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Homework Help: Sine substitution for integration

  1. Aug 1, 2010 #1
    Hello!
    This is a quick question more to do with understanding.

    When using a sine substitution in an integral, such as:

    [tex] \int \sqrt{a^2-x^2} dx[/tex]

    Using the substitution

    [tex]x = a sin{t}[/tex]

    Don't you 'lose' some information? Because the range of values for x can be from neg. inf. to pos. inf.. But sin (t) can only take values between -1 and +1. Since a is a constant, it would do little to alter the range of values for sin t. Thus, if x = 4: 4 = a sin (t), and if a is less than 4, then the equation doesn't work.

    Does this make any sense? I don't understand whether I should be cautious or not when making substitutions of any kind; since here it seems not to matter even though they do not cover the same range of values.
    Any help appreciated, I can re-write if it's not particularly clear, as it's hard to communicate what I mean, as I don't think I fully understand what my concerns are!
    Thanks in advance.
     
  2. jcsd
  3. Aug 1, 2010 #2
    If you're dealing strictly with real valued integrals, then x has to be between +/- a, which is covered by a*sin(t).
     
  4. Aug 2, 2010 #3
    Spasiba Raskolnikov!
    So if I had a definite integral outside of the range of values that a sin (t) can have, then I must employ some other form of substitution?
    Thanks again,
    nobahar.
     
  5. Aug 2, 2010 #4
    No, this should always work.

    Let [tex] x = asin(t) [/tex]

    Then

    [tex] \int \sqrt{a^2-x^2} dx = \int \sqrt{a^2 - a^2sin^2(t)} \ acos(t) dt = a^2 \int \sqrt{1 - sin^2(t)} \ cos(t) dt = a^2 \int cos^2(t)dt. [/tex]

    Just use the double angle identity from there to finish up.
     
    Last edited: Aug 2, 2010
  6. Aug 2, 2010 #5
    Oh, and if you're wondering what inspired us to use the substitution x = a sint, then consider the right triangle with hypotenuse of length a and one leg of length x. Then the other leg is sqrt(a^2 - x^2). If you let t be the angle opposite leg x, then x = a sint. This is a very common "trick" that you should learn to use when your integrand doesn't seem to be easy normally and resembles some form of the pythagorean theorem or some application of a trig function.
     
  7. Aug 2, 2010 #6
    Thanks again raskalnikov.
    Sorry, but I am still having difficulty with the range of values.
    Using the above example, if I have a definite integral where x > a for one or both limits of the integral, then I cannot find a corresponding value for t. Under this circumstance, must I employ a different trigonometric identity?
     
  8. Aug 2, 2010 #7

    Mark44

    Staff: Mentor

    For this indefinite integral, [tex] \int \sqrt{a^2-x^2} dx[/tex]

    if you're working with a definite integral with this integrand, and if x > a or x < -a, your integrand will be undefined in the reals.
     
  9. Aug 2, 2010 #8
    Thanks Mark44.
    Unfortunately, and rather annonyingly, that falls beyond what I know at the moment.
    So it is still possible? Because my calculator will not provide an output for
    [tex]arcsin(\frac{x}{a})[/tex]
    where x/a is greater than 1. Thus I'm guessing there is something else you have to do? Although from what you've said I doubt I could follow it.
    Thanks for all the help.
     
  10. Aug 2, 2010 #9

    vela

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    I doubt that. It's basic algebra which you should have seen before calculus, and if you don't know this, you might as well learn it now.

    If x>a or x<-a, is the quantity inside the square root, a2-x2, positive or negative? What does that tell you about the allowed values of x?
     
  11. Aug 2, 2010 #10
    Thanks for the response. I fear you have to much confidence in me Vela!
    If x>a or x<-a then the value will be negative on both occasions (due to being squared) inside the square root, resulting in an imaginary number...
    Is this correct?
     
  12. Aug 2, 2010 #11

    vela

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    Yes, the inside will be negative. The result of the square root, however, depends on what set of numbers you're working in. Usually, you're working with only real numbers, in which case, the square root is undefined for |x|>a. The integral would therefore be defined only when x is restricted to the range -a≤x≤a. That's why the substitution x=a sin t is sufficient for solving this problem. In fact, it's quite nice that it automatically accounts for the restriction that |x|≤a.

    If you allow complex numbers as well, you can lift the restriction on x and try to find an answer, but there's quite a bit of math you need to learn about complex functions before you'll be able to do the integral correctly.
     
    Last edited: Aug 2, 2010
  13. Aug 3, 2010 #12
    Ohhhhhhhh. That's excellent! Many thanks vela, that's helped alot.
    Thankyou to all of you, Raskalnikov, Mark44 and vela. The help is much appreciated!
     
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