Deriving this "familiar arccosine form" of integral

In summary, the expression cos(theta) represents the arc-cosine of the angle between the vector representing the object's motion and the plane containing the object and the observer. The expression can be developed using the geometry of an ellipse, which provides insight into the original physics where it is being applied.
  • #1
Odious Suspect
43
0
I posted a question about this yesterday, but realized I had made a stupid mistake in my derivation.
Orbital dynamics: "The familiar arc-cosine form"
That error has been corrected. I still have a deeper question. I believe this expression can be developed using the geometry of an ellipse in a way which adds insight into the original physics where it is being applied.

Some 30 odd years ago I saw something of this nature done in an introductory physics course.

Is anybody here familiar with such a development?

$$-\int \frac{1}{\sqrt{a+2 bx-hx^2}} \, dx=\frac{1}{\sqrt{h}}\arccos \left[\frac{b-hx}{\sqrt{a+b^2h}}\right]$$
 
Physics news on Phys.org
  • #2
I am not sure what to do with your post, do you want to prove the above equation or what?
 
  • #3
No. I can prove it. I want a geometric development. That is to say, I believe the constants and variables in the expression ##cos \theta = \frac{b-hx}{\sqrt{a+b^2h}}## have some geometric significance pertaining to an ellipse.
 
  • #5
That only addresses the form I really am familiar with. Notice that the form I am asking about is more complicated.
 
  • #6
Odious Suspect said:
That only addresses the form I really am familiar with. Notice that the form I am asking about is more complicated.

Yes, I know that you are asking about something more complicated.
Might it be the case that the method used in the simple case can be generalized? I don't know. I offered a possible starting point.
Sorry, if I didn't provide you with exactly what you seek.
 
Last edited:
  • #7
Sorry about being so abrupt. I get grouchy when I can't figure something out.

I'm guessing this might be something to do with rotation of coordinates to eliminate the "cross product" term.

In some universe, the denominator in the cosine expression represents a radius.
 

Related to Deriving this "familiar arccosine form" of integral

1. What is the "familiar arccosine form" of integral?

The "familiar arccosine form" of integral is a commonly used form of the integral of the inverse cosine function. It is written as ∫√(1-x^2)dx, and is used to evaluate integrals involving inverse trigonometric functions.

2. How is the "familiar arccosine form" of integral derived?

The "familiar arccosine form" of integral is derived by using the substitution x = cosθ, which transforms the integral into the form ∫√(1-cos^2θ)sinθdθ. This can then be simplified using trigonometric identities to give ∫sinθdθ, which can be easily evaluated.

3. Why is the "familiar arccosine form" of integral useful?

The "familiar arccosine form" of integral is useful because it allows us to evaluate integrals involving inverse trigonometric functions, which can be difficult to solve using other methods. It also has many applications in physics, engineering, and other fields.

4. Can the "familiar arccosine form" of integral be used for any inverse trigonometric function?

No, the "familiar arccosine form" of integral can only be used for integrals involving the inverse cosine function. Other inverse trigonometric functions, such as inverse sine and inverse tangent, have their own specific forms of integration.

5. Are there any limitations to using the "familiar arccosine form" of integral?

Yes, the "familiar arccosine form" of integral can only be used for integrals with a square root of a quadratic expression in the integrand. It is not applicable for integrals with higher degree roots or more complicated expressions. Additionally, the substitution x = cosθ may not always be the most efficient method for evaluating a given integral.

Similar threads

Orbital dynamics: "The familiar arc-cosine form"
    • Mechanics
Replies
1
Views
858
A Help needed with derivation: solving a complex double integral
    • General Math
Replies
1
Views
787
Derivative of an integral function
    • Calculus
Replies
3
Views
2K
MHB Surface Area Formula for \(z = x^2 + y^2\) Derivation
    • Calculus
Replies
2
Views
2K
I Why Is the Derivative of the Function Set as a Constant in Classical Dynamics?
    • Classical Physics
Replies
3
Views
1K
MHB What went wrong when applying the FTOC to maximize arc length?
    • Calculus
Replies
2
Views
5K
How Do You Derive the Axes of a Kepler Orbit from Its Equation?
    • Introductory Physics Homework Help
Replies
1
Views
1K
I Atmospheric Density vs Altitude in an O'Neil Cylinder
    • Classical Physics
Replies
1
Views
663
A Quantization isn't fundamental
    • Beyond the Standard Models
5
Replies
163
Views
23K
Integral with quadratic denominator^n
    • Calculus and Beyond Homework Help
Replies
2
Views
2K

Hot Threads

Recent Insights

Back
Top