let A be non empty closed_interval Subset of REAL; :: thesis: ( A = [.0,(PI * (3 / 2)).] implies integral ((sin (#) cos),A) = 1 / 2 )
assume A = [.0,(PI * (3 / 2)).] ; :: thesis: integral ((sin (#) cos),A) = 1 / 2
then ( upper_bound A = PI * (3 / 2) & lower_bound A = 0 ) by Th37;
then integral ((sin (#) cos),A) = (1 / 2) * (((cos . 0) * (cos . 0)) - ((cos . (PI * (3 / 2))) * (cos . (PI * (3 / 2))))) by Th90
.= (1 / 2) * (((cos . (0 + (2 * PI))) * (cos . 0)) - ((cos . (PI * (3 / 2))) * (cos . (PI * (3 / 2))))) by SIN_COS:78
.= (1 / 2) * ((1 * 1) - (0 * 0)) by SIN_COS:76, SIN_COS:78 ;
hence integral ((sin (#) cos),A) = 1 / 2 ; :: thesis: verum