Calculus: Early Transcendentals, Chapter 5, 5.2, Section 5.2, Problem 10

You need to use the midpoint rule to approximate the interval. First, you need to find Delta x, such that:
Delta x = (b-a)/n
The problem provides b=pi/2 , a=0 and n = 4, such that:
Delta x = (pi/2-0)/4 = pi/8
Hence, the following 4 intervals of length pi/8 are: [0,pi/8], [pi/8,pi/4], [pi/4,(3pi)/8], [(3pi)/8,pi/2].
Now, you may evaluate the integral such that:
int_0^pi/2 cos^4 x dx = Delta x(f((0+pi/8)/2) + f((pi/8+pi/4)/2) + f((pi/4+(3pi)/8)/2) + f(((3pi)/8+pi/2)/2))
int_0^pi/2 cos^4 x dx = pi/8(f(pi/16) + f(3pi/16) + f(5pi/16) + f(7pi/16))
int_0^pi/2 cos^4 x dx = pi/8(cos^4(pi/16) + cos^4 (3pi/16)+ cos^4(5pi/16) + cos^4(7pi/16)))
cos(pi/8) = cos((pi/4)/2) => cos^2((pi/4)/2) = (1 + cos(pi/4))/2 => cos^4((pi/4)/2) = ((1 + cos(pi/4))^2)/4
cos^4((pi/4)/2) = ((1 + sqrt2/2)^2)/4
cos^4((pi/4)/2) = ((2 + sqrt2)^2)/16
cos(pi/16) = cos((pi/8)/2) => cos^2((pi/8)/2) = (1 + sqrt((1 + sqrt2/2)/2))/2
cos^4(pi/16) =(0.9619)^2 = 0.6637
cos(3pi/16) = -cos(pi/16) => cos^4(3pi/16) = cos^4(pi/16) = 0.6637
cos(5pi/16) = -cos(pi/16) => cos^4(3pi/16) = cos^4(pi/16) = 0.6637
cos(7pi/16) = -cos(3pi/16) = -(-cos(pi/16) ) => cos^4(7pi/16) = cos^4(pi/16) = 0.6637
int_0^pi/2 cos^4 x dx = pi/8( 0.6637 + 0.6637 + 0.6637 + 0.6637)
int_0^pi/2 cos^4 x dx = 4*pi/8(0.6637)
int_0^pi/2 cos^4 x dx = pi/2*(0.6637)
int_0^pi/2 cos^4 x dx = 0.3318*pi
Hence, approximating the definite integral, using the midpoint rule, yields int_0^pi/2 cos^4 x dx = 0.3318*pi.