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Integration using inverse trig indentities?

  1. Feb 17, 2015 #1
    1. The problem statement, all variables and given/known data
    1.[tex]\int{\frac{sinx}{1+cos^{2}x}} \, dx[/tex]
    2.[tex]\int{\frac{1}{13-4x+x^2}} \, dx[/tex]

    2. Relevant equations
    Inverse trig identities.

    3. The attempt at a solution
    For the first one, I'm not too sure about what to do with the [tex]sinx[/tex] on the numerator and i have tried u-substitution to no avail (different answer). For the second one, I tried factorising but only had solutions for [tex]x\inℂ[/tex].
     
  2. jcsd
  3. Feb 18, 2015 #2
    You can maybe try to start with the substitution u = cos(x) and reforming the integral. A second substitution will be necessary after the first substitution.
     
  4. Feb 18, 2015 #3
    Second substitution is not necessary. It can be done in that first one.
     
  5. Feb 18, 2015 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    For the second integral start by completing the square in the denominator.
     
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