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Homework Help: Substituting Integration with trigonometric function

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


    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

    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]

  9. May 20, 2012 #8


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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.

    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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