how do you intergrate (1-2cost)^2

Question

(1-2cost)^2 dt with boundaries of 5pi/3 and pi/3

use this integral to find  the exact value of the shaded area.

Answer 1

QUESTION: how do you integrate (1-2cost)^2  with boundaries of 5pi/3 and pi/3.
use this integral to find  the exact value of the shaded area.

I = int (1 – cost)^2 dt
I = int 1 – 2cost + cos^2t dt

Using the identity cos2t = 2.cos^2t - 1

I = int 1 – 2cost + (1/2)(1 + cos2t) dt
I = int 3/2 – 2cost + (1/2)cos2t dt
I = 3t/2 – 2sint + (1/4)sin2t

Using the limits t = pi/3 and t = 5.pi/3

I = {15.pi/6 – 2.sin(5.pi/3) + (1/4)sin(10.pi/3)} – {3.pi/6 – 2.sin(pi/3) + (1/4)sin(2.pi/3)}
I = {15.pi/6 + sqrt(3) – sqrt(3)/8} – {3.pi/6 – sqrt(3) + sqrt(3)/8}
I = 12.pi/6 +7.sqrt(3)/8 + 7.sqrt(3)/8
I = 2.pi + 7.sqrt(3)/4