let A be closed-interval Subset of REAL ; :: thesis: ( A = [.0 ,PI .] implies integral (sin (#) cos ),A = 0 )
assume A = [.0 ,PI .] ; :: thesis: integral (sin (#) cos ),A = 0
then ( upper_bound A = PI & lower_bound A = 0 ) by Th37;
then integral (sin (#) cos ),A = (1 / 2) * (((cos . 0 ) * (cos . 0 )) - ((cos . PI ) * (cos . PI ))) by Th90
.= (1 / 2) * (((cos . (0 + (2 * PI ))) * (cos . 0 )) - ((cos . PI ) * (cos . PI ))) by SIN_COS:83
.= (1 / 2) * ((1 * 1) - ((- 1) * (- 1))) by SIN_COS:81, SIN_COS:83 ;
hence integral (sin (#) cos ),A = 0 ; :: thesis: verum