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Homework Help: Integrate sin (1+ cos ^2 x) ? Please help!

  1. Dec 6, 2008 #1
    First of all, hi I'm new here my name is crisanna. I stumbled upon this site across the web and realised this 's a great site!

    Anyway , here 's my question. Does anyone know how to integrate sin (1 +cos ^2 x) ?

    I tried the method integrate by parts but I got stucked. Below is my attempt :
    u= 1+ cos^2 x

    du/dx = 1/2 x + 1/2 sinxcosx + c ( TBH i dont even know if this part is correct :bugeye: )
     
    Last edited: Dec 7, 2008
  2. jcsd
  3. Dec 6, 2008 #2
    x's are missing.

    It is not correct. Use the chain rule carefully.
     
  4. Dec 7, 2008 #3

    Gib Z

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    Try a substitution rather than integrating by parts.
     
  5. Dec 7, 2008 #4
    Tried again and i got du/dx = -2sinxcosx using the chain rule. Was that correct??

    then I was about to use the substitution method..
     
  6. Dec 24, 2008 #5

    Gib Z

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    Hopefully that next substitution was u= cos x. =]
     
  7. Dec 24, 2008 #6
    Is it
    [tex]\sin \left( 1 + \cos^2 x\right)[/tex]

    or
    [tex]\sin x \cdot \left(1 + \cos^2 x\right)[/tex]

    Quite a difference!
     
  8. Dec 24, 2008 #7
    Substitute

    [tex] 1 + cos^2x = u [/tex]

    Now you will get a denominator of -sin(2x).

    But we [tex] 1 + cos^2x = u [/tex] hence find x in terms of arccos something. Using the rules of inverse trigo find 2x in terms of arcsin something.

    Now apply parts Twice
     
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