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    Can someone give me some help with this one please.

    2sin x cos (x/2)

    I've written cos (x/2) in terms of x (1/4)rt(cos x +1), but don't know where to take it from there.

    By differentiating the answer, I think it can be written as 4 cos^2 (x/2) sin (x/2). No idea how to get there though!

    Thanks for any help.
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    You can get to 4cos^2(x/2)sin(x/2) by using the identity sin(2x) = 2sin(x)cos(x).

    As for integrating it, this looks like one of those in which you have to apply integration by parts twice and derive an equation with the original integral, although I'm not certain on that. Does that sound familar at all to you?
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    (Original post by Dragon)
    You can get to 4cos^2(x/2)sin(x/2) by using the identity sin(2x) = 2sin(x)cos(x).

    As for integrating it, this looks like one of those in which you have to apply integration by parts twice and derive an equation with the original integral, although I'm not certain on that. Does that sound familar at all to you?
    Sorry to sound really dim here, but how can we use sin(2x) = 2sin(x)cos(x)

    when we have 2sin x?
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    Make the substitution x = x/2 in the identity to get sin(x) = 2sin(x/2)cos(x/2)}
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    i've used the product to sum formula : 2sinAcosB=sin(A+B)+sin(A-B)
    can anyone please verify my answer using this method?
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    Ah yes, that's a much better way of doing it than what I suggested.
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    (Original post by Dragon)
    You can get to 4cos^2(x/2)sin(x/2) by using the identity sin(2x) = 2sin(x)cos(x).

    As for integrating it, this looks like one of those in which you have to apply integration by parts twice and derive an equation with the original integral, although I'm not certain on that. Does that sound familar at all to you?
    (Original post by kingsclub)
    i've used the product to sum formula : 2sinAcosB=sin(A+B)+sin(A-B)
    can anyone please verify my answer using this method?
    You don't need IBP or the factor formulae. OP, from this stage, consider the substitution u=\cos (\frac{x}{2}).
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    Thanks for the help everyone. I did it by expressing sin x as 2 sin (x/2) cos (x/2). Thanks for the substitution idea though, I'm sure that would work great as well.
 
 
 
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