let A be closed-interval Subset of REAL ; :: thesis: ( A = [.0 ,(PI / 2).] implies integral cos ,A = 1 )
assume A = [.0 ,(PI / 2).] ; :: thesis: integral cos ,A = 1
then ( upper_bound A = PI / 2 & lower_bound A = 0 ) by Th37;
then integral cos ,A = 1 - (sin . 0 ) by Th39, SIN_COS:82
.= 1 - (cos . ((PI / 2) - 0 )) by SIN_COS:83
.= 1 - 0 by SIN_COS:82 ;
hence integral cos ,A = 1 ; :: thesis: verum