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Calculus II Trig Sub

  1. Mar 4, 2009 #1
    I'm very unsure on how to do this equation...
    Someone PLEASE HELP!!!

    [tex]\int[/tex]x[tex]^{3}[/tex][tex]\sqrt{}x^{2}+9[/tex]



    i do know that
    if you...
    let x = 3sec [tex]\vartheta[/tex]
    dx = 3 sec [tex]\vartheta[/tex] tan [tex]\vartheta[/tex] du
     
  2. jcsd
  3. Mar 4, 2009 #2
    sqrt of x^2? you sure you wrote that right?
     
  4. Mar 4, 2009 #3
    you wouldn't substitute in a secant, a tangent would do much better, when you sub in you have to figure out what you should put in the place of that x^3 too!
     
  5. Mar 4, 2009 #4
    yeah. its the integral of x^3 times the sqrt of x^2 + 9
     
  6. Mar 4, 2009 #5
    so would it be 3tan^3 theta times the sq rt of 3tan^2theta +9?
     
  7. Mar 4, 2009 #6
    and you need d(theta) in there too, plus remember you have to raise the entire term to the power
     
  8. Mar 4, 2009 #7
    would d(theta) be 3sec(theta)tan(theta) ?
     
  9. Mar 4, 2009 #8
    Remember your letting x = 3tan (theta).

    x^3 means 9tan^3 (theta) and the other term becomes sqrt [ 9tan^2 (theta) + 9 ]

    Factor out a square root of 9 out of the square root and use the identity: 1 + tan^2 x = sec^2 x to "simplify" your problem.

    You should be left with an integration of a sec to the third and a tangent to the third.
     
  10. Mar 4, 2009 #9
    if you let x = 3tan (theta)

    what is the derivative in terms of d (theta) ?
     
  11. Mar 4, 2009 #10
    is this correct so far?

    9 [tex]\int[/tex] tan[tex]^{3}[/tex] [tex]\theta[/tex] [tex]\sqrt{[/tex] sec [tex]^{2}[/tex] [tex]\theta[/tex]
     
  12. Mar 4, 2009 #11
    Close.

    Like lumbar said, you forgot the d(Theta).

    if you let x = 3tan(Theta)

    derive the above equation in terms of d(Theta)

    you should get:

    dx = <derivative of right hand equation>
     
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