Integration of √(tanx)

[the_jackal]

Can anyone solve this function:
y = √(tan(x))

I have tried this using Matlab but i am not pleased with the solution i got.

Thanks

 

[kennyg]

y=√tanx
y²=tanx
2ydy=(sec²x)dx=(1+tan²x)dx=(1+y^4)dx
dx=2ydy/1+y^4,
so∫√tanx•dx=∫y•2ydy/1+y^4=∫(2y²/1+y^4)dy
=∫[(y²+1)/1+y^4]dy+∫[(y²-1)/1+y^4]dy
=∫[(1+y½)/(y½+y²)]dy+∫[(1-y½)/(y½+y²)]dy
= A + B

A:
Let t=y-1/y,dt=(1+y½)dy
and y½+y²=t²+2
A=∫1/(t²+2)dt=1/√2tan-1(t/√2)=1/√2tan-1[(y-1/y)/√2]

B:
Let g=y+1/y,dg=(1-y½)dy
and y½+y²=g²-2
B=∫1/(g²-2)dg=???
I forget
Consult a calculus integration table,
you will find the formulus,then
"A+B+ a constant C" is the solution!

 

[elnenez]

Re: Integration of √(tanx)

\[\frac{-2 tan ^{-1}\left(1-\sqrt{2} \tan ^{\frac{1}{2}}(x)\right)+2 \tan ^{-1}\left(\sqrt{2} tan ^{\frac{1}{2}}(x)+1\right)}{2 \sqrt{2}}\]

\[+\frac{\log \left(-tan(x)+\sqrt{2} tan ^{\frac{1}{2}}(x)-1\right)-\log \left(tan (x)+\sqrt{2} \tan ^{\frac{1}{2}}(x)+1\right)}{2 \sqrt{2}}+Constant\]

 

[the_jackal]

Thanks kennyg and elnenez for the solution. I have used matlab to solve this and i am getting a totally different solution.