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

  1. Apr 23, 2016 #1
    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
     
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  3. Apr 23, 2016 #2

    Ssnow

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    Hi, but ##y## is a fixed number or is a function ##y(x)##?
     
  4. Apr 23, 2016 #3

    Math_QED

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  5. Apr 23, 2016 #4
    y is just a fixed number the question says the answer to the integral is a function of y
     
  6. Apr 23, 2016 #5

    Ssnow

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    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## ...
     
  7. Apr 23, 2016 #6

    Ssnow

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    remember that ##\cos{x}=\frac{1-t^{2}}{1+t^{2}}##...
     
  8. Apr 23, 2016 #7
    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]
     
  9. Apr 23, 2016 #8
    think ive got it, if i call the original integral Q so
    [tex](2-\frac{2}{y+1})Q+\frac{2}{y+1} [/tex]
     
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