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Understanding integration with trig identities, and absolute value

  1. Jul 20, 2014 #1
    1. The problem statement, all variables and given/known data

    In integration, we are allowed to use identities such as [itex]sinx = \sqrt{1-cos^2x}[/itex]. Why does that work, and why doesn't make a difference in integration? Graphing [itex]\sqrt{1-cos^2x}[/itex] is only equal to sinx on certain intervals such as[itex](0, \pi) [/itex]and [itex](2\pi, 3\pi)[/itex]. More correctly, shouldn't we use the absolute value of [itex]\sin\left({x}\right)[/itex]?

    [itex]sin^2x = 1 - cos^2x[/itex]
    [itex]|sinx| = \sqrt{1 = cos^2x}[/itex]
    or defined piecewisely = {[itex]\sin\left({x}\right)[/itex] in regions where it is above the x-axis, and -[itex]\sin\left({x}\right)[/itex] in regions where x is below the x-axis.

    Is it possible to even truly isolate "[itex]sin\left({x}\right)[/itex]" from
    [itex]sin^2x = 1 - cos^2x[/itex]? It seems as the |[itex]sin\left({x}\right)[/itex]| is the closest we can to isolate it.

    Sorry if I may seem confusing, but the concept of absolute value still confuses me.

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 20, 2014 #2
    Can you provide an example? All I feel like I can tell you right now is that you're right, sin(x) is not sqrt(1-cos^2(x)) and that you cannot make that claim in integration either, but I don't know the whole story.
     
  4. Jul 20, 2014 #3
    Well, I was just thinking about this, nothing more. Actually, I don't think this pertains too much to integration, since no one in the right mind would use that substitution. But how would someone isolate sinx from cos^2x + sin^2x = 1, there seems to always be an absolute value in the way. And I really don't think schools teach absolute value well, or at all. I've been taught that the square root of something results in +/-, but it wouldn't make sense that sinx = +/- sqrt(1-cos^2x).

    But in generally, in integration, I usually see the books ignoring the absolute value. For example, [itex]∫ x^5\sqrt{1-x^3} dx [/itex], you can make the substitution [itex] 1 - x^3 = z^2[/itex]. The square root of [itex]z^2[/itex] would be the absolute value of z, but they just use integrate z instead.

    Also, in [itex]∫\frac{\sqrt{x-x^2}}{x^4} dx [/itex], if you use the substitution [itex]x = \frac{1}{u} [/itex], then you'll end up with [itex]-∫ \sqrt{\frac{1}{u^2}(u-1)}u^2 du [/itex], and the [itex]\frac{1}{u^2}[/itex] simply becomes [itex]\frac{1}{u} [/itex]. Also, when calculating arc lengths, such as [itex]y^3 = 8x^2[/itex] from 1 to 8, sqrt(x^(-2/3)) becomes x^(-1/3).

    I apologize for the long message, I got addicted to use LaTeX.
     
    Last edited: Jul 20, 2014
  5. Jul 22, 2014 #4
    Anyone have any insights as to why this is true?
     
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