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Book on multivariable calculus

  1. Dec 23, 2004 #1
    I am reading a book on multivariable calculus for my course and I have tried the question :

    [tex]

    \int_{0}^{\frac{\pi}{4}} dx \int_{0}^{Sec(x)} y^3 dy

    [/tex]

    apparently the answer is 1/3..I have TRIED to get this answer... yet i cannot yield 1/3....help!!!

    (anyone from the UK...I have a good A Level in maths {up to P3 for Pure and M3 for mechanics, grade A...i know iterated integrals but this one stumps me}...arse :))
     
    Last edited: Dec 23, 2004
  2. jcsd
  3. Dec 23, 2004 #2
    I worked it out and got 1/3. Show your work so that I can show you where you've gone wrong.
     
  4. Dec 23, 2004 #3

    quasar987

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    Are you using the formula

    [tex]\int sec^n u = \frac{1}{n-1}tgu \ sec^{n-2}u+\frac{n-2}{n-1}\int sec^{n-2}udu[/tex]

    ?

    Using this on [itex]sec^{4}x[/itex], I get 1/3 too.
     
    Last edited: Dec 23, 2004
  5. Dec 23, 2004 #4

    learningphysics

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    You can do the problem with integration by parts.
     
  6. Dec 24, 2004 #5
    I have never seen that formula before ever :)...

    naturally i established the first integral, and ended up with

    [tex]

    \frac{1}{4}\int_{0}^{\frac{\pi}{4}}sec^4 (x) dx
    [/tex]
    however beyond this i begin to lame up... doing it by parts usually ends up in misery and i usually end up going around in circles.

    [tex]

    \frac{1}{4}\int_{0}^{\frac{\pi}{4}}(sec^2 (x))(sec^2 (x)) dx
    [/tex]

    Would this suffice in the first instance, and take it from parts there on in..

    BK Flamer :cry:
     
    Last edited: Dec 24, 2004
  7. Dec 24, 2004 #6

    learningphysics

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    Yes. Set [tex]u=sec^2(x)[/tex] and [tex]dv=sec^2(x)dx[/tex]

    You only have to do integration by parts this one time... continue, and tell us where you get stuck...
     
  8. Dec 24, 2004 #7

    t!m

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    If sec is raised to an even power, you can simply do it using substitution and identities. That's how I did it, and also got 1/3. You're second step is correct, now let u=tan(x). Anything?
     
  9. Dec 24, 2004 #8
    Doing it as we speak....hang on guys... :D
     
  10. Dec 24, 2004 #9
    ok so i've said :

    [tex]

    u = \sec^2(x) \quad \quad
    \frac{du}{dx} = 2\sec^2 (x) \tan(x) \quad \quad

    \frac {dv}{dx} = sec^2 (x) \quad \quad
    v = \tan (x)

    [/tex]

    So therefore, by parts, the integral {excluding constants and what not for simplicities sake}

    [tex]

    \int (sec^2 (x))(sec^2 (x)) dx = sec^2 (x)tan (x) - 2\int sec^2 (x) (tan^2 (x)) dx

    [/tex]

    Or have I made a right boob-up?
     
    Last edited: Dec 24, 2004
  11. Dec 24, 2004 #10

    t!m

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    I believe you meant

    [tex]
    u = \sec^2(x)
    [/tex]

    But you derived it correctly, so I'll assume that was a typo. Otherwise, your setup looks right to me. I'm gonna go ahead and post how I would have done it without integrating by parts, since parts seems to be the preferred method right now.
     
    Last edited: Dec 24, 2004
  12. Dec 24, 2004 #11

    learningphysics

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    Everything is correct. Just integrate [tex]sec^2(x)tan^2(x)dx[/tex]. Have a look at it for a while... hint: use the fact that [tex]sec^2(x)[/tex] is the derivative of [tex]tan(x)[/tex]
     
  13. Dec 24, 2004 #12
    So i try to evaluate that other integral, by parts too... the :

    [tex]
    \int sec^2 (x) (tan^2 (x)) dx[/tex]

    if someone would give me a guide as to how many times im going to need to do this repeated parts thingy then I wont be so scared ;) :rofl:

    [tex]u = \tan^2(x) \quad \quad\frac{du}{dx} = 2\tan (x) \sec^2(x) \quad \quad\frac {dv}{dx} = sec^2 (x) \quad \quad v = \tan (x)[/tex]

    Ok so far?!?!...i still get the feeling that its not going to terminate anytime soon
     
    Last edited: Dec 24, 2004
  14. Dec 24, 2004 #13

    learningphysics

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    Not integration by parts here: I'll do a variable substitution here: let [tex]y=tan(x)[/tex] so [tex]dy=sec^2(x)dx[/tex] so, I can rewrite your integral as:

    [tex]\int y^2dy[/tex]

    which becomes

    [tex](1/3)y^3=(1/3)tan^3(x)[/tex]
     
  15. Dec 24, 2004 #14
    Oooh yeah!! :D ha god dammit im a fool... Could someone show it to me by parts too, cos it was that method that was murdering me.

    :rolleyes: :tongue:
     
  16. Dec 24, 2004 #15

    t!m

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    Without integrating by parts, using only a trig identity and variable substitution:

    [tex]
    \frac{1}{4}\int_{0}^{\frac{\pi}{4}}(sec^2 (x))(sec^2 (x)) dx
    =\frac{1}{4}\int_{0}^{\frac{\pi}{4}}(tan^2 (x)+1)(sec^2 (x)) dx
    [/tex]

    Letting [tex]u=tan(x); du=sec^2(x)dx[/tex]

    We have
    [tex]
    \frac{1}{4}\int(u^2 +1)du
    =\frac{1}{4}(\frac{1}{3}u^3 +u|
    =(\frac{1}{12}tan^3 (x)+\frac{1}{4}tan(x)|_{0}^{\frac{\pi}{4}}
    =\frac{1}{3}
    [/tex]

    Seems easier to me.
     
  17. Dec 24, 2004 #16
    Most definatly...

    HOWEVER :D...

    due to my sloppy notation and consumption of beer, i was missing out something that was so so simple.

    After careful considering, i arrived at the integral :


    [tex]

    \frac{1}{4}(sec^2 (x) tan (x) - \frac{2}{3} tan^3 (x))

    [/tex]

    And when the limits are put in I do believe it yields 1/3...

    I am therefore a boob!!! :rofl:
     
  18. Dec 24, 2004 #17

    learningphysics

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    You're right! I didn't see the identity.
     
    Last edited: Dec 24, 2004
  19. Dec 24, 2004 #18

    learningphysics

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    That's alright. I'm a boob too! :rolleyes: Hopefully us boobs can still find the right answer going about it the hard way!
     
  20. Dec 24, 2004 #19
    Thanks all who gave advice...much appreciate

    Merry christmas all...

    :biggrin: :biggrin: :rofl:
     
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