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
( f . x = 1 / ((cos . x) ^2 ) & cos . x <> 0 ) ) & dom tan = Z & Z = dom f & f | A is continuous holds
integral f,A = (tan . (sup A)) - (tan . (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
( f . x = 1 / ((cos . x) ^2 ) & cos . x <> 0 ) ) & dom tan = Z & Z = dom f & f | A is continuous holds
integral f,A = (tan . (sup A)) - (tan . (inf A))
let f be PartFunc of REAL ,REAL ; :: thesis: ( A c= Z & ( for x being Real st x in Z holds
( f . x = 1 / ((cos . x) ^2 ) & cos . x <> 0 ) ) & dom tan = Z & Z = dom f & f | A is continuous implies integral f,A = (tan . (sup A)) - (tan . (inf A)) )
assume A1: ( A c= Z & ( for x being Real st x in Z holds
( f . x = 1 / ((cos . x) ^2 ) & cos . x <> 0 ) ) & dom tan = Z & Z = dom f & f | A is continuous ) ; :: thesis: integral f,A = (tan . (sup A)) - (tan . (inf A))
then A2: ( f is_integrable_on A & f | A is bounded ) by INTEGRA5:10, INTEGRA5:11;
A3: tan is_differentiable_on Z by A1, INTEGRA8:33;
A4: for x being Real st x in dom (tan `| Z) holds
(tan `| Z) . x = f . x

proof

let x be Real; :: thesis: ( x in dom (tan `| Z) implies (tan `| Z) . x = f . x )
assume x in dom (tan `| Z) ; :: thesis: (tan `| Z) . x = f . x
then A5: x in Z by A3, FDIFF_1:def 8;
then (tan `| Z) . x = 1 / ((cos . x) ^2 ) by A1, INTEGRA8:33
.= f . x by A1, A5 ;
hence (tan `| Z) . x = f . x ; :: thesis: verum

end;

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