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    how do i intergrate

    sin2xcosx ?

    thanks
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    There's a thread just below this one of exactly the same nature.
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    sin2x = 2sinxcosx

    => (2) INT (sinx)(cosx)^2 dx

    = (-2) INT (-sinx)(cosx)^2 dx

    Now use the fact that: INT [f '(x)].[f(x)]n dx = [1/(n+1)][f(x)]n+1 + c

    = (-2/3)cos3 + c
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    (Original post by sarah12345)
    how do i intergrate

    sin2xcosx ?

    thanks
    is this perceived as being the hardest integral or what - quite a few threads asking about that particular question :rolleyes:

    use fact that:INT [f '(x)].[f(x)]n dx = [1/(n+1)][f(x)]n+1 + c, as said by mockel - its basically recognition - a skill you'll pick up with practice

    pk
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    i got one quite complicated
    intergate cosx*sqrt(cos2x)
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    (Original post by n_e_m)
    i got one quite complicated
    intergate cosx*sqrt(cos2x)
    Write cos2x = 1 - 2sin^2 x and then sub u = sinx

    Then write u = 1/rt(2) sint say
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    (Original post by n_e_m)
    i got one quite complicated
    intergate cosx*sqrt(cos2x)
    cos2x = 1 - 2sin^2x

    Let u = sinx -> 2u^2 = 2sin^2x -> 1 - 2u^2 = 1 - 2sin^2x
    Also: du/dx = cosx -> dx = 1/cosx du

    -> Int. (cosx).Sqrt[cos2x] dx = Int. cosx.Sqrt[1 - 2u^2].(1/cosx) du. = Int. Sqrt(1 - u^2) du. = Int. (1 - u^2)^(1/2) du. = [-1/(2u)](2/3)(1 - u^2)^(3/2) + k = -[1/(3u)](1 - u^2)^(3/2) + k = -[1/(3sinx)][cos^2x]^(3/2) + k = -[1/(3sinx)][cos^3x] + k = (-1/3){[cos^2x.cosx]/sinx} + k = (-1/3)(cos^2x/tanx) + k
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    many thanx...
    i wonder if this type of complicated integration could come up on P5 paper?
 
 
 
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