How do I find the area inside the cardioid #r=1+costheta#?

#4pi#

Explanation:

We will integrate the area differential for a polar function:
#dA = d(1/2 r^2 theta) = r theta d r + 1/2 r^2 d theta#

We can integrate this just #d theta#:
#dA = (r theta (dr)/(d theta) + 1/2 r^2) d theta #

This yields
#A = int\ \ dA = int_0^(2pi) ([1 + cos theta] * theta * (- sin theta) + 1/2 (1 + cos theta)^2) d theta #
#A = int_0^(2pi) [3/4 + 1/4 cos2theta + cos theta - theta sin theta - 1/2 theta sin 2theta] d theta #

This integral is pretty simple, with a few integrations by parts.

The ultimate answer turns up to be #A = 4pi#.