let A be non empty closed_interval Subset of REAL; :: thesis: for Z being open Subset of REAL st A c= Z & Z c= dom (ln * sec) & Z = dom tan & tan | A is continuous holds
integral (tan,A) = ((ln * sec) . (upper_bound A)) - ((ln * sec) . (lower_bound A))
let Z be open Subset of REAL; :: thesis: ( A c= Z & Z c= dom (ln * sec) & Z = dom tan & tan | A is continuous implies integral (tan,A) = ((ln * sec) . (upper_bound A)) - ((ln * sec) . (lower_bound A)) )
assume A1: ( A c= Z & Z c= dom (ln * sec) & Z = dom tan & tan | A is continuous ) ; :: thesis: integral (tan,A) = ((ln * sec) . (upper_bound A)) - ((ln * sec) . (lower_bound A))
then A2: ( tan is_integrable_on A & tan | A is bounded ) by INTEGRA5:10, INTEGRA5:11;
A3: ln * sec is_differentiable_on Z by A1, FDIFF_9:18;
A4: for x being Real st x in Z holds
cos . x <> 0

proof

let x be Real; :: thesis: ( x in Z implies cos . x <> 0 )
assume x in Z ; :: thesis: cos . x <> 0
then x in dom sec by A1, FUNCT_1:11;
hence cos . x <> 0 by RFUNCT_1:3; :: thesis: verum

end;

A5: for x being Element of REAL st x in dom ((ln * sec) `| Z) holds
((ln * sec) `| Z) . x = tan . x

proof

let x be Element of REAL ; :: thesis: ( x in dom ((ln * sec) `| Z) implies ((ln * sec) `| Z) . x = tan . x )
assume x in dom ((ln * sec) `| Z) ; :: thesis: ((ln * sec) `| Z) . x = tan . x
then A6: x in Z by A3, FDIFF_1:def 7;
then A7: cos . x <> 0 by A4;
((ln * sec) `| Z) . x = tan x by A1, A6, FDIFF_9:18
.= tan . x by A7, SIN_COS9:15 ;
hence ((ln * sec) `| Z) . x = tan . x ; :: thesis: verum

end;

dom ((ln * sec) `| Z) = dom tan by A1, A3, FDIFF_1:def 7;
then (ln * sec) `| Z = tan by A5, PARTFUN1:5;
hence integral (tan,A) = ((ln * sec) . (upper_bound A)) - ((ln * sec) . (lower_bound A)) by A1, A2, A3, INTEGRA5:13; :: thesis: verum