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Integrating cot^2 2x dx. Rep for help. watch

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    Hi
    Can anyone integrate cot^2 2x and show me how you did it please? The answer in the book is -x - 1/2 cot x + c.

    Rep for the first/most helpful answer

    Rosie <== crap at p3
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    (Original post by crana)
    Hi
    Can anyone integrate cot^2 2x and show me how you did it please? The answer in the book is -x - 1/2 cot x + c.

    Rep for the first/most helpful answer

    Rosie <== crap at p3
    Hang on a sec... I've got the answer.
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    Dividing

    sin^2(2x) + cos^2(2x) = 1

    by sin^2(2x), we get

    1 + cot^2(2x) = cosec^2(2x).

    So we want to integrate -1 + cosec^2(2x). Most A-level formula booklets (eg, http://www.aqa.org.uk/admin/library/gce-maths-bklet.pdf) tell you that the derivative of cot(x) is -cosec^2(x). By the chain rule it follows that the derivative of cot(2x) is -2 cosec^2(2x). So the integral of cosec^2(2x) is -(1/2) cot(2x) + constant.

    (int) cot^2(2x) dx
    = (int) -1 + cosec^2(2x) dx
    = -x - (1/2) cot(2x) + constant.
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    (Original post by crana)
    Hi
    Can anyone integrate cot^2 2x and show me how you did it please? The answer in the book is -x - 1/2 cot x + c.

    Rep for the first/most helpful answer

    Rosie <== crap at p3
    ∫cot²(2x) dx

    = ∫(cos²(2x))/(sin²(2x)) dx

    = ∫cos(2x) * (cos(2x))/(sin²(2x)) dx

    = 1/2 * ∫cos(2x) * (sin(2x))' * (sin(2x))^(-2) dx

    =...
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    ∫cot^2(2x) dx

    let u = 2x

    ∫cot^2(u) dx

    By definition:

    cot^2(u) = cosec^2(u) -1

    so:

    ∫cosec^2(u) -1 dx

    du/dx = 2

    so dx = du/2

    (1/2)∫cosec^2(u) -1 du

    given in the formula book:

    "derivative of cot(@) is -cosec^2(@)"
    (ie cosec^2(@) integrates to -cot(@) )

    so integral goes to:

    (1/2).(-cot(u)) - u + C

    = -(1/2)cot(2x) - x + C

    Hope this helps
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    (Original post by zizero)
    ∫cot²(2x) dx

    = ∫(cos²(2x))/(sin²(2x)) dx

    = ∫cos(2x) * (cos(2x))/(sin²(2x)) dx

    = 1/2 * ∫cos(2x) * (sin(2x))' * (sin(2x))^(-2) dx

    =...
    f(x) = cos(2x) ; g'(x) = (sin(2x))' * (sin(2x))^(-2)
    f'(x) = 2sin(2x) ; g(x) = -(sin(2x))^(-1) = -1/(sin(2x))

    ...
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    (Original post by zizero)
    f(x) = cos(2x) ; g'(x) = (sin(2x))' * (sin(2x))^(-2)
    f'(x) = 2sin(2x) ; g(x) = -(sin(2x))^(-1) = -1/(sin(2x))

    ...
    So, we get: -0,5*(cos(2x))/(sin2x) - 0,5*2*∫sin(2x)/sin(2x) dx

    = -1/2 * cot2x - x + k

    Cheers everyone

    ill distribute some rep as soon as I can again
    rosie
 
 
 
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