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    Show that the area of the finite region in the positive quadrant bounded by the curve = cos^3 x and the curve y = sin^£ x and the y-axis is given by (5(2)^1/2 - 4)/6

    what is the point where they cross

    y=sin^3 x -sin^3x
    i get to this:
    y=(cosx - sinx)(cos^2 x + sin^2 x + cosxsinx)
    (cosx - sinx)(1 + cosxsinx)

    where do i go from here
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    (Original post by Custardcream000)
    Show that the area of the finite region in the positive quadrant bounded by the curve = cos^3 x and the curve y = sin^£ x and the y-axis is given by (5(2)^1/2 - 4)/6

    what is the point where they cross

    y=sin^3 x -sin^3x
    i get to this:
    y=(cosx - sinx)(cos^2 x + sin^2 x + cosxsinx)
    (cosx - sinx)(1 + cosxsinx)

    where do i go from here
    You could expand the brackets and then integrate.

    It helps to know what \dfrac{d}{dx}(\cos^3x) is, and similarly for the sine.
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    (Original post by ghostwalker)
    You could expand the brackets and then integrate.

    It helps to know what \dfrac{d}{dx}(\cos^3x) is, and similarly for the sine.
    Im trying to find the point where they intersect before i integrate to find area so thats what i meant where do i go from here in order to find where they intersect?
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    (Original post by Custardcream000)
    Im trying to find the point where they intersect before i integrate to find area so thats what i meant where do i go from here in order to find where they intersect?
    Thought you'd already sorted that, as the method is completely different.

    When they intersect then the two y values are equal

    \cos^3x=\sin^3x

    So,

    \tan^3x=1

    ...
 
 
 
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