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    ∫ 1/(3+5cosx) dx

    Can someone please show me how to do this as i'm having a brain freeze!
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    Try using the substitution t = tan(x/2) ....
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    (Original post by mockel)
    Try using the substitution t = tan(x/2) ....
    In which case

    cosx = (1-t^2)/(1+t^2)

    and

    dx = 2dt/(1+t^2)
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    am i supposed to know what substitution to use as the book didnt give that?!
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    Well, whenever you see something of that form (i.e. trig functions) and nothing obvious comes to mind, then you should use this.
    Guess that's not too much help...
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    It might be because this is assumed knowledge from P4? Its worth knowing what dx comes to instead of having to work it out each time.
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    (Original post by jmzcherry)
    am i supposed to know what substitution to use as the book didnt give that?!
    I remember asking the same question about this same substitution. It appears in an example somewhere in Heinemann edexcel P5, I think, but it would be useful if they had some of these standard substitutions in the summaries at the end of the chapters.

    As an alternative, print a copy of dvs's notes from this thread:

    http://www.thestudentroom.co.uk/t116149.html

    Aitch
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    The substitution is somewhere in the formula book. In the trigonometry section with all trig functions represented in terms of t.

    Its used as said in complicated trig integration when you cant think what to do
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    how about the integration of 1/(1+x^4)
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    (Original post by ~*Joe*~)
    how about the integration of 1/(1+x^4)
    Easy.

    Aitch
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    (Original post by Aitch)
    Easy.

    Aitch

    From

    http://www.calc101.com/webMathematica/integrals.jsp

    by the way...

    Aitch
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    Ah, now I can see my mistake with that one...

    ...looking at it
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    (Original post by ~*Joe*~)
    how about the integration of 1/(1+x^4)
    1 + x^4 = (1 + sqrt[2]x + x^2)(1 - sqrt[2]x + x^2)

    Partial fractions will get you to what Aitch posted.
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    (Original post by Aitch)
    Easy.

    Aitch
    Why cant you do:

    ∫1/(1+x^4) dx
    ∫1/(1+(x²)²) dx

    => (arctan(x²))/2x + C

    ?
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    Isn't that like total illegals?
    I could then ask, "why can't you do INT 1/(1+x^2) dx, as ln(x^2) ?"...

    Wait, why did you divide by 2x?
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    (Original post by Feria)
    Why cant you do:

    ∫1/(1+x^4) dx
    ∫1/(1+(x²)²) dx

    => (arctan(x²))/2x + C

    ?
    When you do that you're basically using the substitution u=x² incompletely.
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    Oh yeah, I forgot there wasnt any x initially... just me making stupid observational mistakes... ignore me :P
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    (Original post by Womble548)
    The substitution is somewhere in the formula book. In the trigonometry section with all trig functions represented in terms of t.
    It seems to be in the AQA formula book but not in the Edexcel one. (Unless I've missed it somehow.)

    How strange.

    Aitch
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    Aitch, it definitely is.
    Look in the Trigonometric Identities section, and under all the sin(A±B) bumph, you get:
    t = tan(A/2) : sinA = 2t/(1+t²) , cosA = (1-t²)/(1+t²)

    They don't give you dx, but that's pretty quick to derive.
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    (Original post by mockel)
    Aitch, it definitely is.
    Look in the Trigonometric Identities section, and under all the sin(A±B) bumph, you get:
    t = tan(A/2) : sinA = 2t/(1+t²) , cosA = (1-t²)/(1+t²)

    They don't give you dx, but that's pretty quick to derive.
    This is what I've got (in paper form), and I still can't find the above! I can find them in the AQA version OK. Perhaps I'm going mad!

    http://www.edexcel.org.uk/VirtualCon...cification.pdf

    I've learned them anyway...

    Aitch
 
 
 
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