We have: #int_(frac(pi)(2))^(0) (frac(1 + cos(2 t))(2)) d t#
First, let's evaluate the integral:
#Rightarrow int (frac(1 + cos(2 t))(2)) dt = int (frac(1)(2) (1 + cos(2 t))) d t#
Take the constant out:
#Rightarrow int (frac(1 + cos(2 t))(2)) d t = frac(1)(2) int (1 + cos(2 t)) d t#
Apply the sum rule:
#Rightarrow int " " (frac(1 + cos(2 t))(2)) d t = frac(1)(2) (int(1) d t + int(cos(2 t)) d t)#
#Rightarrow int " " (frac(1 + cos(2 t))(2)) dt = frac(1)(2) (t + int(cos(2 t) d t))#
Let's substitute #u = 2 t Rightarrow frac(d u)(d t) = 2 Rightarrow d t = frac(1)(2) d u#:
#Rightarrow int " " (frac(1 + cos(2 t))(2)) dt = frac(1)(2) (t + int(cos(u) cdot frac(1)(2) d u))#
Take the constant out:
#Rightarrow int " " (frac(1 + cos(2 t))(2)) dt = frac(1)(2) (t + frac(1)(2) (int(cos(u)) d u))#
#Rightarrow int " " (frac(1 + cos(2 t))(2)) dt = frac(1)(2) (t + frac(1)(2) sin(u))#
Replacing #u# with #2 t#:
#Rightarrow int " " (frac(1 + cos(2 t))(2)) dt = frac(1)(2) (t + frac(1)(2) sin(2 t))#
Now, let's compute the boundaries:
#Rightarrow int_(frac(pi)(2))^(0) (frac(1 + cos(2 t))(2)) d t = (frac(1)(2) (0 + frac(1)(2) sin(2 cdot 0))) - (frac(1)(2) (frac(pi)(2) + frac(1)(2) sin(2 cdot frac(pi)(2))))#
#Rightarrow int_(frac(pi)(2))^(0) (frac(1 + cos(2 t))(2)) d t = (frac(1)(2) (frac(1)(2) cdot 0)) - (frac(1)(2) (frac(pi)(2) + frac(1)(2) cdot 0))#
#Rightarrow int_(frac(pi)(2))^(0) (frac(1 + cos(2 t))(2)) d t = 0 - frac(pi)(4)#
#therefore int_(frac(pi)(2))^(0) (frac(1 + cos(2 t))(2)) d t = - frac(pi)(4)#
