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    Integration of 4sin2x.cos2x dx

    Not sure how to do this but I tried by substitution and got -cos^2(2x)+c

    u=cos2x
    du/dx=-2sin2x
    dx/du=-1/2sin2x
    sin2x(dx/du)=-1/2
    But arrived at the wrong answer is there a quicker qay of doing this?
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    I would let u= sin2x

    du/dx = 2cos2x

    du = dx*2cos2x.
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    Another way of doing this question is to notice that sin(4x)=sin(2x+2x)=2sin(2x)cos(2 x) then using that to make a trig substitution.

    However, the answer you got is actually correct. If you try my method you get (-1/2)cos(4x) + k which can be manipulated to show that (-1/2)cos(4x) differs from -cos^2(2x) by a constant and hence they are equivilant answers to the question.
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    (Original post by Luppy021)
    Integration of 4sin2x.cos2x dx

    Not sure how to do this but I tried by substitution and got -cos^2(2x)+c

    u=cos2x
    du/dx=-2sin2x
    dx/du=-1/2sin2x
    sin2x(dx/du)=-1/2
    But arrived at the wrong answer is there a quicker qay of doing this?




    Uploaded with ImageShack.us

    There are several ways of doing this problem and they might get answers that LOOK different.


    EDIT

    Why the rep?
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    Yeah I get it now cheers all
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    Your answer is correct. I think it will differ by a constant from the book answer.
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    You could also use integration by parts
 
 
 
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