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Integration Help

  1. Nov 16, 2007 #1
    In a diff. Equation im doing I have to integrate

    [tex]

    \sin ^2 x\sec x


    [/tex]


    which I can remember how to do.

    What is the best method for this?
     
  2. jcsd
  3. Nov 16, 2007 #2
    Sec(x) = 1/cos x => (sin x)^2 / cos x

    Notice how one of them is the derivative of the other.
     
  4. Nov 16, 2007 #3
    with respect to x I'm guessing?

    [tex]\int sin^2 x sec x dx=\int \frac{sin^2 x}{cos x} dx[/tex]

    just do a substitution. you probably didn't see the substitution right away with the secant, best to change to sin and cos first
     
  5. Nov 16, 2007 #4
    [tex]

    \int {\frac{{\sin ^2 x}}{{\cos x}}} dx

    [/tex]

    let u=sinx du=coxdx

    I don't see this working becuase i have (1/cosx)dx not cosx dx....

    Am I looking at the wrong substitution?

    Rob
     
  6. Nov 16, 2007 #5
    or if let u=cosx im left with an exta sinx after substitution...
     
  7. Nov 16, 2007 #6

    rock.freak667

    User Avatar
    Homework Helper

    Try using sin^2x=1-cos^2x and then look for the integral of secx
     
  8. Nov 16, 2007 #7
    .............
     
  9. Nov 16, 2007 #8
    not sure what your getting at there

    [tex]
    \[
    \int {\frac{{\sin ^2 x}}{{\cos x}}} dx = \int {\frac{{1 - \cos ^2 x}}{{\cos x}}dx = ?}
    \]
    [/tex]
     
  10. Nov 16, 2007 #9
    I just tried integration by parts and I end up going in a circle...
    don't know why this one is stumping me so much.
     
  11. Nov 16, 2007 #10
    hm..didn't notice that sin^2 x....

    what rock.freak.667 meant was to split up that fraction and you'll have integral of sec x -cos x. so you'd have to find the integral of sec x.
     
  12. Nov 16, 2007 #11
    [tex]\[
    \begin{array}{l}
    \int {\sin ^2 x} \sec xdx = \int {(1 - \cos ^2 x)\sec xdx = \int {\sec x - \sec x\cos x\cos xdx} } \\
    = \int {\sec x - \cos xdx} \\
    \end{array}
    \]
    [/tex]

    Does this work?
     
  13. Nov 16, 2007 #12
    yes, now you'd need to find the integral for sec x
     
  14. Nov 16, 2007 #13
    which i think its just the ln abs(secX + tanX)
     
  15. Nov 16, 2007 #14
    Thanks guys.
     
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