How Do You Solve the Integral of 1/(y+cos(x))^2?

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Discussion Overview

The discussion revolves around solving the integral of \( \frac{1}{(y+\cos(x))^2} \) after initially solving \( \frac{1}{y+\cos(x)} \) using a substitution method. Participants explore various approaches to tackle the integral, including integration by parts and substitutions, while clarifying the nature of the variable \( y \).

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Technical explanation

Main Points Raised

  • One participant describes solving the integral \( \frac{1}{y+\cos(x)} \) using the substitution \( t=\tan(x/2) \) and arrives at a specific expression.
  • Another participant questions whether \( y \) is a fixed number or a function of \( x \), clarifying that \( y \) is indeed a fixed number.
  • Several participants suggest using the substitution \( t=\tan{\frac{x}{2}} \) to simplify the integral further.
  • One participant notes the relationship \( \cos{x}=\frac{1-t^{2}}{1+t^{2}} \) as part of the substitution process.
  • A participant expresses that they have derived two integrals, one being twice the original and the other expressed in terms of \( t \) and \( y \).
  • Another participant summarizes their findings, relating the original integral \( Q \) to a new expression involving \( y \) and \( Q \).

Areas of Agreement / Disagreement

Participants generally agree on the approach of using the substitution \( t=\tan(x/2) \) and that \( y \) is a fixed number. However, there is no consensus on the final form of the integral or the correctness of the derived expressions, as participants are still exploring different methods and results.

Contextual Notes

Participants have not resolved all mathematical steps, and there are indications of uncertainty regarding the manipulation of integrals and the relationships between variables. The discussion reflects ongoing exploration rather than definitive conclusions.

Needhelp62
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First part of the question was to work out the integral 1/(y+cos(x)) between x=0 and x=pi/2 by using the substitution t=tan(x/2).
Got this to be \frac{2}{\sqrt{y^2-1}}arctan(\sqrt{\frac{y-1}{y+1}})
The next question says HENCE find integral with the same limits of \frac{1}{(y+cos(x))^2}
Ive tried using by parts to maybe get something minus the first integral but i can't get it to work. This exam is known for being full of horrible questions so anything is possible here
 
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Hi, but ##y## is a fixed number or is a function ##y(x)##?
 
Ssnow said:
Hi, but ##y## is a fixed number or is a function ##y(x)##?
y is just a fixed number the question says the answer to the integral is a function of y
 
ok, in the risolution you must treat ##y## as a number and try with the substitution ##t=\tan{\frac{x}{2}}##. You will obtain an integral that is the ratio of two polynomials in ##t## ...
 
remember that ##\cos{x}=\frac{1-t^{2}}{1+t^{2}}##...
 
Ssnow said:
remember that ##\cos{x}=\frac{1-t^{2}}{1+t^{2}}##...
Thank you I get how the original integral i used now.I have two integrals with the first being twice the original and the second being
\frac{2t^2}{(y+1)+t^2(y-1)}
 
Needhelp62 said:
Thank you I get how the original integral i used now.I have two integrals with the first being twice the original and the second being
\frac{2t^2}{(y+1)+t^2(y-1)}
think I've got it, if i call the original integral Q so
(2-\frac{2}{y+1})Q+\frac{2}{y+1}
 

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