Need help on integration question i found on net

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Discussion Overview

The discussion revolves around a specific integration problem encountered by a participant, focusing on various methods of solving the integral. Participants explore substitution techniques and partial fraction decomposition, with an emphasis on different approaches to arrive at the solution.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses confusion about the solution to the integral, suggesting a substitution of kx² = gsinθ.
  • Another participant proposes using the substitution x√(k/g) = tanh(u) instead of sin, indicating a different approach to the problem.
  • A third participant mentions that using a denominator of kx² + g would lead to an arctan answer, while the current method should yield an atanh answer.
  • A participant shares their work on the integral, using partial fractions to derive a solution involving hyperbolic functions, specifically mentioning the use of tanh⁻¹.
  • Another participant agrees with the use of partial fractions but suggests that a substitution involving tanh could lead to a quicker solution.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to solve the integral, as multiple approaches are proposed and discussed. There is acknowledgment of different techniques, but no agreement on a single preferred method.

Contextual Notes

Some participants' methods depend on specific substitutions and transformations, which may not be universally applicable. The discussion includes various assumptions about the integral's form and the conditions under which different solutions may be valid.

Who May Find This Useful

Students or individuals seeking assistance with integration techniques, particularly those involving hyperbolic functions and substitution methods, may find this discussion beneficial.

Keval
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Question is in orange, answer is in black.
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I got no idea how they got this answer :\

The way I am trying is using a substition of [itex]kx^2 = gsin\theta[/itex]

Just the one question i can't get my head aroung in this exercise step by step method would be appreciated
 
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Hi Keval! :smile:

Not sin, but tanh … try substituting x√(k/g) = tanhu :wink:
 
With denominator instead k x^2 + g you get an arctan answer, right? Use the same steps on this one and get an atanh answer.
 
someone check this out for me ??

[itex]-\frac{1}{k} \int \frac{dx}{x^2-\frac{g}{k}}[/itex]


to simplify i let[itex]\frac{g}{k}=m^2[/itex] to get


[itex]-\frac{1}{k}\int \frac{dx}{x^2-m^2}=-\frac{1}{k}\int \frac{dx}{(x-m)(x+m)}[/itex]


Using partial fractions i got
[itex]\int \frac{dx}{x^2-m^2}=\int \frac{\frac{1}{2m}}{x-m}dx+\frac{-\frac{1}{2m}}{x+m}dx=[/itex]



[itex]\frac{1}{2m}\ln|x-m|-\frac{1}{2m}\ln|x+m|=\frac{1}{2m}\ln \left| \frac{x-m}{x+m}\right|=[/itex]

[itex]-\frac{1}{m}\cdot \frac{1}{2}\ln \left|\frac{x+m}{x-m} \right|=-\frac{1}{m}\tanh^{-1}\left( \frac{x}{m}\right)[/itex]

[itex]\frac{1}{km}\tanh^{-1}\left( \frac{x}{m}\right)=\frac{1}{\sqrt{gk}}\tanh^{-1}\left(\sqrt{\frac{k}{g}}x \right)[/itex]
 
Hi Keval! :wink:

Yes, your partial fractions method is fine. :smile:

But it would be far quicker to start "let x√(k/g) = tanhu, dx√(k/g) = sech2u du" …

try it! :smile:
 

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