, 2007

### Gib Z

Don't give up if the work gets a little complex btw. Personally I tend to think it's a sad sign when a simple looking integral gets really ugly, but this one gets REALLY ugly.

4. Sep 21, 2007

### chaoseverlasting

It doesnt get ugly. Just keep at it. It just a bit long is all.

5. Sep 21, 2007

### CompuChip

And you probably need a little identity about cos(arctan(z)).

6. Sep 21, 2007

### Gib Z

"a bit long" is somewhat an understatement =P

Don't click on this link if you want the integral done by yourself!!:

http://mcraefamily.com/MathHelp/CalculusIntegralTableOfIntegralsSqrtTan.htm [Broken]

Last edited by a moderator: May 3, 2017
7. Sep 22, 2007

### zachnorious

Thank u all very much. :)
With ur suggestions I got it until
$$2\int(\frac{u^2}{1+u^4}du$$

but then, stuck :p

I'm goin to see the solution now, thnx Gib for the link

8. Dec 7, 2009

### murshid_islam

I got
$$2\int\frac{u^2}{1+u^4}du = 2\int\frac{u^2}{\left(u^2 + \sqrt{2}u + 1\right)\left(u^2 - \sqrt{2}u + 1\right)}du$$

the partial fraction expansion doesn't look good. :(

9. Dec 7, 2009

### Redbelly98

Staff Emeritus
Mod's note:
Misleading nonsense has been deleted. PF regrets the confusion this may have caused.

murshid_islam, I'll suggest starting a new thread as your best chance of getting help, since this thread was from 2 years ago. Again, we regret the misleading, confusing posts that were made here.

10. Dec 10, 2009

Thanks.