let A be closed-interval Subset of REAL ; :: thesis: 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 ) . (sup A)) - ((tan + sec ) . (inf A))
let Z be open Subset of REAL ; :: thesis: 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 ) . (sup A)) - ((tan + sec ) . (inf A))
let f be PartFunc of REAL ,REAL ; :: thesis: ( 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 ) . (sup A)) - ((tan + sec ) . (inf A)) )
assume A1: ( 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 ) ; :: thesis: integral f,A = ((tan + sec ) . (sup A)) - ((tan + sec ) . (inf A))
then A2: ( f is_integrable_on A & f | A is bounded ) by INTEGRA5:10, INTEGRA5:11;
B: for x being Real st x in Z holds
( 1 + (sin . x) <> 0 & 1 - (sin . x) <> 0 ) by A1;
then A3: tan + sec is_differentiable_on Z by A1, Th12;
A4: for x being Real st x in dom ((tan + sec ) `| Z) holds
((tan + sec ) `| Z) . x = f . x

proof

let x be Real; :: thesis: ( x in dom ((tan + sec ) `| Z) implies ((tan + sec ) `| Z) . x = f . x )
assume x in dom ((tan + sec ) `| Z) ; :: thesis: ((tan + sec ) `| Z) . x = f . x
then A5: x in Z by A3, FDIFF_1:def 8;
then ((tan + sec ) `| Z) . x = 1 / (1 - (sin . x)) by A1, B, Th12
.= f . x by A1, A5 ;
hence ((tan + sec ) `| Z) . x = f . x ; :: thesis: verum

end;

dom ((tan + sec ) `| Z) = dom f by A1, A3, FDIFF_1:def 8;
then (tan + sec ) `| Z = f by A4, PARTFUN1:34;
hence integral f,A = ((tan + sec ) . (sup A)) - ((tan + sec ) . (inf A)) by A1, A2, A3, INTEGRA5:13; :: thesis: verum