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

[justwild]

Homework Statement

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

 

[tiny-tim]

hi justwild!

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

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

show us what you got ​

 

[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

 

[tiny-tim]

hi justwild!

yes, that looks right

now use partial fractions

(factoring (1 + z⁴) = (1 + iz²)(1 - iz²) = (1 + e^iπ/4)timesetc )

 

[justwild]

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

 

[SammyS]

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

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.

 

[Ray Vickson]

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

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

 

[tiny-tim]

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

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)​