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Definite Intergral (subtuition)

  1. Apr 30, 2007 #1
    1. The problem statement, all variables and given/known data
    Evaluate:
    [tex] \displaystyle \int_{\pi/6}^{\pi/2} \frac{\cos(z)}{\sin^{9}(z)}\, dz[/tex]

    2. Relevant equations
    [tex]u=sin(z)\longrightarrow\space\,du=cos(z)dz[/tex]

    3. The attempt at a solution
    After making the substitution and simplifying I got the[tex]\displaystyle \int_a^b g(u)\,du[/tex] where:
    g(u) =[tex]\frac{1}{u^9}[/tex]
    a =1
    b = .5
    Then I do [tex]\displaystyle\int_{.5}^{1}\frac{1}{u^9}\,du\longrightarrow\frac{-1}{8sin(z)^8}[/tex]
    I evaluated that on the interval [.5,1] but I got wrong answer, I know I went wrong somewhere.
     
    Last edited: Apr 30, 2007
  2. jcsd
  3. Apr 30, 2007 #2

    Dick

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    Did you evaluate z from [.5,1] or u? You should evaluate u over that interval or if you want to change back to z, then use the z limits.
     
  4. Apr 30, 2007 #3
    Ah,
    I think I evaluated:
    [tex]\frac{-1}{8sin(z)^8}[/tex]
    z from .5 to 1.
    so I evaluate u first?
     
  5. Apr 30, 2007 #4

    Dick

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    You already changed the z limits to u limits. You may as well just evaluate -1/(8*u^8) between them.
     
  6. Apr 30, 2007 #5
    Ah thanks,
    One more question for you
    How would I go about evaluating:
    [tex]\displaystyle \int_0^1 x^2\sqrt{7 x + 9}\,dx[/tex]
    what sub would help me out?
     
  7. Apr 30, 2007 #6

    Dick

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    It would help you a lot. You would get sqrt(u) times a polynomial in u. And then everything would turn into fractional powers. Try it! You'll like it!
     
  8. Apr 30, 2007 #7
    LOL.

    [tex]\displaystyle \int_0^1 x^2\sqrt{7 x + 9}\,dx[/tex]
    [tex]u=7x+9\longrightarrow\,du=7dx[/tex]
    So
    [tex]\displaystyle \int_{9}^{16} x^2\sqrt{u}\,du[/tex]
    What do I do to rid of the x^2!
     
    Last edited: Apr 30, 2007
  9. Apr 30, 2007 #8

    Dick

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    Solve for x in terms of u. BTW what happened to the 1/7 as in dx=du/7.?
     
  10. Apr 30, 2007 #9
    [tex]\frac{1}{7}\displaystyle \int_{9}^{16} x^2\sqrt{u}\,du[/tex]
    Oh you mean that one?:wink:

    So do you mean:
    [tex]\frac{d}{du}=2u[/tex]?
    then[tex]\displaystyle \int_{9}^{16} 2u\sqrt{u}\,du[/tex] ??
     
  11. Apr 30, 2007 #10

    Dick

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    Yes, I mean that one. But on the other question I mean why not put x=(u-9)/7? So x^2=(u-9)^2/49. Then expand and multiply by u^(1/2)?
     
  12. Apr 30, 2007 #11
    Ahhh I see what you mean:
    [tex]\frac{1}{7}\displaystyle \int_{9}^{16} x^2\sqrt{u}\,du[/tex]
    [tex]x=(u-9)/7[/tex]So:
    [tex]\frac{1}{7}\displaystyle \int_{9}^{16} (\frac{u-9}{7})^2\sqrt{u}\,du[/tex]
    [tex]\frac{1}{7}\displaystyle \int_{9}^{16} (\frac{(u-9)^2}{49})\sqrt{u}\,du[/tex]
    I then can expand and evaluate from there, correct?
     
  13. Apr 30, 2007 #12

    Dick

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    Sure then you'll just get stuff like u^(5/2) etc. Easy to handle for one of your obvious abilities! Sorry, gotta go. zzzzzzz.
     
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