Integrate 1/(y+cos(x))^2 dx

  • #1

Main Question or Discussion Point

First part of the question was to work out the integral [tex]1/(y+cos(x))[/tex] between x=0 and x=pi/2 by using the substitution [tex]t=tan(x/2).[/tex]
Got this to be [tex]\frac{2}{\sqrt{y^2-1}}arctan(\sqrt{\frac{y-1}{y+1}})[/tex]
The next question says HENCE find integral with the same limits of [tex]\frac{1}{(y+cos(x))^2}[/tex]
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
 

Answers and Replies

  • #2
Ssnow
Gold Member
509
149
Hi, but ##y## is a fixed number or is a function ##y(x)##?
 
  • #4
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
 
  • #5
Ssnow
Gold Member
509
149
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## ...
 
  • #6
Ssnow
Gold Member
509
149
remember that ##\cos{x}=\frac{1-t^{2}}{1+t^{2}}##...
 
  • #7
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
[tex]\frac{2t^2}{(y+1)+t^2(y-1)}[/tex]
 
  • #8
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
[tex]\frac{2t^2}{(y+1)+t^2(y-1)}[/tex]
think ive got it, if i call the original integral Q so
[tex](2-\frac{2}{y+1})Q+\frac{2}{y+1} [/tex]
 

Related Threads on Integrate 1/(y+cos(x))^2 dx

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
12
Views
690K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
18
Views
268K
Replies
2
Views
2K
  • Last Post
Replies
6
Views
7K
Replies
11
Views
2K
Replies
5
Views
17K
Replies
4
Views
799
Top