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Integral of tanhx.

  1. Nov 10, 2007 #1
    What's wrong with my solution?

    [tex]\int \sqrt {\tanh x}dx[/tex]

    for [tex]u = \tanh x\rightarrow du = \frac {dx}{\cosh^2x}[/tex]

    [tex]\int \sqrt {\tanh x}dx = \int\sqrt {u}\cosh^2x dx = \int\frac {\sqrt {u}dx}{1 - u} = \int \frac {du}{2(1 + \sqrt {u})} - \int \frac {du}{2(1 - \sqrt {u})}[/tex]

    for [tex]v = (1 + \sqrt {u})\rightarrow dv = \frac {du}{2\sqrt {u}}[/tex]and for [tex]z = (1 - \sqrt {u})\rightarrow dz = - \frac {du}{2\sqrt {u}}[/tex]

    [tex]\int\frac {dv(v - 1)}{v} + \int\frac {dz(1 - z)}{z} = v - \ln v + \ln z - z[/tex]

    [tex]\int \sqrt {\tanh x}dx = \ln (\frac {1 - \sqrt {\tanh x}}{1 + \sqrt {\tanh x}}) + 2\sqrt {\tanh x}[/tex]
  2. jcsd
  3. Nov 10, 2007 #2
    [tex]I=\int \sqrt{tanhx}dx[/tex]



    [tex]I=\int \sqrt{u}cosh^2xdu=\int \frac{\sqrt{u}du}{1-u^2}[/tex]

    Notice, that [tex]cosh^2x=\frac{1}{1-tanh^2x}[/tex]

    Then, let [tex]t=\sqrt{u}[/tex] [tex]\frac{dt}{du}=\frac{1}{2\sqrt{u}}=\frac{1}{2t}[/tex].

    [tex]I=2\int \frac{t^2dt}{1-t^4}=\int \frac{dt}{1-t^2} + \int \frac{dt}{1+t^2}[/tex]

    So, your problem is: when you substitute dx you write the same dx in integral. But it is wrong! You must write du :)

    Also look at [tex]cosh^2x=\frac{1}{1-tanh^2x}[/tex]
    Last edited: Nov 10, 2007
  4. Nov 11, 2007 #3
    it would be faster to make a u^2 substitution
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