How to Solve the Integral of cot(x)/sin(x)?

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    Integral
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Discussion Overview

The discussion revolves around the integral of cotangent divided by sine, specifically the expression \(\int \frac{\cot x}{\sin x}\,dx\). Participants explore various methods for solving this integral, including substitution and integration by parts, while sharing their experiences and frustrations with the problem.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Technical explanation

Main Points Raised

  • One participant expresses difficulty in solving the integral \(\int \frac{\cot x}{\sin x}\,dx\) and seeks assistance.
  • Another participant suggests expressing the integral in terms of sine and cosine, leading to the form \(\int \frac{\cos x}{\sin^2 x}\,dx\) and proposes a substitution \(u = \sin x\).
  • A subsequent reply indicates that the substitution leads to the solution \(-\csc x + C\), reflecting on the simplicity of the approach.
  • Some participants share their frustrations about missing straightforward solutions, indicating a common experience in tackling integrals.
  • One participant proposes using integration by parts, referencing the ILATE rule to guide the choice of functions for integration.
  • A participant questions another about their educational background, suggesting a personal connection based on their guesses about the other’s level of study.

Areas of Agreement / Disagreement

Participants generally agree on the methods for approaching the integral, but there are multiple proposed techniques and no consensus on a single preferred method. The discussion remains open-ended with various suggestions and personal experiences shared.

Contextual Notes

Some participants express uncertainty about their approaches and the effectiveness of different methods, indicating that the discussion is exploratory and not definitive.

Ornum
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Hello again people, thank you for the help I received on the last integral I posted here but never got around to replying to, the help was much appreciated. However I've encountered another integral which for some reason I just cannot seem to solve, being:

[itex]\int \frac{\cot x}{\sin x}\,dx[/itex]

If anyone can aid me in solving this I would be very glad. Thanks in advance.
 
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This might help you to see it:

[tex] {d \over {dx}}({1 \over {f(x)}}) = {{ - f'(x)} \over {f(x)^2 }}[/tex]
 
Ornum said:
Hello again people, thank you for the help I received on the last integral I posted here but never got around to replying to, the help was much appreciated. However I've encountered another integral which for some reason I just cannot seem to solve, being:

[itex]\int \frac{\cot x}{\sin x}\,dx[/itex]

If anyone can aid me in solving this I would be very glad. Thanks in advance.
In many situations, it's really simple just to express everything in terms of sine and cosine:

[itex]\int \frac{\cos x}{\sin ^2 x}\,dx[/itex]

Let [itex]u = \sin x[/itex], therefore [itex]dx = du/\cos x[/itex]. Making the substitution:

[itex]\int \frac{du}{u^2} = -u^{-1} + C = -\csc x + C[/itex]
 
Thank you very much, now you've just shown me I cannot believe how I missed that, however that tends to be how it always is. :wink:
 
I always get frustrated when something was staring me in the face but I couldn't see it. That's why I thought a hint may be of more help that the 'full' solution.

Paul. :wink:
 
I'd always get frustrated when I couldn't see something staring me in the face. That's why I thought a hint may be of more help that the 'full' solution.

Paul. :wink:
 
or u can write the above integrl as

INTEGRAL OF cosx cosec(sqr)x

take cosec(sqr)x as 2nd function and integrate by parts ...gives u answer instantly...
 
Some general advice: after you have an integral in a "standard looking form" (not even a standard form, for you wouldn't be reading my advice if it were in a standard form) try integration by parts keeping the ILATE rule in mind

I = inverse trigonometric function
L = logarithmic function
A = algebraic function
T = trigonometric function
E = exponential function

This order gives you an idea of which function to chose as u and which to chose v, when you wish to evaluate the integral [tex]\int u dv[/tex].

[tex] \int udv = uv - \int vdu[/tex]

By the way, Dr. Brain are you from India? My guess is that you're in class 11/12. Correct me if I am wrong ;-)
 

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