let A be closed-interval Subset of REAL ; for Z being open Subset of REAL
for f being PartFunc of REAL ,REAL st A c= Z & ( for x being Real st x in Z holds
( 1 + (sin . x) <> 0 & 1 - (sin . x) <> 0 & f . x = 1 / (1 + (sin . x)) ) ) & dom (tan - sec ) = Z & Z = dom f & f | A is continuous holds
integral f,A = ((tan - sec ) . (upper_bound A)) - ((tan - sec ) . (lower_bound A))
let Z be open Subset of REAL ; for f being PartFunc of REAL ,REAL st A c= Z & ( for x being Real st x in Z holds
( 1 + (sin . x) <> 0 & 1 - (sin . x) <> 0 & f . x = 1 / (1 + (sin . x)) ) ) & dom (tan - sec ) = Z & Z = dom f & f | A is continuous holds
integral f,A = ((tan - sec ) . (upper_bound A)) - ((tan - sec ) . (lower_bound A))
let f be PartFunc of REAL ,REAL ; ( A c= Z & ( for x being Real st x in Z holds
( 1 + (sin . x) <> 0 & 1 - (sin . x) <> 0 & f . x = 1 / (1 + (sin . x)) ) ) & dom (tan - sec ) = Z & Z = dom f & f | A is continuous implies integral f,A = ((tan - sec ) . (upper_bound A)) - ((tan - sec ) . (lower_bound A)) )
assume that
A1:
A c= Z
and
A2:
for x being Real st x in Z holds
( 1 + (sin . x) <> 0 & 1 - (sin . x) <> 0 & f . x = 1 / (1 + (sin . x)) )
and
A3:
dom (tan - sec ) = Z
and
A4:
Z = dom f
and
A5:
f | A is continuous
; integral f,A = ((tan - sec ) . (upper_bound A)) - ((tan - sec ) . (lower_bound A))
A6:
f is_integrable_on A
by A1, A4, A5, INTEGRA5:11;
A7:
for x being Real st x in Z holds
( 1 + (sin . x) <> 0 & 1 - (sin . x) <> 0 )
by A2;
then A8:
tan - sec is_differentiable_on Z
by A3, Th39;
A9:
for x being Real st x in dom ((tan - sec ) `| Z) holds
((tan - sec ) `| Z) . x = f . x
dom ((tan - sec ) `| Z) = dom f
by A4, A8, FDIFF_1:def 8;
then
(tan - sec ) `| Z = f
by A9, PARTFUN1:34;
hence
integral f,A = ((tan - sec ) . (upper_bound A)) - ((tan - sec ) . (lower_bound A))
by A1, A4, A5, A6, A8, INTEGRA5:10, INTEGRA5:13; verum