Integration of Trigonometric fucntions Watch

Custardcream000
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#1
Report Thread starter 2 years ago
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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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ghostwalker
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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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Custardcream000
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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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ghostwalker
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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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