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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?

    Problem solved!
    Last edited: Aug 12, 2010
  2. jcsd
  3. Aug 12, 2010 #2


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    I think you mean arcsinh rather than arcsin, don't you?
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