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    integrate the square root of tan x?
    if so how?
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    Make the substitution y2 = tan x

    When you've done this, rearrange the numerator to be

    2y2 = 2(1+y2) - 2

    And go from there.
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    how do you go from there ?
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    y = √tanx

    ∫ (tanx)1/2.dx
    y² = tanx
    2y.dy/dx = sec²x
    dy/dx = (sec²x)/2y
    dy/dx = (sec²x)/(2√(tanx))
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    (Original post by Dekota)
    y = √tanx

    ∫ (tanx)1/2.dx
    y² = tanx
    2y.dy/dx = sec²x
    dy/dx = (sec²x)/2y
    dy/dx = (sec²x)/(2√(tanx))
    Elegant solution.
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    (Original post by Sumeet)
    Elegant solution.
    What solution? :p:
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    It is not solution yet. It's not that easy. This will leave you at the integration below:
    \int{\frac{2y^2}{y^4 +1}}dy
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    Since I had nothing better to do...

    Let I = ∫ sqrt[tan(x)] dx. Then using Cexy's sub, we transform the integral into:
    I = 2 ∫ u^2/(1+u^4)

    Some tricks are in order.
    I = 2 ∫ (u^2 + 1 - 1)/(1 + u^4) du
    = 2 ∫ (u^2 + 1)/(u^4 + 1) du - 2 ∫ 1/(1 + u^4) du
    = 2J - 2H

    Where:
    J = ∫ (u^2 + 1)/(u^4 + 1) du
    = ∫ (1 + 1/u^2)/(u^2 + 1/u^2) du

    Setting v=u-1/u with dv/dv=1+1/u^2, we get:
    J = ∫ dv/(v^2 + 2) = (1/sqrt2) arctan(v/sqrt2) + C

    And:
    H = ∫ 1/(1 + u^4) du

    Now note that (1 + u^4) = (1+u^2)^2 - 2u^2 = (1+u^2-sqrt(2)u)(1+u^2+sqrt(2)u).

    So you can split the integrand in H using partial fractions. Eventually, and I mean eventually, you will be able to integrate H. And hence you can find ∫sqrt[tan(x)] dx.

    Alternatively
    I = 2 ∫ u^2/(1+u^4) du
    could be split into something prettier using a dirty trick:
    I = ∫ [(u^2 + 1) + (u^2 - 1)]/(1+u^4) du
    = ∫ (u^2 + 1)/(u^4 + 1) du + ∫ (u^2 - 1)/(u^4 + 1) du
    = ∫ (1 + 1/u^2)/(u^2 + 1/u^2) du + ∫ (1 - 1/u^2)/(u^2 + 1/u^2) du
    = J + H*

    We already did J, and H* can be done similarly if we use the substituion v=u+1/u instead.
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    (Original post by BCHL85)
    It is not solution yet. It's not that easy. This will leave you at the integration below:
    \int{\frac{2y^2}{y^4 +1}}dy
    OK sorry... I'm easily satisfied. It looked so neat and tidy I thought it must be correct and finished.
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    dvs, I was stopped at H, I couldn't figure out how to solve H
 
 
 
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