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

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SUMMARY

The discussion centers on the necessity of trigonometric integration for the function \( y = \sqrt{25 - x^2} \), which represents the upper semi-circle of radius 5. Participants emphasize that traditional power rule integration is not applicable due to the presence of a radical involving a variable, necessitating trigonometric substitution. Key indicators for using trigonometric integration include the presence of quadratic expressions under radicals or in denominators, as well as the failure of simpler techniques like substitution. The conversation concludes that while trigonometric substitution is clever and effective, numerical integration is also a viable alternative.

PREREQUISITES
  • Understanding of trigonometric functions and identities
  • Familiarity with integration techniques, particularly substitution
  • Knowledge of the Pythagorean theorem and its application in calculus
  • Basic properties of integrals, including the power rule
NEXT STEPS
  • Study the method of trigonometric substitution in integrals
  • Learn about numerical integration techniques for functions involving radicals
  • Explore the application of the Pythagorean theorem in calculus problems
  • Investigate common pitfalls in applying integration rules, such as the power rule
USEFUL FOR

Students and educators in calculus, particularly those focusing on integration techniques, as well as mathematicians interested in the application of trigonometric functions in solving integrals.

fateswarm
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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"?
 
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fateswarm said:
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"?

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.
 
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Ray Vickson said:
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.

Thank you very much. I think I can go on from there; investigating it further.
 
fateswarm said:
What is the thing that tells me "hang on there, this requires trigonometric integration"?

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.
 
fateswarm said:
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"?

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.
 
fateswarm said:
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?
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.
fateswarm said:
What is the thing that tells me "hang on there, this requires trigonometric integration"?
Most likely because simpler techniques such as substitution can't be used.
 
fateswarm said:
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"?

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 ;) ).
Mark44 said:
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.

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?
 
Raze said:
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)
Yes, that's exactly what I mean.
Raze said:
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?
Yes.

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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