Revisited

Now I want to solve this integral by using the process of integration by substitution. I'll start by taking the constant out of the square root function and I get

(Eq'n 1)

Next I want to eliminate the square root. The form of the function under the radical gives me an almost obvious substitution that enables me to do just that: I define x/a=iSinθ, so that dx/a=iCosθdθ, and I get

(Eq'n 2)

I then invert the substitution,
θ=Sin^{-1}(x/ia), and get

(Eq'n 3)

which we got in the previous derivation. But we also know that

(Eq'n 4)

which we use in the following application.

Example Application:

The Catenary

In the x-z plane gravity operates in the
negative z-direction and the x-axis, running horizontally, defines ground level.
We set up a vertical pole of height z_{0} at the point x_{0} and
set up a second vertical pole of height z_{1} at the point x_{1}.
Then we take a perfectly limp cord of uniform density along its length and of
length L>x_{1}-x_{0} and attach its ends to the tops of the
poles, allowing it to hang free under its own weight. We want to derive a
description of the cord's location,
point by point, as a function z(x) on the domain between x_{0} and x_{1}.

We say that each element ds of the cord's length has the weight wds. At any point on the cord the straight line tangent to the cord makes an angle ϕ with the x-axis. At each and every point on the cord the tension T within the cord must point in the direction parallel to the tangent line at that point and, because the cord does not move, the forces acting on the cord at each and every point must balance. Thus we infer two facts:

1. The vertical component of the tension must change along the cord in a way that balances the weight of each differential segment of the cord; that is,

(Eq'n 5)

2. The horizontal component of the tension does not change; that is,

(Eq'n 6)

Equation 6 tells us immediately that

(Eq'n 7)

in which C represents a constant. We divide Equation 5 by that constant and then, noting that division by a constant commutes with the operation of differentiation, we substitute from Equation 7 to get

(Eq'n 8)

Referring to our desired function z(x) and the angle that the cord makes with the x-axis at any given point, we have

(Eq'n 9)

in which I have defined the upper-case zee for convenience in what follows. We also describe the length of the differential segment of the cord as

(Eq'n 10)

We thus have Equation 8 in the form

(Eq'n 11)

Dividing that equation by the radical and integrating gives us

(Eq'n 12)

If we solve that equation for Z, multiply the result by dx, and integrate, we get

(Eq'n 13)

which describes our cord as a catenary. The integration
constants A, B, and C must take values that give the curve the correct values
z(x_{0})=z_{0}, z(x_{1})=z_{1}, and length L,
which we calculate from

(Eq'n 14)

in which I have replaced Z with the corresponding hyperbolic
sine and used Cosh^{2}ξ-Sinh^{2}ξ=1
to set up the integration in the second line. I leave the actual determination
of A, B, and C to the interested reader.

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