Back to Contents

    The solution of this integral is almost trivial. We begin by replacing the sine by the formula expressing it as a function of imaginary exponentials,

(Eq'n 1)

Next we multiply both the numerator and the denominator by eix to get

(Eq'n 2)

Now we can use the chain rule with y=eix and the knowledge, already deduced, that

(Eq'n 3)

to obtain

(Eq'n 4)

    If we multiply and divide the argument of the logarithm by e-ix/2, we get

(Eq'n 5)

For the sake of the act of calculation I exploit the fact that lnAB=lnA+lnB to rewrite that equation as

(Eq'n 6)

I then absorb ln(i) into the constant of integration, which drops out in the definite integral, and get

(Eq'n 7)

which we find in the Table of Integrals, usually under ∫(cscx)dx.

Example Application:

Energy Dissipated by a Skidding Wheel

    Imagine that we have drawn a circle of radius r0 with its center coinciding with the origin of an x-y grid. Imagine further that we have attached a long, straight, and slender arm to a pivot built upon that center, thereby making the arm free to revolve about the circle. We use θ to denote the angle between the arm and the positive part of the x-axis, with positive angles measured in the counterclockwise sense as per convention.

    Next we construct a straight line parallel to the x-axis and tangent to the circle at θ=π/2. We mount on the arm a wheel whose attachment is free to slide along the arm. The wheel rolls freely in the direction parallel to the arm. If the wheel moves in the direction perpendicular to the direction defined by the arm, it skids across the underlying surface and encounters a velocity-independent frictional force of


(Eq'n 8)

in which mg represents the weight pressing the wheel against the underlying surface and k represents the coefficient of friction. We now include on the arm a mechanism that constrains the wheel to move only on the tangent line described above.

    We want to calculate the amount of energy that the wheel dissipates through friction as the arm swings from θ1 to θ2. We know from basic trigonometry that the wheel lies a distance


(Eq'n 9)

from the arm's pivot. As the arm swings through the incremental angle dθ, the wheel rolls a distance dr (which we ignore) and skids a distance rdθ. Those distances comprise the sides of a right triangle whose hypotenuse lies on the tangent line and describes the actual motion of the wheel. We calculate the amount of dissipated energy by multiplying the frictional force by the incremental distance rdθ and integrating the product. We get

(Eq'n 10)


Back to Contents