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    (Original post by posthumus)
    What answer did you get ?
    -3.5
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    (Original post by otrivine)
    wow! wait, can i now just use substitution, my integration by parts was the same as urs but do not get then how you or why you considered I?:confused:
    You don't have to use I, it just makes it easier.

    You let I = \displaystyle\int e^{3x} cos 2x \ dx

    You have:

    \displaystyle\int e^{3x} cos 2x \ dx = \dfrac{1}{3}e^{3x} \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x - \dfrac{4}{9} \displaystyle\int e^{3x} \cos 2x dx

    If you notice, you have the same function to be integrated on both sides. So you can replace them both with "I":

    I = \dfrac{1}{3}e^{3x} \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x -\dfrac{4}{9}I

    Then, because I = \displaystyle\int e^{3x} cos 2x \ dx (i.e. the question) you just make "I" the subject of the formula.
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    (Original post by otrivine)
    they got 12x/y but should there not be a y with 12x?
    If it's 12x/y, then it's 12x/y... the question is still doable.
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    (Original post by posthumus)
    Its quite unlikely to come up if you ask me - I don't think it has ever before... where did you get the question from ?
    It's never come up, no, and a question like that probably won't to be honest. I got it from my maths teacher who got it from a really old (like 10-15 years old) A-Level maths textbook.
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    (Original post by usycool1)
    It's never come up, no, and a question like that probably won't to be honest. I got it from my maths teacher who got it from a really old (like 10-15 years old) A-Level maths textbook.
    I just realised that the textbook I have has a whole section on integrals of the form e^{ax}cos(bx) and e^{ax}sin(bx) .
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    (Original post by brittanna)
    I just realised that the textbook I have has a whole section on integrals of the form e^{ax}cos(bx) and e^{ax}sin(bx) .
    I see! I don't have any questions like that in my textbook
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    someone please help me? Name:  maths.jpg
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    (Original post by samfreak)
    someone please help me? Name:  maths.jpg
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    Nah solutionbank is right

    (Original post by brittanna)
    I just realised that the textbook I have has a whole section on integrals of the form e^{ax}cos(bx) and e^{ax}sin(bx) .
    which book you using i dont see it in the green book?
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    (Original post by samfreak)
    someone please help me? Name:  maths.jpg
Views: 95
Size:  51.5 KBName:  maths.jpg
Views: 95
Size:  51.5 KB
    No.

    Notice that the limits of the integral changed.

    \displaystyle \int_a^b u \mathrm{ \ } dx =  -\int_b^a u \mathrm{ \ } dx
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    (Original post by justinawe)
    No.

    Notice that the limits of the integral changed.

    \displaystyle \int_a^b u \mathrm{ \ } dx =  -\int_b^a u \mathrm{ \ } dx
    Hi, in connecting rates, what is r/h in triangle? is this a formula for finiding what ? thanks
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    (Original post by justinawe)
    No.

    Notice that the limits of the integral changed.

    \displaystyle \int_a^b u \mathrm{ \ } dx =  -\int_b^a u \mathrm{ \ } dx
    where have they chnaged?
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    (Original post by QwertyG)
    Nah solutionbank is right



    which book you using i dont see it in the green book?
    I have an older one I think. It says:
    Advanced Maths A2 core for edexcel.

    It's this one here:

    Name:  a2-core-mathematics-for-edexcel-2076-p.jpg
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    (Original post by samfreak)
    where have they chnaged?
    On the second line (the line beginning with A_1=) it was \int^0_\frac{pi}{4} y dx whereas on the line after it was \int^\frac{pi}{4}_0 y dx
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    (Original post by brittanna)
    On the second line (the line beginning with A_1=) it was \int^0_\frac{pi}{4} y dx whereas on the line after it was \int^\frac{pi}{4}_0 y dx
    Hi In cone why do we use r/h ? and what is it in triangle
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    (Original post by otrivine)
    Hi In cone why do we use r/h ? and what is it in triangle
    I don't think I know what you mean. Have you get an example of a question that uses it?
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    *cries*
    had to do a C4 paper as a test in class, i knew what paper it was gona be and got 51%

    ha hahahalgnsrlkgnsrin
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    (Original post by brittanna)
    I don't think I know what you mean. Have you get an example of a question that uses it?
    http://www.examsolutions.net/maths-r...tutorial-3.php

    this tutor uses that r/h formula? thanks
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    (Original post by otrivine)
    http://www.examsolutions.net/maths-r...tutorial-3.php

    this tutor uses that r/h formula? thanks
    It's just the way similar shapes are defined (from GCSE). So in this case, the ratio of the height to the radius is the same in each of the triangles he looked at. I guess triangles would be the same as it was a triangle he was looking at in the cone question.
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    (Original post by brittanna)
    It's just the way similar shapes are defined (from GCSE). So in this case, the ratio of the height to the radius is the same in each of the triangles he looked at. I guess triangles would be the same as it was a triangle he was looking at in the cone question.
    oh so what does r/h calculate ?
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    (Original post by otrivine)
    oh so what does r/h calculate ?
    It's just the ratio of of the length and height of the triangle, it doesn't represent any physical quantity. But because the two triangles are mathematically similar, it gives us an extra equation we can use.

    Do you remember doing similar triangles at GCSE?
 
 
 
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