**P3 Integration Questions** Watch

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devesh254
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#1
Report Thread starter 13 years ago
#1
1) Find the integral of (cosx) / [4-(sinx)^2]

(cosx over 4 minus sin squared x)

2) Use integration to find the area enclosed by the ellipse, which has the parametric equations:

x = 5cost y = 2sint 0<=t<2pi

3) For a certain curve dy/dx = (4y) / (3x-1) and the point P(2,1) lies on the curve.

Express y in terms of x.


ANSWERS:

1) 1/4 ln[(2 + sinx)/(2 - sinx)] + C

2) 10pi

3) y = [(3x - 1)/5]^(4/3)


What I did:

1) I figured out that the denominator was (2 + sinx)(2-sinx) but was stuck after that.

2) I got -10pi...I know that u ignore the - signs in areas but still I should be getting 10pi...

3) Stuck with the constant...got confused with it...so all i've done so far is a differential equation, and I integrated both sides to get:

1/4 ln|4y| = 1/3 ln|3x - 1| + C


Thx in advance!
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Gaz031
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#2
Report 13 years ago
#2
(Original post by devesh254)
[COLOR=Red][B]1) Find the integral of (cosx) / [4-(sinx)^2]

(cosx over 4 minus sin squared x)
Use u=sinx. The integral will be transformed into:
INT 1/(4-u^2) du = INT 1/(2-u)(2+u) du which you can do using partial fractions.

2) Use integration to find the area enclosed by the ellipse, which has the parametric equations:

x = 5cost y = 2sint 0<=t<2pi
If you sketch it out you'll see the ellipse has 4 equal areas in the 4 quadrants. The area in the 1st quadrant is given by:
INT (2sint).(-5sint) dt with lower limit pi/2 and upper limit 0. [The lower limit is pi/2, as x=0 at t=pi/2.]
So the total area is given by:
INT 40(sint)^2 dt with lower limit 0, upper limit pi/2 (noticed i've just swapped the limits and remove the minus sign from the integral).
INT 20(1-cos2t) dt [0,pi/2].
[20t-10sin2t] [0,pi/2]
=10pi

3) For a certain curve dy/dx = (4y) / (3x-1) and the point P(2,1) lies on the curve. Express y in terms of x.
INT (1/4).(1/y) dy = INT 1/(3x-1) dx
(1/4)lny=(1/3)ln(3x-1) + C
(2,1) on curve: (1/4)ln1=(1/3)ln5 + C
But ln1=0, so C=-(1/3)ln5
(1/4)lny=(1/3)ln(3x-1)-(1/3)ln5
lny=(4/3)ln[(3x-1)/5]
lny=ln[(3x-1)/5]^(4/3)
y=[(3x-1)/5]^(4/3)

Edit: Interesting, we have the same insurance as well as the same firm.
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devesh254
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#3
Report Thread starter 13 years ago
#3
thx again mate...

and oh yeh...kool...lol...:laugh:
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