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Substituting Integration with trigonometric function

  1. May 20, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]\int_{1}^{5}\sqrt{4-(x-3)^2}dx[/itex]

    2. Relevant equations



    3. The attempt at a solution
    Not really a homework. Just curious.

    Let [itex]x-3=2sin u[/itex]
    [itex]x=2sin u+3[/itex]

    Now limit change,
    when x=1, u = arcsin (-1)
    when x=5, u = arcsin (1)

    Since there are infinitely many solution of arcsin (-1) and arcsin (1), How do we know which ones to choose?

    Thank you very much
     
    Last edited: May 20, 2012
  2. jcsd
  3. May 20, 2012 #2
    [itex]\int_{1}^{5}\sqrt{4-(x-3)^2}dx[/itex]

    You have to write itex for the Latex to work, not tex. And the ending tag is /itex with a forward slash, not \itex with a backward slash.

    Also, show your attempt at a solution or nobody will help you.
     
  4. May 20, 2012 #3
    There are not infinitely many solutions to [itex]sin^{-1}(-1)[/itex] and [itex]sin^{-1}(1)[/itex]. The range of [itex]sin^{-1}(x)[/itex] is [itex][-\frac{π}{2},\frac{π}{2}][/itex].

    EDIT: While trigonometric substitution is a valid way to solve this definite integral, you can also solve the problem by thinking about what the graph of [itex]y=\sqrt{4-(x-3)^{2}}[/itex] is, and remembering that a definite integral is the area between the curve and the x-axis.
     
    Last edited: May 20, 2012
  5. May 20, 2012 #4
    Why is that? can't it be [3∏/2, 5∏/2]? or let's say [-∏/2 , 5∏/2] ?
     
  6. May 20, 2012 #5
    I know. I can solve this integral by modelling it as a semi circle and find the area.

    But here, I am curious about the trigonometric substitution.
     
  7. May 20, 2012 #6
    I'm not sure "why" exactly. But if I had to guess I'd say just because. That's the way the inverse sine function was defined. It was given a restricted range. I'm not sure you need a reason why since it's in the definition of the inverse sine function.

    EDIT: You can use [itex]\frac{3π}{2}[/itex] as your lower limit of integration and [itex]\frac{5π}{2}[/itex] as your upper limit if you want. You'll get the same answer since [itex]4cos^{2}u[/itex], the integrand you end up with after the substitution, has a period of [itex]π[/itex].
     
    Last edited: May 20, 2012
  8. May 20, 2012 #7

    Ray Vickson

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    When you change variables to [itex] x = 3 + 2 \sin(t)[/itex] you can see directly that [itex]x[/itex] ranges from 1 to 5 when [itex]t[/itex] ranges from [itex] -\pi/2 \text{ to } \pi/2. [/itex] We don't need to bother with different branches of the arcsin function; in fact, we can forget about arcsin completely----just look at the original requirements.

    BTW: it is perfectly OK to use "[tex ]" and "[/tex ]" (no spaces); that will give you a displayed result, like [tex] \int_1^5 \sqrt{4 - (x-3)^2} \, dx, [/tex] instead of the in-line result you get when using "[itex ]" and "[/itex ]", which would write the same expression as [itex] \int_1^5 \sqrt{4 - (x-3)^2} \, dx.[/itex]

    RGV
     
  9. May 20, 2012 #8

    SammyS

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    BTW, there are not infinitely many solutions for arcsin(-1) and arcsin(1).

    The arcsine function is a perfectly well-defined function for which arcsin(-1) = -π/2 and arcsin(1) = π/2 .

    However, each of the following two equations do have infinitely many solutions.
    [itex]1=2\sin(u)+3[/itex]

    [itex]5=2\sin(u)+3[/itex]​
    As Ray V. mentioned, you don't need to consider all of those solutions for this problem.
     
  10. May 21, 2012 #9
    Alright. Thank you so much, you guys.
     
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