Evaluating Integral Involving Arcsin, Arccos, and Arctan

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In summary, the conversation is discussing how to evaluate the line integral of a vectorial function over a closed path parametrized by g(t)=(sint, cost, sin(2t)), and whether there are any singularities present. The solution involves isolating any singularities and evaluating line integrals over circles with radius h approaching zero. It is also mentioned that an exact differential over a closed path is always 0.
  • #1
kingwinner
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Evaluate
∫(x^2+arcsinx) dx + (arccosy)dy + (z^2+arctanz)dz
C
where C is parametrized by g(t)=(sint, cost, sin(2t)), 0<t<2pi



I tried doing it directly, but it gets really horrible and I don't think I can integrate the resulting function, is there a trick or short cut to this question?

Thank you!
 
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  • #2
Calculate its rotational of the vectorial function, if its zero then try calculating the singularities inside C(where the function is infinite) Then isolate the singularities with circles with radius h with h going to zero. Evaluate the line integrals of those circles and there is your awnser. :)
If u get stuck check http://www.tubepolis.com
 
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  • #3
Are there any singularities? If we're doing it "directly", why does it matter whether there are singularities or not?

Will any theorem help?
 
  • #4
That is a closed path and Stingray788 is simply referring to the fact that an exact differential over a close path is always 0. That is clearly an exact differential because the coefficient of dx depends only on x, the coefficient of dy depends only on y, the coefficient of dz depends only on z.
 

1. What are the basic properties of arcsin, arccos, and arctan functions?

The arcsin, arccos, and arctan functions are inverse trigonometric functions that can be used to find the angle in a right triangle given the ratio of sides. The arcsin function is the inverse of the sine function, the arccos function is the inverse of the cosine function, and the arctan function is the inverse of the tangent function.

2. How do you evaluate integrals involving arcsin, arccos, and arctan?

To evaluate integrals involving arcsin, arccos, and arctan, you can use the substitution method. This involves using a trigonometric identity to rewrite the integral in terms of one of the inverse trigonometric functions. Then, you can use the inverse trigonometric function to find the value of the integral.

3. Can you provide an example of evaluating an integral involving arcsin, arccos, and arctan?

Sure, one example is evaluating the integral ∫(x^2)/(1+x^2) dx. Using the substitution method, we can rewrite this integral as ∫(1-1/(1+x^2)) dx. Then, we can use the fact that arctan'(x)=1/(1+x^2) to evaluate the integral. The final result is ∫(1-1/(1+x^2)) dx = x-arctan(x) + C.

4. Are there any special cases to consider when evaluating integrals involving arcsin, arccos, and arctan?

Yes, when evaluating integrals involving arccos and arctan, it is important to consider the range of these functions. The range of arccos is typically restricted to [0, π], while the range of arctan is restricted to (-π/2, π/2). This means that when using the substitution method, you may need to adjust the limits of integration to match the range of the inverse trigonometric function being used.

5. How is the evaluation of integrals involving arcsin, arccos, and arctan related to real-world applications?

The evaluation of integrals involving arcsin, arccos, and arctan can be applied to various real-world situations, such as calculating the volume of a solid of revolution, determining the center of mass of an object, and solving problems in physics and engineering. These functions are also used in computer graphics and navigation systems to determine angles and distances.

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