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    Hello ,
    thanks for all help in advance!


    How do we integrate this :

    1/cosx


    I thought that is was going to be ln "mod" cosx "mod"

    But then it is not........:confused::confused::confused::confused: how to integrate this, there are no rules in out edexcel c4 book on this...then why does the exam board expect us to know how to integrate this?
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    If you multiply by (sec x + tan x) / (sec x + tan x) the integrand becomes

    ( (sec x)^2 + sec x tan x ) / (sec x + tan x) dx

    Let u = tan x + sec x, then du = (sec x)^2 + sec x tan x dx

    The result of the substitution is 1/u du. Integrate to get ln (abs(u)) +C

    = ln( sec x + tan x) +C
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    (Original post by laurawoods)
    Hello ,
    thanks for all help in advance!


    How do we integrate this :

    1/cosx


    I thought that is was going to be ln "mod" cosx "mod"

    But then it is not........:confused::confused::confused::confused: how to integrate this, there are no rules in out edexcel c4 book on this...then why does the exam board expect us to know how to integrate this?
    Multiply top and bottom by

    (\sec x + \tan x)

    to get

    \displaystyle \int \dfrac{\sec x (\sec x + \tan x)}{\sec x + \tan x} dx

    This is not obvious, but you can now apply a standard log result.
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    (Original post by Indeterminate)
    Multiply top and bottom by

    (\sec x + \tan x)

    to get

    \displaystyle \int \dfrac{\sec x (\sec x + \tan x)}{\sec x + \tan x} dx

    This is not obvious, but you can now apply a standard log result.
    hello, there sorry i still dont seem to be understanding where did we get "secx+tanx " from and wouldnt they cacel each other out?
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    (Original post by laurawoods)
    hello, there sorry i still dont seem to be understanding where did we get "secx+tanx " from and wouldnt they cacel each other out?
    Well, yes they would, but then you'd end up at square one.

    Expand the numerator to find that it is the derivative of the denominator. You can then write down the answer.

    Oh, and we don't get it from anywhere special as it's one of those things that works.
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    (Original post by Indeterminate)
    Well, yes they would, but then you'd end up at square one.

    Expand the numerator to find that it is the derivative of the denominator. You can then write down the answer.

    Oh, and we don't get it from anywhere special as it's one of those things that works.
    hello sorry i am not very clever tto understand that ...pls can u help me to understand where the sex + tan x thing is coming from ...where did u pull it out suddenly from... oh i am not understanding...
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    here...Name:  secx.png
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Size:  8.7 KB
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    (Original post by laurawoods)
    hello sorry i am not very clever tto understand that ...pls can u help me to understand where the sex + tan x thing is coming from ...where did u pull it out suddenly from... oh i am not understanding...
    Well, first of all,

    \dfrac{\sec x + \tan x}{\sec x + \tan x} = 1

    and multiplying something by 1 doesn't change it.

    It just so happens that if you multiply \sec x by this version of 1, you get something that integrates to a log function.
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    (Original post by patterson)
    here...Name:  secx.png
Views: 86
Size:  8.7 KB
    doesnt say in the C4 EDEXCEL BOOK though!
    i am going to literally break out in tears now!
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    (Original post by Indeterminate)
    Well, first of all,

    \dfrac{\sec x + \tan x}{\sec x + \tan x} = 1

    and multiplying something by 1 doesn't change it.

    It just so happens that if you multiply \sec x by this version of 1, you get something that integrates to a log function.
    sorry ..byt i dont really know where this secx+tanx comes from IN THE FIRST PLACE. i simply dont seem to be getting it .... where does it come from , oh god?
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    An alternative method:

    \displaystyle \frac{1}{\cos x} = \frac{\cos x}{\cos^2 x} = \frac{\cos x}{1-\sin^2 x} = \frac{\cos x}{(1+\sin x)(1-\sin x)}

    Then substitute u=\sin x and split up using partial fractions.
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    (Original post by laurawoods)
    doesnt say in the C4 EDEXCEL BOOK though!
    i am going to literally break out in tears now!
    What board do you use?

    I doubt you'd be expected to integrate sec(x) from scratch in a C4 exam.
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    (Original post by notnek)
    What board do you use?

    I doubt you'd be expected to integrate sec(x) from scratch in a C4 exam.
    hello there, we are using the EDEXCEL board,,,which one for u ?
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    It's just one of those things that you aren't expected to know unless you're already familiar with it. It does make life a whole lot easier, though.
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    (Original post by laurawoods)
    sorry ..byt i dont really know where this secx+tanx comes from IN THE FIRST PLACE. i simply dont seem to be getting it .... where does it come from , oh god?
    It's just a trick that works, but you could try the substitution

    u = \tan(\frac{x}{2})

    to convince yourself.
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    (Original post by Indeterminate)
    It's just a trick that works, but you could try the substitution

    u = \tan(\frac{x}{2})

    to convince yourself.
    hello i am doing the edexcel board , do you think in the exam they will expect us to dervie this thing from scratch ?

    thanks ))
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    (Original post by laurawoods)
    hello there, we are using the EDEXCEL board,,,which one for u ?
    \displaystyle \int \sec x \ dx = \ln |\sec x +\tan x|+c

    This is in the Edexcel formula book.
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    (Original post by laurawoods)
    hello i am doing the edexcel board , do you think in the exam they will expect us to dervie this thing from scratch ?

    thanks ))
    Not a chance, but if only they would :rolleyes:
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    (Original post by aznkid66)
    It's just one of those things that you aren't expected to know unless you're already familiar with it. It does make life a whole lot easier, though.
    hello i am doing the edexcel board , do you think in the exam they will expect us to dervie this thing from scratch ?
    thanks
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    (Original post by notnek)
    \displaystyle \int \sec x \ dx = \ln |\sec x +\tan x|+c

    This is in the Edexcel formula book.
    sorry for all the disturbances caused... i discovered it now! thank you very much guys!
 
 
 
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