let A be 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:83
.= (1 / 2) * ((1 * 1) - (0 * 0 )) by SIN_COS:81, SIN_COS:83 ;
hence integral (sin (#) cos ),A = 1 / 2 ; :: thesis: verum