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Integral Help

  1. Mar 26, 2008 #1
    I've been having trouble with the following three integrals. If anybody could give me a hint on how to get started, that would be greatly appreciated.

    Problem 1.
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex][tex]\frac{dx}{1+cos^2(x)}[/tex]

    2. The attempt at a solution

    I've tried writing it in these ways:

    [tex]\int\frac{dx}{2-sin^2(x)}[/tex]

    [tex]\int\frac{dx}{(secx+cosx)(cosx)}[/tex]

    Let u=cosx
    [tex]\int\frac{-du}{\sqrt{1-u^2}(1+u^2)}[/tex]

    So far I've gotten nowhere. I've also tried to make the integral look like the integral for arctanx. I realize that I need to have a -sinx in the numerator but don't know how to make it work out.

    Problem 2:
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex][tex]\frac{dx}{sec^2(x)+tan^2(x)}[/tex]

    2. The attempt at a solution

    I've tried writing it like:

    [tex]\int[/tex][tex]\frac{cos^2(x)}{1+sin^2(x)}[/tex]

    U-substitution and trig-substition don't seem to work.

    Problem 3:
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex][tex]\frac{dx}{x^3+1}[/tex]

    2. The attempt at a solution

    In this problem I actually got somewhere:

    I split the denominator into two polynomials and used partial fractions.

    I ended with:

    [tex]\frac{1}{3}[/tex]ln(x+1) - [tex]\frac{1}{3}[/tex][tex]\int[/tex][tex]\frac{(x-2)dx}{x^2-x+1}[/tex]

    Then:

    [tex]\frac{1}{3}[/tex]ln(x+1) - [tex]\frac{1}{6}[/tex][tex]\int[/tex][tex]\frac{(2x-4)dx}{x^2-x+1}[/tex]

    [tex]\frac{1}{3}[/tex]ln(x+1) - [tex]\frac{1}{6}[/tex][tex]\int[/tex][tex]\frac{(2x-1)dx}{x^2-x+1}[/tex] - [tex]\int[/tex][tex]\frac{-3dx}{x^2-x+1}[/tex]

    [tex]\frac{1}{3}[/tex]ln(x+1) - [tex]\frac{1}{6}[/tex]ln(x^2-x+1)+3[tex]\int[/tex][tex]\frac{dx}{x^2-x+1}[/tex]

    Now I'm stuck.... don't know what to do.

    Any help would be greatly appreciated!
     
  2. jcsd
  3. Mar 26, 2008 #2

    Gib Z

    User Avatar
    Homework Helper

    For the first two, trying [tex]t= \tan (x/2)[/tex] should work nicely =]

    Third one, for the final integral, complete the square in the bottom and get it into the standard arctan form.
     
  4. Mar 27, 2008 #3
    Hey thanks for the help.

    For the integral:

    [tex]\int[/tex][tex]\frac{dx}{1+cos^2(x)}[/tex]

    I use u=tan(x/2) so, after much simplification, I get...

    [tex]\int[/tex][tex]\frac{2(1+u^2)du}{(1+u^2)^2+(1-u^2)^2}[/tex]

    [tex]\int[/tex][tex]\frac{(1+u^2)du}{(1+u^4)}[/tex]

    Now I don't know exactly what I should do. If I try trig substituion, I end up with..

    [tex]\frac{1}{2}[/tex][tex]\int[/tex][tex]\frac{d\theta}{\sqrt{tan\theta}}[/tex] + [tex]\frac{1}{2}[/tex][tex]\int[/tex][tex]\frac{tan\thetad\theta}{2\sqrt{tan\theta}}[/tex]

    Now I don't really know what I should do from here...
     
  5. Mar 28, 2008 #4

    Gib Z

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    Homework Helper

    You should check your simplification, i get [tex]\int \frac{ 2(u^2+1)}{1+ (u^2-1)^2} } du [/tex] which can be done with some more partial fractions.
     
  6. Mar 28, 2008 #5

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi noblerare! :smile:

    Yes, Gib Z's tan(x/2) method is a good one, which you should definitely remember, and it certainly works in this case.

    But in this case, the square means that tanx might work even better (or cotx).

    Alternatively:

    Hint: mutiply both top and bottom by cosec^2(x) … now what does that remind you of … ? :smile:
     
  7. Mar 28, 2008 #6
    Yay! I finally got it. Thank you so much, Gib Z and tiny-tim!
     
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