# Parametric Intergration question

## 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 :\

CompuChip
Homework Helper
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.

Defennder
Homework Helper
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

Defennder
Homework Helper
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.

tiny-tim
Homework Helper
$$\int f(x) dx = \int f(t) \frac{dx}{dt} \ dt$$

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 :)

CompuChip
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 $du$.