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    (Original post by *dave*)
    I have NEVER answered a question about arctanx or arccosx or all of those. I swear to god if those come up i am buggered.
    It won't come up, but if you are interested the differentiation of such functions is fairly simple. For example, if you want to find the derivive of arcsin(x) you would do this:

    y = arcsin(x)
    sin(y) = x

    Then, differentiating w.r.t. x

    cos(y).dy/dx = 1
    dy/dx = 1/cos(y)

    Now, we have

    sin^2(y) + cos^2(y) = 1
    cos^2(y) = 1 - sin^2(y)
    cos(y) = sqrt[ 1 - sin^2(y) ]

    and since sin(y) = x we have

    cos(y) = sqrt( 1 - x^2 )

    therefore

    dy/dx = d( arcsin(x) )/dx = 1/sqrt(1-x^2)

    If you are interested you can use similar methods to find the derivitives of arccos(x) and arctan(x).
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    (Original post by fishpaste)
    Much respect to you for getting that far. Well done.
    BTW, I answered the question you posted here: http://www.uk-learning.net/t46254.html.
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    Here is another nasty integral if anybody wants to spend their time churning out the algebra:

    INT 1/(1+x^3) dx
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    (Original post by mikesgt2)
    Here is another nasty integral if anybody wants to spend their time churning out the algebra:

    INT 1/(1+x^3) dx
    Any hints?
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    (Original post by Bhaal85)
    Any hints?
    Well, you will need to factorise the denominator. However, you should know that this integral is a bit beyond P3 since you will need arctan integrals.
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    (Original post by mikesgt2)
    Well, you will need to factorise the denominator. However, you should know that this integral is a bit beyond P3 since you will need arctan integrals.
    I only went up to P3, what is x^3+1 factorised? Is there a rule for factorising cubics quickly? I had to use some P2 stuff, and equate like terms:

    1/[(x+1)(x^2-x+1)] ?
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    (Original post by mikesgt2)
    Well, you will need to factorise the denominator. However, you should know that this integral is a bit beyond P3 since you will need arctan integrals.

    I got as far as 1/3ln(x+1) + 1/3int.(2-x)/(x^2-x+1)

    (although its probably littered with mistakes as I'm too tired to work effectively at this hour).
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    (Original post by Bhaal85)
    I only went up to P3, what is x^3+1 factorised? Is there a rule for factorising cubics quickly? I had to use some P2 stuff, and equate like terms:

    1/[(x+1)(x^2-x+1)] ?
    Then use partial fractions.
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    (Original post by mikesgt2)
    Then use partial fractions.
    1/3(x+1)^-1 + 2/3(x^2-x+1)^-1

    Is it assumed that x=sinx? I have never been told that before, the only reason I mention this is that you said something about arcsin?
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    (Original post by Bhaal85)
    1/3(x+1)^-1 + 2/3(x^2-x+1)^-1

    Is it assumed that x=sinx? I have never been told that before, the only reason I mention this is that you said something about arcsin?
    Im not sure your partial fractions are correct. The numerator of the second fraction should have an x-coefficient.
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    (Original post by Ralfskini)
    I got as far as 1/3ln(x+1) + int.(2-x)/(x^2-x+1)

    (although its probably littered with mistakes as I'm too tired to work effectively at this hour).
    This is correct so far, but I think the integral should also be multipled by 1/3.

    (Original post by Bhaal85)
    1/3(x+1)^-1 + 2/3(x^2-x+1)^-1

    Is it assumed that x=sinx? I have never been told that before, the only reason I mention this is that you said something about arcsin?
    Your partial fractions are not correct, you need to find constants A,B,C such that

    1/[(x+1)(x^2-x+1)] = A/(x+1) + (Bx+C)/(x^2-x+1)
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    (Original post by mikesgt2)
    This is correct so far, but I think the integral should also be multipled by 1/3.


    Your partial fractions are not correct, you need to find constants A,B,C such that

    1/[(x+1)(x^2-x+1)] = A/(x+1) + (Bx+C)/(x^2-x+1)
    I think this must be more advanced BE. I just assumed that I could do:

    A/(x+1) + B/(x^2-x+1) alas, not that easy then. Oh well.
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    Peter, non linear partial fractions DO come up, make sure you're familiar with them, along with things like quadratic denominators.

    E.g.

    1/(x-1)(x+3)^2 into partial fractions.
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    (Original post by fishpaste)
    Peter, non linear partial fractions DO come up, make sure you're familiar with them, along with things like quadratic denominators.

    E.g.

    1/(x-1)(x+3)^2 into partial fractions.
    I am aware that that would be in the form:

    1/(x-1)(x+3)^2 = A/(x-1) + B/(x+3) + C/(x+3)^2

    However:

    1/[(x+1)(x^2-x+1)] looked dissimiliar as the quadratic is inside the bracket, which is why I treated it the same, this is somewhat confusing.
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    (Original post by mik1a)
    Some difficult P1, as I can't ask questions I don't understand!

    1. A curve has the equation:

    x = 4y^2 - 9
    Find the area between the curve and the line x = 1

    Find the minimum value of x, and the value of y that gives this.

    When y = 17, and c is set to 9, find the equations of the normal to the curve and find where it meets the x axis.
    Total area is twice area under curve from x=-9 to x=1.

    A=∫{-9 -> 1}y dx
    = ½ ∫ (x+9)^½ dx
    = (1/2)(2/3)[(x+9)^(3/2)]{-9 -> 1}
    = (1/3){(10)^(3/2) - (0)^(3/2)}
    = (1/3)(10)^(3/2)
    = 10.54
    2A = 21.08
    =======

    From the graph,
    min x = -9 @ y=0
    ======= ===

    I'm not sure what you mean by 'c', but

    at y=17, x = 4*17² - 9 = 1147

    gives P(1147,17), a point on the curve.

    dy/dx = 1/(dx/dy) = 1/(8y)
    => m = -8y, for normal
    ========

    (y-y1) = m(x-x1)
    y - 17 = -8*17(x - 1147)
    y + 136x = 156,009, eqn of normal at P
    =============

    This normal meets x-axis, when y=0, giving
    x = 156,009/17
    x = 9,177
    ======
    Attached Images
     
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    (Original post by mikesgt2)
    I get this nasty thing:

    [ ln(x^2 + 2x + 2) - ln(x^2 - 2x + 2) + 2arctan(x+1) + 2arctan(x-1) ]/16 + C
    It seems that it is necessary to use the substitutions (x+1)^2 = tan(beta) and (x-1)^2 = tan(alpha), right?
    But does it appear in P3?
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    (Original post by keisiuho)
    It seems that it is necessary to use the substitutions (x+1)^2 = tan(beta) and (x-1)^2 = tan(alpha), right?
    But does it appear in P3?
    No.
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    (Original post by Bhaal85)
    No.
    It doesnt appear in P3 or the substitutions arent needed?
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    (Original post by keisiuho)
    It doesnt appear in P3 or the substitutions arent needed?
    Yes.
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    (Original post by Bhaal85)
    Yes.
    Postcount + 1

    no you wont need to do anything with inverse trig functions in p3 (edexcel)
 
 
 
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