# Calculus/Parametric Integration

 ← Parametric Differentiation Calculus Parametric and Polar Equations/Parametric Equations/Exercises → Parametric Integration

## Introduction

Because most parametric equations are given in explicit form, they can be integrated like many other equations. Integration has a variety of applications with respect to parametric equations, especially in kinematics and vector calculus.

{\displaystyle {\begin{aligned}x&=\int x'(t)\mathrm {d} t\\y&=\int y'(t)\mathrm {d} t\end{aligned}}}

So, taking a simple example, with respect to t:

${\displaystyle y=\int \cos(t)\mathrm {d} t=\sin(t)+C}$

## Arc length

Consider a function defined by,

${\displaystyle x=f(t)}$
${\displaystyle y=g(t)}$

Say that ${\displaystyle f}$ is increasing on some interval, ${\displaystyle [\alpha ,\beta ]}$. Recall, as we have derived in a previous chapter, that the length of the arc created by a function over an interval, ${\displaystyle [\alpha ,\beta ]}$, is given by,

${\displaystyle L=\int _{\alpha }^{\beta }{\sqrt {1+(f'(x))^{2}}}\mathrm {d} x}$

It may assist your understanding, here, to write the above using Leibniz's notation,

${\displaystyle L=\int _{\alpha }^{\beta }{\sqrt {1+\left({\frac {\mathrm {d} y}{\mathrm {d} x}}\right)^{2}}}\mathrm {d} x}$

Using the chain rule,

${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}={\frac {\mathrm {d} y}{\mathrm {d} t}}\cdot {\frac {\mathrm {d} t}{\mathrm {d} x}}}$

We may then rewrite ${\displaystyle \mathrm {d} x}$,

${\displaystyle {\frac {\mathrm {d} x}{\mathrm {d} t}}\mathrm {d} t}$

Hence, ${\displaystyle L}$ becomes,

${\displaystyle L=\int _{\alpha }^{\beta }{\sqrt {1+\left({\frac {\mathrm {d} y}{\mathrm {d} t}}\cdot {\frac {\mathrm {d} t}{\mathrm {d} x}}\right)^{2}}}{\frac {\mathrm {d} x}{\mathrm {d} t}}\mathrm {d} t}$

Extracting a factor of ${\displaystyle \left({\frac {\mathrm {d} t}{\mathrm {d} x}}\right)^{2}}$,

${\displaystyle L=\int _{\alpha }^{\beta }{\sqrt {\left({\frac {\mathrm {d} t}{\mathrm {d} x}}\right)^{2}}}{\sqrt {\left({\frac {\mathrm {d} x}{\mathrm {d} t}}\right)^{2}+\left({\frac {\mathrm {d} y}{\mathrm {d} t}}\right)^{2}}}{\frac {\mathrm {d} x}{\mathrm {d} t}}\mathrm {d} t}$

As ${\displaystyle f}$ is increasing on ${\displaystyle [\alpha ,\beta ]}$, ${\displaystyle {\sqrt {\left({\frac {\mathrm {d} t}{\mathrm {d} x}}\right)^{2}}}={\frac {\mathrm {d} t}{\mathrm {d} x}}}$, and hence we may write our final expression for ${\displaystyle L}$ as,

${\displaystyle \int _{\alpha }^{\beta }{\sqrt {\left({\frac {\mathrm {d} x}{\mathrm {d} t}}\right)^{2}+\left({\frac {\mathrm {d} y}{\mathrm {d} t}}\right)^{2}}}\mathrm {d} t}$

### Example

Take a circle of radius ${\displaystyle R}$, which may be defined with the parametric equations,

${\displaystyle x=R\sin \theta }$
${\displaystyle y=R\cos \theta }$

As an example, we can take the length of the arc created by the curve over the interval ${\displaystyle [0,R]}$. Writing in terms of ${\displaystyle \theta }$,

${\displaystyle x=0\implies \theta =\arcsin \left({\frac {0}{R}}\right)=0}$
${\displaystyle x=R\implies \theta =\arcsin \left({\frac {R}{R}}\right)=\arcsin(1)={\frac {\pi }{2}}}$

Computing the derivatives of both equations,

${\displaystyle {\frac {\mathrm {d} x}{\mathrm {d} \theta }}=R\cos \theta }$
${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} \theta }}=-R\sin \theta }$

Which means that the arc length is given by,

${\displaystyle L=\int _{0}^{\frac {\pi }{2}}{\sqrt {(-R\sin \theta )^{2}+R^{2}\cos ^{2}\theta }}\mathrm {d} \theta }$

By the Pythagorean identity,

${\displaystyle L=R\int _{0}^{\frac {\pi }{2}}\mathrm {d} \theta =R{\frac {\pi }{2}}}$

One can use this result to determine the perimeter of a circle of a given radius. As this is the arc length over one "quadrant", one may multiply ${\displaystyle L}$ by 4 to deduce the perimeter of a circle of radius ${\displaystyle R}$ to be ${\displaystyle 2\pi R}$.