Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 12

int_0^(pi/2) (2-sin (theta))^2 d theta
First, expand the integrand.
=int_0^(pi/2) (2-sin (theta))(2-sin (theta)) d theta
= int _0^(pi/2) (4-4sin (theta)+sin^2 (theta)) d theta
For the last term of the integrand, apply the trigonometric identity cos(2theta) = 1 - 2sin^2 (theta) .
= int_0^(pi/2) (4-4sin(theta) + 1/2 - cos(2theta)/2) d theta
= int_0^(pi/2) (9/2-4sin theta -(cos(2theta))/2) d theta
= int_0^(pi/2) 9/2 d theta - int_0^(pi/2) 4sin theta d theta - int_0^(pi/2) (cos(2 theta))/2 d theta
Apply the integral formulas int adx = ax , int sin xdx = -cosx and int cos x dx = sinx .
= (9/2theta + 4cos(theta) - (sin (2theta))/4 )|_0^(pi/2)
= (9/2*pi/2 + 4cos(pi/2) - sin (2*pi/2)/4) - (9/2*0+4cos(0) - (sin (2*0))/4)
=((9pi)/4+4*0+0/4)-(0+4*1-0/4)
=(9pi)/4 - 4

Therefore, int _0^(pi/2) (2-sin(theta))^2 d theta = (9pi)/4 - 4 .