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Math Help - Plz tell me how to integrate this

  1. #1
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    Post Plz tell me how to integrate this

    integrate sqrt(tan x)
    I substituted tan x = u^2
    then i got x = arc tan u^2
    By substituting i got 2u^2/(1+u^4)
    now what????
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  2. #2
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    Quote Originally Posted by Beenish View Post
    integrate sqrt(tan x)
    I substituted tan x = u^2
    then i got x = arc tan u^2
    By substituting i got 2u^2/(1+u^4)
    now what????
    Now use partial fractions. Note that

    u^4 + 1 = u^4 - 2u^2 + 2u^2 + 1 = (u^4 - 2u^2 + 1) - 2 u^2

     = (u^2 - 1)^2 - 2u^2 = (u^2 - 1 - \sqrt{2} u) (u^2 - 1 + \sqrt{2} u)

    (u^2 - \sqrt{2} u - 1) (u^2 + \sqrt{2} u - 1)

    and these two quadratics are irreducible. The ensuing algebra is your pleasant job to do.


    Of interest: integrate Sqrt[Tan[x]] - Wolfram|Alpha

    (Be sure to click on Show steps).
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  3. #3
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     \int \frac{ x^2}{x^4 + 1} ~dx


     = \frac{1}{2} \int \frac{ (x^2 - 1) + ( x^2 + 1)}{ x^4 + 1}~dx

     = \frac{1}{2} \int \frac{ x^2 - 1 }{x^4 + 1} ~dx + \frac{1}{2} \int \frac{ x^2 + 1}{x^4 + 1}~dx

     = \frac{1}{2} \int \frac{ 1 - \frac{1}{x^2} }{x^2 + x^{-2}}~dx + \frac{1}{2} \int \frac{ 1 + \frac{1}{x^2} }{x^2 + x^{-2}}~dx

     = \frac{1}{2}\int \frac{d ( x  + \frac{1}{x} )}{ (x  + \frac{1}{x} )^2 - 2  } +  \frac{1}{2}\int \frac{d ( x  - \frac{1}{x} )}{ (x  - \frac{1}{x} )^2 + 2  }


    Recall the formulae

     \int \frac{du}{u^2 + a^2 } = \frac{1}{a} \tan^{-1}(\frac{u}{a})

      \int \frac{du}{u^2 - a^2 } = \frac{1}{a} \tanh^{-1}(\frac{u}{a})


    \frac{1}{2}\int \frac{d ( x  + \frac{1}{x} )}{ (x  + \frac{1}{x} )^2 - 2  } +  \frac{1}{2}\int \frac{d ( x  - \frac{1}{x} )}{ (x  - \frac{1}{x} )^2 + 2  }

     = \frac{1}{2\sqrt{2}}\tanh^{-1}(\frac{ x  + \frac{1}{x} }{\sqrt{2}}) + \frac{1}{2\sqrt{2}}\tan^{-1}(\frac{ x  - \frac{1}{x} }{\sqrt{2}}) + C

     = \frac{1}{2\sqrt{2}}\tanh^{-1}(\frac{ x^2   + 1 }{\sqrt{2}x}) + \frac{1}{2\sqrt{2}}\tan^{-1}(\frac{ x ^2 - 1}{\sqrt{2}x}) + C
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  4. #4
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    Thank u

    Thanx a lot. It was very helpful answer.
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