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Integrating Cos(X)Sin(X) three ways. watch

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    Well, as you can tell, its another test i've got today, so my maths ability has dissapeared.
    Please can someone remind me how to integrate this using three methods?
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    what do you mean by three methods?

    a) by parts
    b) using the identity sin(x)cos(x)=sin(2x)/2
    c) changing cos(x) = √(1-sin²(x)) and using a substitution u = sin(x)

    ???????????????????????????????? ??????????
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    (Original post by elpaw)
    what do you mean by three methods?

    a) by parts
    b) using the identity sin(x)cos(x)=sin(2x)/2
    c) changing cos(x) = √(1-sin²(x)) and using a substitution u = sin(x)

    ???????????????????????????????? ??????????
    Thanks.
    Thats exactly what i meant.
    My brain switches off before an exam, so i couldn't remember.
    And i did them last thing yesterday!
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    hmm i need to get back up there with the maths - it's all slipped away from me this year... i couldn't even remember one method of doing that!
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    Just thought i'd give these a go.....

    I = ∫sin(x)cos(x)dx

    (a) by parts

    u = cos(x) v' = sin(x)
    u' = -sin(x) v = -cos(x)
    => I = -cos²(x) -∫sin(x)cos(x)dx = -cos²(x) - I
    => 2I = -cos²(x)
    => I = -cos²(x)/2 +C == sin²(x)/2 +C'

    (b) using the identity sin(x)cos(x)=sin(2x)/2

    I = ∫sin(2x)/2 dx
    = -1/4 cos(2x) +C
    (you can go on):
    I = -1/4 (cos²(x) - sin²(x)) + C
    = -1/4 (2cos²(x) - 1) + C
    = -cos²(x)/2 + C' == sin²(x)/2 + C''

    hence (b) == (a)

    (c) changing cos(x) = √(1-sin²(x)) and using a substitution u = sin(x)

    I = ∫√(1-sin²(x))sin(x)dx

    u = sin(x) => du = cos(x) dx = √(1-u²) dx
    => dx = du/√(1-u²)

    I = ∫√(1-u²)u/√(1-u²) du
    = ∫u du
    = u²/2 + C
    = sin²(x)/2 +C == -cos²(x)/2 +C'

    hence (c) == (b) == (a)
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    surely the easiest way is let u=sinx, then the integral becomes u.du which is 1/2sin^2x + C
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    (Original post by Ralfskini)
    surely the easiest way is let u=sinx, then the integral becomes u.du which is 1/2sin^2x + C
    The easiest way is to do it by inspection, recognising that it's sort of in the form f'(x)f(x). You can just integrate as you would a normal function, checking that you get the negatives correct.

    [Integral]sin(x)cos(x) = (1/2)sin^2(x) + C

    OR

    [Integral]sin(x)cos(x) = -(1/2)cos^2(x) + D

    Ben
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    jus use the substitution u = sinx

    then du = cosx dx

    ∫u du

    u^2 * (1/2)

    (sin^2 x )/2
 
 
 
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