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Homework Help: Computation of integrals.

  1. Nov 13, 2006 #1

    MathematicalPhysicist

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    i need to solve/prove the next two integrals:
    [tex]\int\frac{dx}{u^2+u+4}[/tex]
    and i need to show that:
    [tex]\int_{0}^{\pi}\sqrt{1+sinx}dx=4[/tex]
    the problem is that i have a clue to substitute u=sinx and then sin(pi)=0=sin0 so the integral should be equal zero, is it not?
    ofcourse the integrand becomes: sqrt(1+u)/sqrt(1-u^2)
     
  2. jcsd
  3. Nov 13, 2006 #2
    For the first you can use the method of partial fractions.

    You're right about the second. With the given limits, the integral is equal to zero.
     
  4. Nov 13, 2006 #3
    substitute u = 1 + sin x. Are you sure the integral is equal to 4, not -4?
     
  5. Nov 13, 2006 #4
    But the plot (area) of the function sqrt(1+sin(x)) from 0 to pi seems to be non-zero!
     
  6. Nov 13, 2006 #5
    It's non-zero.
     
  7. Nov 13, 2006 #6

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    neutrino, how would i use partail fractions here?
    i mean i need to decompose u^2+u+4 into a product of terms, but i have complex roots here.
     
  8. Nov 13, 2006 #7
    complete the square.
     
  9. Nov 13, 2006 #8

    MathematicalPhysicist

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    you mean something like this: u^2+u+4=(u-2)^2+5u
    i still dont get an appropiate term to integrate.
     
  10. Nov 13, 2006 #9
    More like (u +0.5)^2 + 15/4
     
  11. Nov 13, 2006 #10

    MathematicalPhysicist

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    ok, thanks.
    btw, what about the second integral does it equal zero or it really does equal 4?
     
  12. Nov 13, 2006 #11
    u^2+u+4= (u + 0.5)^2 + 15/4.

    edit: too slow, the second integral should equal minus -4, I guess they're defining is it as area so you just need the modulus.
     
  13. Nov 13, 2006 #12
    Not -4. I just put the function through the Integrator and substituted the values, and I got 4. This graph is completely above the x-axis.
     
  14. Nov 13, 2006 #13
    Btw, you will need to know what the derivative of the inverse tangent is.
     
  15. Nov 13, 2006 #14
    Oops, I'm missing/added a minus somewhere. Didn't have the common senese to think about the graph :).
     
  16. Nov 15, 2006 #15

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    wait a minute, then integral does converge to 4, care to explain how, where did i get it wrong?
     
  17. Nov 15, 2006 #16

    dextercioby

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    For the integral try the sub

    [tex] \tan\frac{x}{2}=t [/tex]

    Daniel.
     
  18. Nov 15, 2006 #17

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    but what's wrong with the substitution that im given a hint to use here?
    i.e
    sinx=u?
     
  19. Nov 15, 2006 #18

    HallsofIvy

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    If you let u= sin(x) then du= cos(x)dx so your integral will involve something like [itex]\frac{du}{cos(x)}[/itex] (with the cos(x) converted to u of course) but [itex]cos(\pi/2)= 0[/itex] so that is not a valid substitution.
     
  20. Nov 15, 2006 #19

    NateTG

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    You need to be careful about which square root you're using: The integrand should be:
    [tex]\frac{\sqrt{1+u}}{\pm \sqrt{1-u^2}}[/tex]
    The trick is that you'll be using one square root from [itex]x=0[/itex] to [itex]x=\frac{\pi}{2}[/itex] and the other root from [itex]x=\frac{\pi}{2}[/itex] to [itex]x=0[/itex].
     
    Last edited: Nov 15, 2006
  21. Nov 15, 2006 #20

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    ok, i understand the trick here, i havent seen this point x=pi/2 as a "bad" point, thank you for the pointers.
     
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