let A be non empty closed_interval Subset of REAL; :: thesis: ( A = [.0,((PI * 3) / 2).] implies integral (cos,A) = - 1 )
assume A = [.0,((PI * 3) / 2).] ; :: thesis: integral (cos,A) = - 1
then ( upper_bound A = (PI * 3) / 2 & lower_bound A = 0 ) by Th37;
then integral (cos,A) = (- 1) - (sin . 0) by Th39, SIN_COS:76
.= (- 1) - (sin . (0 + (2 * PI))) by SIN_COS:78
.= (- 1) - 0 by SIN_COS:76 ;
hence integral (cos,A) = - 1 ; :: thesis: verum