Integrating csc(x): Easier Than You Think

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Discussion Overview

The discussion revolves around the integration of the cosecant function, specifically \(\int \csc(x) \, dx\). Participants explore various methods and formulas for solving this integral, including the use of the half-angle tangent substitution and algebraic manipulations.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant expresses uncertainty about how to integrate \(\csc(x)\) and feels that the solution should be straightforward.
  • Another participant suggests using the substitution \(t = \tan(x/2)\) as a method to approach the integral.
  • A different participant questions the specific formula mentioned and attempts to clarify the differentiation involved in the substitution.
  • Further clarification is provided on the transformation of \(\csc(x)\) using trigonometric identities and substitution, leading to a logarithmic result.
  • One participant seeks clarification on a specific equality presented in the discussion regarding the transformation of the integral.
  • Another method is introduced involving multiplying by a form of one to simplify the integral, leading to a logarithmic expression.
  • Participants engage in further simplification and algebraic manipulation to clarify the integration process.
  • One participant expresses gratitude for the assistance received and reflects on their understanding of the topic.

Areas of Agreement / Disagreement

Participants present multiple methods for integrating \(\csc(x)\), and while some methods are discussed positively, there is no consensus on a single approach or resolution to all questions raised.

Contextual Notes

Some participants express confusion about specific steps in the integration process, indicating potential gaps in understanding or assumptions that may not have been fully articulated.

DrKareem
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haven't found a way of doing it so far. I have a feeling that it's extremely easy, and I'm missing how to do it somehow :/
 
Last edited:
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do you mean cosec(x), the cosecans ? If so, just use the t=tan(x/2) formula's...

marlon
 
I'm not sure what formula you're talking about.

[tex]\int \csc(x) dx[/tex]

If you take [tex]t=tan(\frac{x}{2})[/tex],

you'd get:

[tex]\frac {d}{dx} \tan(x)= \frac{1}{2}.sec^2(x)[/tex]

Not sure how to go from there...
 
Last edited:
What marlon meant, is the following:
[tex]csc(x)=\frac{1}{\sin(x)}=\frac{\cos^{2}(\frac{x}{2})+\sin^{2}(\frac{x}{2})}{2\sin(\frac{x}{2})\cos(\frac{x}{2})}=\frac{1+tan^{2}(\frac{x}{2})}{2tan(\frac{x}{2})}[/tex]

Substitute [tex]u=tan(\frac{x}{2})[/tex]
This implies:
[tex]\frac{du}{dx}=\frac{1}{2}\frac{1}{\cos^{2}(\frac{x}{2})}=\frac{1}{2}(u^{2}+1)[/tex]
Or:
[tex]dx=\frac{2du}{u^{2}+1}[/tex]
Hence, we have:
[tex]\int{csc(x)}dx=\int\frac{du}{u}=ln|u|+C=ln|tan(\frac{x}{2})|+C[/tex]
 
I'm not sure how you did this equality:


[tex]\frac{\cos^{2}(\frac{x}{2 })+\sin^{2}(\frac{x}{2})}{2\sin(\frac{x}{2})\cos(\ frac{x}{2})}=\frac{1+tan^{2}(\frac{x}{2})}{2tan(\frac{x}{2})}[/tex]


Can you please clarify?

Other than that, it's all clear, thank you very much.
 
Last edited:
[itex]\int \csc x = \int \csc x \left(\frac{\csc x - \cot x}{\csc x - \cot x}\right) = \int \frac{du}{u} = \ln |csc x - cot x|[/itex]
(is that what we're talking about?)
 
Last edited:
yes, nice method :)

Still i would like someone to explain the question of my last post.
Thank you :)
 
separate the fractions and simplify
[itex]\frac{\cos u}{2\sin u}+ \frac{\sin u}{2\cos u} =[/itex]
[itex]\frac{1}{2\tan u} + \frac{tan u}{2}[/itex]
get a common denominator and you're done.
 
Still i would like someone to explain the question of my last post.

Just divide by cos^2(x/2) in both the numerator and the denominator of the LHS of the expression in question and it will drop straight out.
 
  • #10
Yes, excellent, so know i know i am stupid hehe :)

Thanks a lot for your help guys :)
 

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