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    Hi,

    just unsure on how to deal with this integral as it deals with 3x instead of 2x, which is what I am used to, and do not know the identities needed.

    Integrate: 4 . cos^2[3x]

    Many Thanks

    Streety
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    cos6x = 2cos^2(3x) - 1
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    (Original post by streetyfatb)
    Hi,

    just unsure on how to deal with this integral as it deals with 3x instead of 2x, which is what I am used to, and do not know the identities needed.

    Integrate: 4 . cos^2[3x]

    Many Thanks

    Streety
    cos2x = cos^2x - sin^2x = cos^2x - (1-cos^2x) = 2cos^2x - 1
    cos^2(x) = 0.5(cos2x + 1)
    therefore
    cos^2(3x) = 0.5(cos6x + 1)
    4cos^2[3x] =2cos6x + 2
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    (Original post by streetyfatb)
    Hi,

    just unsure on how to deal with this integral as it deals with 3x instead of 2x, which is what I am used to, and do not know the identities needed.

    Integrate: 4 . cos^2[3x]

    Many Thanks

    Streety
    Hello,

    I just joined and came across your question; i havn't done it yet, but i think the idea is to use the cos idedtity:

    cos 2x = 2cos^2 x - 1

    letting x = 3x, which gives teh LHS as (cos 6x +1)/2 as being = 2 cos^2 x

    i can't see anything wrong with that, lol
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    (Original post by streetyfatb)
    Hi,

    just unsure on how to deal with this integral as it deals with 3x instead of 2x, which is what I am used to, and do not know the identities needed.

    Integrate: 4 . cos^2[3x]

    Many Thanks

    Streety
    Hello,

    I just joined and came across your question; i havn't done it yet, but i think the idea is to use the cos idedtity:

    cos 2x = 2cos^2 x - 1

    letting x = 3x, which gives teh LHS as (cos 6x +1)/2 as being = 2 cos^2 3x

    i think that the idea
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    déja vu
 
 
 
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