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    q. Integrate sinxcosx/(cos2x+3)^1/2

    i got sin2x/2*(cos2x+3)^-1/2

    then -cos2x/4*(cos2x+3)^-1/2

    then -(cos2x+3)^1/2 /4*0.5*2

    hence I GOT, -(cos2x+3)^1/2 /4

    but the answer is -(cos2x+3)^1/2 /2 :/
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    Did you remember to divide by \frac{1}{2}?

    (If you use a u-sub of "u = cos2x" you end up having to divide by \frac{1}{2}, you could of missed it out...)
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    (Original post by soutioirsim)
    Did you remember to divide by \frac{1}{2}?
    yh 0.5.....
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    i got the same answer as you
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    (Original post by soutioirsim)
    Did you remember to divide by \frac{1}{2}?

    (If you use a u-sub of "u = cos2x" you end up having to divide by \frac{1}{2}, you could of missed it out...)
    im nt using tht method of u du etc. im jus integrating a section making the trig function then 4rm there.....group and intregrate. i dnt get wt i did wrong in my method lol :/
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    (Original post by cccdude)
    i got the same answer as you
    lol! wt method....reverse chain rule etc.?
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    nah its not by substituion.. its through standard patters

    let Y= (cos2x+3)^1/2

    Differentiate to give -sin2x/(cos2x+3)^-1/2

    Using the identity sin2x-2sinxcosx you will have to divide the function by 2 to get back to the original equation
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    sin2x=2sinxcosx

    Made an error in previous post
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    (Original post by cccdude)
    nah its not by substituion.. its through standard patters

    let Y= (cos2x+3)^1/2

    Differentiate to give -sin2x/(cos2x+3)^-1/2

    Using the identity sin2x-2sinxcosx you will have to divide the function by 2 to get back to the original equation
    well i jus did it via integrating a section....making it the same trig then integratin 2geva.....:/
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    nah the whole thing with this integration is kinda guess work.. you have the guess the differential as i did which was cos2x+3 to the power of 1/2.

    you then differentiate and the end answer should be something like what you the question was asking but then has to be altered.. in our case we had to divide by a negative half

    understand?
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    (Original post by SHUAIB_AKA_SHU)
    im nt using tht method of u du etc. im jus integrating a section making the trig function then 4rm there.....group and intregrate. i dnt get wt i did wrong in my method lol :/
    I don't get your method sorry

    I got the right answer with a u sub though.
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    its quite complicated but is needed on the edexcel C4 syllabus
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    (Original post by soutioirsim)
    I don't get your method sorry

    I got the right answer with a u sub though.
    well ive kind of being doin my own method and its been workin till nw....gnna do sub rule....have to learn and make notes for it nw -_-
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    RIP extricated
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    Can anybody show me how to integrate (sinx)*(x^2) dx ?
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    (Original post by diane_h)
    Can anybody show me how to integrate (sinx)*(x^2) dx ?
    You need two applications of integration by parts. You make a start. Let u = x^2.
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    u=X^2
    Du/Dx=2X
    DX=DU/2X
    X=SQRT(U)
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    (Original post by diane_h)
    u=X^2
    Du/Dx=2X
    DX=DU/2X
    X=SQRT(U)
    Er ... do you know what integration by parts is diane?
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    (Original post by Mr M)
    Er ... do you know what integration by parts is diane?
    Ahhhh..... THAT'S why I've been going around in circles and getting nowhere. By parts is
    Int(U) dv = uv- INT(V) du?
 
 
 
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