= ∫^(π/2)_(0) (1+сos2x)dx/2=(1/2)* ∫^(π/2)_(0) 1dx+(1/2) ∫ cos2x=
[d(2x)=2dx
dx=d(2x)/2]
=(1/2)x|^(π/2)_(0)+(1/2)*(1/2)*(sin2x)|^(π/2)_(0)=
=(π/4)
2.
d(2x)=2dx
dx=d(2x)/2
(1/2) ∫ ^(3)_(2)d(2x)/(2x)^2-1= (1/2) *(1/2) ln |(2x-1)/(2x+1)|^(3)_(2)=
=(1/4)ln|5/7|-(1/4)ln|3/5|=(1/4)ln|25/21|