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    Any ideas on how to solve?
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    (Original post by ReubenOrion)
    Any ideas on how to solve?
    It's a separable equation. You can separate it into

    \dfrac{2}{4-y^2}.dy = \tan x .dx and hence integrate both sides.

    Seeing as this is a 'hence' question, you should be able to use part (a) to help with this.
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    (Original post by ReubenOrion)
    Any ideas on how to solve?
    well you separate the variables. once the y bits are on the left you will need to use partial fractions to integrate.

    on the right you will have 1/cotx which can be expressed in a different way. they are giving you a clue as to the method of integration by mentioning sec2

    the 2 can either stay on the left or move to the right as 1/2.
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    (Original post by the bear)
    well you separate the variables. once the y bits are on the left you will need to use partial fractions to integrate.

    on the right you will have 1/cotx which can be expressed in a different way. they are giving you a clue as to the method of integration by mentioning sec2

    the 2 can either stay on the left or move to the right as 1/2.
    I have integrated the partial fraction and 1/2 tanx

    and have gotten

    *ln 2+y / 2-y = 1/2 ln sec x + ln k

    After substitution I have gotten

    ln k = -ln sqrt2

    I then tried to put this back into * and solve for sec x^2 however I am unsure where I have gone wrong as the answer is sec^2x = 8+4y/ 2-y
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    (Original post by username3075842)
    I have integrated the partial fraction and 1/2 tanx

    and have gotten

    *ln 2+y / 2-y = 1/2 ln sec x + ln k
    Doesn't look right. The LHS is missing a 1/4 in front of the ln. The RHS seems OK. The constant term looks OK as well.

    Once you see why you need the 1/4 there, you can mult through by 4 and hence \frac{1}{2} \ln |\sec x| turns into 2\ln | \sec x | = \ln \sec^2 x
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    so you get

    ½∫ 1/{2 + y} + 1/{2 - y} dy ={ln|secx|} + c

    ====> ln √ {(2 + y)/(2 - y)} = {ln|secx|} + c

    now find c using the information given....
 
 
 
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