Numerous curves subject to integration commonly take the format \(y = f (x)\). While this representation works for most curves, there are cases where it becomes inconvenient or impossible to express them in that form. It is in such scenarios where parametric coordinates prove to be useful.

## Recap of parametric coordinates

In such cases, we introduce a 'dummy' variable, typically denoted as \(t\). This variable represents an abstract concept that assigns values to either the \(x\) or \(y\) coordinate, without being directly plotted on the graph.

Therefore, instead of having a function in the form \(y = f (x)\), we represent a curve using \(y(t) = g(t)\) and \(x(t) = h(t)\), where \(h\) and \(t\) are are functions that represent the transformation of the ( x ) and ( y ) coordinates respectively.

For example, consider a curve described by \(y(t) = 2t\), \(x(t) = 2t\), where \(0 < t < 2\pi\).

By expressing the parametric curve as \((x(t))^2 + (y(t))^2 = (2\cos{t})^2 + (2\sin{t})^2 = 4\cos^2{t} + \sin^2{t} = 4\), we observe that it actually represents a circle with a radius of 4, or \(x^2 + y^2 = 4\).

## Why does parametric integration work?

Normally, we would evaluate an integral of the form \(\int y(x) \, dx\); however, since our curve is not in the form \(y(x)\), we need to make some changes. We utilize a modified version of the Chain Rule and replace \(dx\) with \(\frac{dx}{dt} \, dt\) (treating it as a fraction for operational purposes, although it's not strictly accurate as \(\frac{dx}{dt}\) is not a traditional fraction). This results in an integral of the form \(\int y(t)\frac{dx(t)}{dt}\,dt\).

Another aspect to consider when dealing with parametric integrals is switching the limits. Suppose we have an integral in the form \(\int^b_a f(x) \, dx\). When using parametric representation, we also need to switch the limits, leading to the integral becoming \(\int^d_c f(t) \frac{dx}{dt}\,dt\), with \(c = x^{-1}(a)\) and \(d = x^{-1}(b)\).

## Examples of parametric integration

Parametric integration can be a tricky topic to grasp, so let's walk through a couple of examples to solidify our understanding.

Suppose we have a parametrically defined curve with ( x(t) = 2 - t ) and ( y(t) = e^t - 1 ). Our goal is to determine the enclosed area between the x-axis, the line ( x = 0 ), and the given curve.

To determine the points where the curve intersects the x-axis and where the line \(x = 0\) intersects the curve, we set the y-value to zero. Solving the equation \(e^t - 1 = 0\) yields \(t = 0\). When \(x = 0\), we find \(t = 2\).

Now, with the limits determined, we can proceed with the integral:

\[\int^0_2 (e^t - 1) \frac{d}{dt}(2 - t) dt = - \int^0_2 (e^t - 1) dt = \int^0_2 (e^t - 1) dt\]

We switch the limits to account for the change in sign.

Upon evaluating the expression, we obtain ([e^t - t]_{t=0}^{t=2} = [(e^2 - 2) - (1-0)] = e^2 - 3). By employing parametric integration, let's determine the area of the circle defined by (x(t) = -3\cos(t)) and (y(t) = 3\sin(t)), where (0 < t < 2\pi).

Using parametric integration, find the area of the circle defined by \(x(t) = -3\cos(t)\), \(y(t) = 3\sin(t)\), with \(0 < t < 2\pi\).

Employing the parametric formula, we have:

\[\int^{2\pi}_0 3\sin(t) \frac{d}{dt} (-3\cos(t)) dt = 9 \int^{2\pi}_0 \sin^2(t) dt\]

We can apply a double angle formula here, such as \(2(t) = \frac{1}{2}(1 - \cos(2t))\).

Substituting, we arrive at \(\frac{9}{2} \int^{2\pi}_0 (1 - \cos(2t)) dt = \frac{9}{2} [t - \frac{1}{2} \sin(2t)]^{t=2\pi}_{t=0} = 9\pi\), as expected for a circle with a radius of 3.

**Exam-style question**

Suppose we have a curve defined parametrically with \(x(t) = 3\cos(4t)\) and \(y(t) = 6 \sin(8t)\), where \(0 < t < \frac{\pi}{8}\).

i) Find any turning points of the curve.

ii) Find the area under the curve.

i) To find a turning point, \(\frac{dy}{dx}\) should be equal. By applying the Chain Rule:

\[\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{dy}{dt}(\frac{dx}{dt})^{-1}\]

Using standard formulas for derivatives of trigonometric functions, we obtain \(\frac{dy}{dt} = 48 \cos(8t)\) and \(\frac{dx}{dt} = -12 \sin(4t)\), resulting in \(\frac{dy}{dx} = \frac{48 \cos(8t)}{-12 \sin(4t)}\).

Setting this equal to zero to find the turning point's \(t\) value, we observe that \(\cos(8t) = 0\).

This implies \(8t = \frac{\pi}{2} + n\pi\) (where \(n \in \mathbb{N}\)), which we simplify to \(t = \frac{\pi}{16} + \frac{n\pi}{8}\).

Among these values, only \(t = \frac{\pi}{16}\) falls within the range \(0 < t < \frac{\pi}{8}\).

Thus, the x-coordinate of the turning point is \(x(\frac{\pi}{16}) = 3\cos(\frac{\pi}{4}) = \frac{3\sqrt{2}}{2}\), and the y-coordinate is \(y(\frac{\pi}{16}) = 6\sin(\frac{\pi}{2}) = 6\).

To determine the limits, we find that \(x(0) = 3\) and \(x(\frac{\pi}{8}) = 0\). Therefore, the area under the curve can be expressed as:

\[\int^0_{\frac{\pi}{8}} y(t) \frac{dx}{dt} dt = \int^0_{\frac{\pi}{8}} 6 \sin(8t)(-12\sin(4t)) dt = 72\int^{\frac{\pi}{8}}_0 \sin(8t) \sin(4t) dt\]

We flip the limits to remove the negative sign. To solve this integral, we can utilize a double angle formula.

Given \(\sin(2x) = 2\sin(x)\cos(x)\), we have \(\sin(8x) = 2 \cdot \sin(4x) \cos(4x)\).

Substituting this, we obtain the integral \(144 \int^{\frac{\pi}{8}}_0 \sin^2(4t)\cos(4t) dt\).

Since \(\int \sin(t) = \cos(t)\), it is beneficial to use substitution for this definite integral.

Let \(u = \sin(4t)\), implying \(\frac{du}{dt} = 4\cos(4t)\) and \(dt = \frac{du}{4\cos(4t)}\).

As this is a definite integral, we also need to change the limits of integration.

Setting \(u_1 = \sin(4 \cdot 0) = 0\) and \(u_2 = \sin(4 \cdot \frac{\pi}{8}) = 1\), we find \(144 \int_0^{\frac{\pi}{8}} \sin^2(4t) \cos(4t) dt = 36 \int^1_0 u^2 du\).

This integral can be evaluated directly:

\(36 \int^1_0 u^2 du = 12[u^3]^{u = 1}_{u = 0} = 12[1-0] = 12\).