- #1
fateswarm
- 18
- 0
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"?
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"?
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.
fateswarm said:What is the thing that tells me "hang on there, this requires trigonometric integration"?
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"?
Because the quantity being raised to the 1/2 power is not x.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?
Most likely because simpler techniques such as substitution can't be used.fateswarm said:What is the thing that tells me "hang on there, this requires trigonometric integration"?
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"?
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.
Yes, that's exactly what I mean.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.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?
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.
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.
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.
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.
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.