# Parametric Intergration question

• thomas49th
In summary, the area of a circle is given by the integral: \int^{\frac{5\pi}{3}}_{\frac{\pi}{3}} (1-2\cos t)^{2}f

## Homework Statement

a curve has parametric equations:
x = t - 2sin t
y = 1 - 2cos t

R is enclosed by the curve and the x axis. Show that the area of R is given by the integral:

$$\int^{\frac{5\pi}{3}}_{\frac{\pi}{3}} (1-2\cos t)^{2}$$

## The Attempt at a Solution

Not sure what to do :\

Suppose that the curve was given in the form y = f(x), then how would you calculate the area? Now do the variable substitution from x to t which is given.

You might want to make use of this:
$$\int f(x) dx = \int f(t) \frac{dx}{dt} \ dt$$
By the way what's this mathematical property called?

ermm mathematical propety?

dx/dt = 1 - 2cost

i have to multiply this by f(t) (how do i found that) and then intergrate that with respect to t?

Thanks

Yes, the integration is done wrt t. Isn't that given in the answer? And as for f(t), you should know that f(x) in the original integral f(x)dx represents a particular variable. That variable must now be expressed in terms of t for the integration to work.

$$\int f(x) dx = \int f(t) \frac{dx}{dt} \ dt$$

Hi thomas49th! I suspect you're confused by the (x) and the (t).

Just write it $$\int f dx = \int f \frac{dx}{dt} \ dt$$

or, in this case, $$\int y dx = \int y \frac{dx}{dt} \ dt$$ ahh yeh it's easy

got it!

cheers :)

The property is called substitution of variables. Probably you've seen it in this form:
Solve $$\int \sin(t) \cos(t) \, dt$$; let $u = \sin(t)$, then $du = \cos(t) \, dt$ so $$\int \sin(t) \cos(t) \, dt = \int u \, du = \frac12 u^2 + C = \frac12 \sin^2(t) + C$$.
You'll notice that in switching from the variable t to the variable u, I replaced $\frac{du}{dt} dt$ by $du$.

If you have never seen this, forget it. Otherwise I hope it clarifies what just happened :)