∫

If I had a factor of Sinx in the
numerator of my integrand, I could solve that integral easily. So I figure that
if I exploit the fact that 1=Sin^{2}x+Cos^{2}x, I might be able
to solve the sine part of the integral by integrating by parts, but I am left
with the intractable cosine-squared part. However, if I could multiply the
squared cosine by b^{2} and then subtract a^{2}, I could factor
the resulting binomial, cancel off one of the factors in the denominator, and
see what I can do with the resulting simpler integral.

Therefore, I begin by multiplying and
dividing the intergral by b^{2}-a^{2}, unpacking the squares of
sine and cosine from the b^{2} term, and simplifying the result:

(Eq'n 1)

I can integrate the first integral on the right side of that equation by parts and get

(Eq'n 2)

The second term on the right of that equation, the cosine integral, cancels the cosine integral in the second term on the right side of Equation 1, so we get

(Eq'n 3)

But we already know the solution of the integral on the right side of that equation, so we have, at last,

(Eq'n 4)

Example Application:

Area of an Orbital Segment

We describe the ellipse that represents a planet's orbit by describing the orbit's radius as a function of the longitude measured from the orbit's primary focus, the angle between the radius vector and the line extending from the sun to the perihelion of the orbit. Thus we have

(Eq'n 5)

in which equation r_{0}
represents the perihelion radius of the orbit, e represents the orbit's
eccentricity, and θ
represents the longitude on the orbit. We want to calculate the area that the
planet's radius vector sweeps out as the planet goes from perihelion (θ=0)
to some arbitrary longitude.

For the minuscule element of area we take an isoceles triangle whose sides have length r and whose base has length rdθ, so we have

(Eq'n 6)

We then make appropriate substitutions from Equation 5 and carry out the integration:

(Eq'n 7)

If we want to calculate the area in the
elliptical segments bounded by longitudes θ_{1}
and θ_{2},
then we need only substitute the coordinates
θ_{1}
and θ_{2}
into Equation 7 separately and then subtract the smaller result from the larger
one.

Johannes Kepler's second law of
planetary motion tells us that the radius vector extending between sun and any
given planet sweeps out equal areas in equal times. We can calculate the rate at
which the radius vector sweeps out area if we multiply together the length of
the radius vector of a given planet's orbit at perihelion r_{0} and the
velocity the planet has at perihelion v_{0}; we get

(Eq'n 8)

If we multiply Equation 8 by the
element of time and integrate both sides of the resulting equation, we find that
we can calculate the time that a planet needs to go from perihelion to some
given longitude by dividing Equation 7 by the product r_{0}v_{0}.

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