Solve Indefinite Integral (x/c) arcsin dx/tan(x)

  • Context: Graduate 
  • Thread starter Thread starter Gmen
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary
SUMMARY

The discussion centers on solving the indefinite integral of (arcsin(x/c) dx) / tan(x), where c is a constant. Participants suggest considering whether a closed form solution using elementary, trigonometric, and transcendental functions is necessary or if a series expansion, such as a Taylor series, would suffice. The consensus is that using a series expansion allows for term-by-term integration, which simplifies the process significantly. Differentiation is emphasized as a more manageable approach compared to direct integration.

PREREQUISITES
  • Understanding of indefinite integrals and their properties
  • Familiarity with arcsin and tan functions
  • Knowledge of Taylor series and polynomial integration techniques
  • Basic calculus concepts, including differentiation and integration
NEXT STEPS
  • Research the properties of arcsin and tan functions in calculus
  • Learn about Taylor series and their applications in integration
  • Explore polynomial integration techniques for complex integrals
  • Study differentiation methods to simplify integration problems
USEFUL FOR

Mathematicians, engineering students, and anyone involved in advanced calculus or mathematical modeling, particularly those working with integrals in applied contexts.

Gmen
Messages
8
Reaction score
0
I would like to know the following indefinite integral:

(INTEGRAL)arcsin(x/c) dx / tan(x). c is a constant.
This is not homework...I found it in fron of me while trying to make a 3d model of an enveloped worm gear. I can explain to anyone who isn't bored.
PLEASE tell me there is a solution...I so feel not :(
 
Physics news on Phys.org
Hey Gmen.

I guess you should probably ask whether you need it in a closed form using elementary, trig, and transcendental functions or whether you could do something like say a series expansion (like a Taylor series) and integrate it term by term.

You can use this and depending on how much error you can tolerate just find the cut-off term and compute all terms till that term.
 
This was a really helpful idea. Differentiating is always easier (and possible!) than integrating, and it becomes just a matter of loveable polynomial integration! Thanks! I shall remember this in difficult integrals to come.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K