# Use the Midpoint Rule with the given value of n to approximate the integral.Use the Midpoint Rule with the given value of n to approximate the integral. Integral 0 to (pi/2) 2cos^(5)x dx ,...

Use the Midpoint Rule with the given value of *n* to approximate the integral.

Use the Midpoint Rule with the given value of *n* to approximate the integral.

Integral 0 to (pi/2) 2cos^(5)x dx , n=4 M4=__________________?

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### 1 Answer

You should evaluate the integral using midpoint rule, hence, you need to split the interval `[0,pi/2]` in n = 4 subintervals of equal width such that:

`[0 , pi/8] , [pi/8 , pi/4] , [pi/4 , 3pi/8] , [3pi/8 , pi/2]`

You need to find the midpoint of each subinterval such that:

`x_1 = (pi/8 - 0)/2 , x_2 = (pi/4 - pi/8)/2 , x_3 = (3pi/8 - pi/4)/2 , x_4 = (pi/2 - 3pi/8)/2`

Evaluating the integral using the midpoint rule, yields:

`int_0^(pi/2) 2 cos^5 x dx = 2(pi/2 - 0)/2(cos^5 ((pi/8 - 0)/2) + cos^5 ((pi/4 - pi/8)/2) + cos^5 ((3pi/8 - pi/4)/2) + cos^5((pi/2 - 3pi/8)/2))`

`int_0^(pi/2) 2 cos^5 x dx = pi/2*(cos^5 (pi/16) + cos^5(3pi/16) + cos^5(5pi/16) + cos^5(7pi/16))`

`cos^2 pi/8 = (1 + cos(pi/4))/2 = (2 + sqrt2)/4 => cos pi/8 = (sqrt(2 + sqrt2))/2`

`cos^2 (pi/16) =(2 + (sqrt(2 + sqrt2)))/4`

`cos^5 (pi/16) = (2 + (sqrt(2 + sqrt2)))^2/32*sqrt(2 + (sqrt(2 + sqrt2)))`

`cos(3pi/16) = cos(pi/16 + pi/8) = cos(pi/16)cos(pi/8) - sin(pi/16)sin(pi/8)`

`cos(3pi/16) = (sqrt(2 + (sqrt(2 + sqrt2)))/2)*(sqrt(2 + sqrt2))/2 - (sqrt(2- (sqrt(2 - sqrt2)))/2)*(sqrt(2- sqrt2))/2`

`cos(5pi/16) = cos(3pi/16 + pi/8) = cos(3pi/16)cos(pi/8) - sin(3pi/16)sin(pi/8)`

`cos(7pi/16) = cos(5pi/16 + pi/8) = cos(5pi/16)cos(pi/8) - sin(5pi/16)sin(pi/8)`

**Hence, evaluating the given integral using the midpoint rule yields `int_0^(pi/2) 2 cos^5 x dx = pi/2*(cos^5 (pi/16) + cos^5(3pi/16) + cos^5(5pi/16) + cos^5(7pi/16)).` **