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

In summary: For example, if you have a function of x raised to a power, and the denominator is a rational function, then the rule does not apply.
  • #1
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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  • #2
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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  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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?
 
  • #8
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##
 

1. Why do we need to use trigonometric integration for sqrt(25-x^2)?

The function sqrt(25-x^2) represents a circle with a radius of 5 centered at the origin. In order to find the area under this curve, we need to use trigonometric integration because the bounds of integration will involve trigonometric functions.

2. Can't we just use regular integration techniques for sqrt(25-x^2)?

No, regular integration techniques are not applicable for a function that involves trigonometric functions. Trigonometric integration is specifically designed to handle functions that contain trigonometric terms.

3. How does trigonometric integration work for sqrt(25-x^2)?

Trigonometric integration involves using trigonometric identities and substitution to rewrite the function in terms of trigonometric functions and then integrating accordingly. In the case of sqrt(25-x^2), we can use the substitution x = 5sin(theta) to simplify the integral.

4. Why is sqrt(25-x^2) considered a "difficult" function to integrate?

Sqrt(25-x^2) is considered difficult to integrate because it involves a combination of algebraic and trigonometric terms. This requires a more advanced technique, such as trigonometric integration, to solve the integral.

5. Are there any real-life applications of using trigonometric integration for sqrt(25-x^2)?

Yes, trigonometric integration is commonly used in physics and engineering to solve problems involving circular motion, such as calculating the work done by a force on a rotating object or determining the period of a pendulum.

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