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Why does sqrt(25-x^2) need trigonometric Integration?

  1. Sep 14, 2013 #1
    What tells me that when I need to integrate it I can't just do (25-x^2)^0.5 and go on from there with common power of x integration? What is the thing that tells me "hang on there, this requires trigonometric integration"?
  2. jcsd
  3. Sep 14, 2013 #2

    Ray Vickson

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    The graph of the equation ##y = \sqrt{25-x^2}## is the upper semi-circle of radius 5, centered at (0,0). So, when you integrate y from x = a to x = b you are computing the area under part of a semi-circle, and that suggests that things like trigonometric functions will be needed.
  4. Sep 14, 2013 #3
    Thank you very much. I think I can go on from there; investigating it further.
  5. Sep 14, 2013 #4
    I usually tell my students that trig sub is something you want to think about trying when you have a quadratic expression in a "bad place"; under a radical or in a denominator. Of course you'd want to look for an easy u-sub or algebraic rearrangement first, but that seems to be a good "trigger". Especially for textbook-style problems in a standard calc course.
  6. Sep 16, 2013 #5


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    Dearly Missed

    It doesn't "require" trigonometric substitution.
    You might, for example, to arbitrary degree of accuracy, use numerical integration instead.

    That being said, using trigonometric integration is very..clever.
  7. Sep 16, 2013 #6


    Staff: Mentor

    Because the quantity being raised to the 1/2 power is not x.

    The basic property is this:
    $$\int x^n dx = \frac{x^{n+1}}{n+1} + C$$

    If you have a function of x raised to the power, but still have dx in the integral, the formula above does not apply.
    Most likely because simpler techniques such as substitution can't be used.
  8. Sep 16, 2013 #7
    Every time I ever see something that looks like ##y = \sqrt{a^2 ± b^2}## the first things that pop into my head is "Pythagorean theorem" and "triangle," and in the context of integration "trig substitution" comes immediately after. For me, I remember all the trig I used in geometry of right triangles, and that triggers me to think trig substitution (sorry bout the pun ;) ).

    Quick question. Do you mean more specifically, the quantity being raised to the 1/2 power is not the DUMMY VARIABLE of the integration? (which is x in this case)

    If it were
    $$\int x^n dx $$

    we'd have the dummy variable (x) being raised to the power and could go ahead and use the rule. Is that right?
  9. Sep 16, 2013 #8


    Staff: Mentor

    Yes, that's exactly what I mean.

    Here's an example where the above does not apply:
    $$ \int (x^2 + 3x + 1)^3dx \neq \frac{(x^2 + 3x + 1)^4}{4} + C$$

    In a similar vein, students often misapply this rule --
    ##\int \frac {dx}x = ln|x| + C##

    to something that is incorrect, such as this:
    ## \int \frac {dx}{x^2 + 1} \neq ln|x^2 + 1| + C##
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