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Integral over [0,2pi]

  1. Aug 12, 2010 #1
    I have to find: [tex]\int_{0}^{2\pi}\sqrt{t^2+2} dt[/tex]

    I found that [tex]\int \sqrt{t^2+2} dt = \frac{t\sqrt{t^2+2}}{2} - arcsin(\frac{t}{\sqrt{2}}) + c[/tex]

    But when I fill in [tex]2\pi[/tex] I get: [tex]\frac{2\pi \sqrt{4\pi ^2+2}}{2}- arcsin(\frac{2\pi }{\sqrt{2}})[/tex]

    but [tex]arcsin(\frac{2\pi }{\sqrt{2}})[/tex] doesn't exist..

    Have I done something wrong?
     
  2. jcsd
  3. Aug 12, 2010 #2

    arildno

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    Well, you found wrongly!

    We have:
    [tex]\int\sqrt{t^{2}+a^{2}}dt=a\int\sqrt{1+(\frac{t}{a})^{2}}dt[/tex].

    Use the substitution:
    [tex]\frac{t}{a}=sinh(u)[/tex], sinh being the hyperbolic sine, rather than the trigonometric sine.
    (Using the identity: [tex]cosh^{2}(u)-sinh^{2}(u)=1[/tex]
    We then get:
    [tex]dt=acosh(u)du[/tex]

    And your integral becomes:
    [tex]a^{2}\int{cosh^{2}(u)}du[/tex]
    This is easily integrated:
    [tex]I=\int{cosh^{2}(u)}du=sinh(u)cosh(u)-\int{sinh^{2}(u)du=sinh(u)cosh(u)-\int{cosh^{2}(u)-1}du[/tex]
    That is:
    [tex]2I=u+sinh(u)cosh(u)[/tex]
    or:
    [tex]I=\frac{u}{2}+\frac{sinh(u)\sqrt{sinh^{2}(u)+1}}{2}[/tex]

    Now, you can readily re-substitute for t/a=sinh(u), gaining:
    [tex]\frac{a^{2}}{2}sinh^{-1}(\frac{t}{a})+\frac{t\sqrt{t^{2}+a^{2}}}{2}[/tex]
     
    Last edited: Aug 12, 2010
  4. Aug 12, 2010 #3
    Thanks! Now I found it!
     
  5. Aug 12, 2010 #4

    arildno

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    I made a sign error in the last square root expression, that has been fixed.
     
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