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    How do i integrate these? Ive tried but havent a clue.


    sin^2(x)



    cos^2(x)


    (Its only an C3 AQA question btw)

    thanks
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    [QUOTE=thereddevil]How do i integrate these? Ive tried but havent a clue.


    sin^2(x)



    cos^2(x)


    (Its only an C3 AQA question btw)

    ve just realised how stupid i was.
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    Note that cos2x = cos^2x - sin^2x. Can you see how to use that in order to integrate them?
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    Use double angle formulas

    cos(2x) = 2[cos(x)]^2 - 1

    cos(2x) = 1 - 2[sin(x)]^2

    rearrange so you've got sin square x and cos square x in terms of cos(2x) and then they integrate easily
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    (Original post by thereddevil)
    How do i integrate these? Ive tried but havent a clue.


    sin^2(x)



    cos^2(x)


    (Its only an C3 AQA question btw)

    thanks
    -1/2(cos^2x)

    1/2(sin^2x)

    Just treat it as normal but becareful with the 2(x)
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    (Original post by studyboy)
    -1/2(cos^2x)

    1/2(sin^2x)

    Just treat it as normal but becareful with the 2(x)
    I dont think its as simple as that.
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    (Original post by studyboy)
    -1/2(cos^2x)

    1/2(sin^2x)

    Just treat it as normal but becareful with the 2(x)
    I think you're confusing sin^2x with sin(2x). The sin^2x notation means the former.
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    (Original post by wrockite)
    Use double angle formulas

    cos(2x) = 2[cos(x)]^2 - 1

    cos(2x) = 1 - 2[sin(x)]^2

    rearrange so you've got sin square x and cos square x in terms of cos(2x) and then they integrate easily
    Core 3 qst so none of that....

    the qst was find volume of rev thru 360

    between y=0 to y=pi (not sure tho)

    y = sin^(-1)x

    so x=siny

    V=pi (Integrate) (siny)^2 dy...............right?

    so basically i think its integrate sin^2(x)
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    When you get to Core4 Edexcel (I don;t know if the books are the sameish for AQA) you find that the integral of Sin(^2)x dx = the integral of (1/2 - 1/2cos2x)dx as:

    Cos2x = 1 - 2Sin(^2)x
    So sin(^2)x = 1/2(1 - Cos2x)
    So the integral of Sin(^2)dx = (1/2 - 1/2Cos2x)dx = 1/2x - 1/4Sin2x + C

    Hope that helps with the Sin(^2)x one
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    The integral of Cos(^2)x = 1/2x + 1/4Sin2x + C but i cannot find the reason as yet, sorry. Hope this is of some help to you.
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    sin squared A = half(1-cos2A)

    cos squared A = half(1 + cos2A)
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    I know all that sin(2x) stuff but i think that i must have used the wrong equation. (my previous post)
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    (Original post by thereddevil)
    I know all that sin(2x) stuff but i think that i must have used the wrong equation. (my previous post)
    you should be able to integrate from what i gave.

    if it helps expand the right hand side (thus getting rid of the brackets) and it may be easier
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    (Original post by hermaphrodite)
    you should be able to integrate from what i gave.

    if it helps expand the right hand side (thus getting rid of the brackets) and it may be easier
    but this is C3 not C4 or FP2
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    give me strength.
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    (Original post by Chewwy)
    give me strength.
    sure why not.........in a maths forum?
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    (Original post by Swayum)
    I think you're confusing sin^2x with sin(2x). The sin^2x notation means the former.
    Oh!! I see :p: :p: :p: My mistake! If this is the case then just use the double angle formula
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    (Original post by thereddevil)
    I dont think its as simple as that.
    I'm sorry, I got confused with something else.:p:
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    This is the most confusing thread I've seen on here... and it's not the maths that's confusing.
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    (Original post by kexy)
    This is the most confusing thread I've seen on here... and it's not the maths that's confusing.
    what?

    lol

    youre confusing!!
 
 
 
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