Can someone help me solve the integral of (tan x)^(1/2)?

[zachnorious]

Can please anyone show me how to solve the

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

Thank you in advance,
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) :(

 

[Dick]

Start with substituting u^2=tan(x) and see where that leads you.

 

[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.

 

[chaoseverlasting]

It doesn't get ugly. Just keep at it. It just a bit long is all.

 

[CompuChip]

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

 

[Gib Z]

It doesn't get ugly. Just keep at it. It just a bit long is all.

"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

 

[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

 

[murshid_islam]

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

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. :(

 

[Redbelly98]

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.

 

[murshid_islam]

Mod's note:
Misleading nonsense has been deleted.

Thanks.

murshid_islam, I'll suggest starting a new thread as your best chance of getting help, since this thread was from 2 years ago.

Thanks again for the advice. I will start a new thread.