what bounds would i use to evaluate this integral?

Question

find the area of the region enclosed by r^2=5cos(5theta)

Answer 1

cos 5t >=0 so -pi/2 <=5t<= pi/2
so -pi/10<=t<=pi/10
as cos is an even function
A =2*5/2 Int (0,pi/10) cos(5t) dt = (sin5t) between p/10 and 0
=sinpi/2=1