# Integrating 1/[1+sq.root(tanx)]

1. Nov 7, 2012

### justwild

1. The problem statement, all variables and given/known data

How to integrate ∫1/(1+√tanx) dx?

2. The attempt at a solution

Tried to substitute √tanx with z, but ultimately getting messed up with large expression

2. Nov 7, 2012

### tiny-tim

hi justwild!
if it's the correct large expression, what's the difficulty?

show us what you got

3. Nov 8, 2012

### justwild

we have,
∫1/[1+√tanx] dx ----------1

let tanx=z$^{2}$

hence by applying derivative on both sides,

sec$^{2}$x dx = 2zdz
$\Rightarrow$ 1+tan$^{2}$x dx=2zdz
$\Rightarrow$ 1+z$^{4}$ dx=2zdz
therefore, dx = [2z/(1+z$^{4}$)]dz

Substituting the above value in expression 1 we have,
∫2z/[{1+z$^{4}$}{1+z}] dz
This is what I got and I could not go on from here

4. Nov 8, 2012

### tiny-tim

hi justwild!

yes, that looks right

now use partial fractions

(factoring (1 + z4) = (1 + iz2)(1 - iz2) = (1 + eiπ/4)timesetc )

5. Nov 8, 2012

### justwild

That is OK but I can't use complex numbers here...

6. Nov 8, 2012

### SammyS

Staff Emeritus
z4 + 1 can be factored into two quadratics.

$z^4+1=(z^2+\sqrt{2}\,x+1)(z^2-\sqrt{2}\,x+1)$

Then try partial fractions.

The overall result of the integration promises to be pretty complicated.

7. Nov 8, 2012

### Ray Vickson

The answer will not be complex; the complex numbers come in conjugate pairs, and when the final answer is obtained the imaginary parts go away. This often happens: we end up with some real solution by going out into the complex plane, doing some manipulations there, and finally arriving back on the real line at the end.

RGV

8. Nov 9, 2012

### tiny-tim

a good hint is to complete the square

z4 + 1 = (z4 + 2z2 + 1) - 2z2

works for any symmetric quartic:

z4 + 2az3 + bz2 + 2az + 1

= (z2 + az + 1)2 - (a2 + 2 - b)z2

= (z2 + (a + √(a2 + 2 - b)z + 1)(z2 + (a - √(a2 + 2 - b)z + 1)​