let A be closed-interval Subset of REAL ; :: thesis: for f being PartFunc of REAL ,REAL
for Z being open Subset of REAL st A c= Z & ( for x being Real st x in Z holds
f . x = ((exp_R . x) * (sin . (exp_R . x))) / ((cos . (exp_R . x)) ^2 ) ) & Z c= dom (sec * exp_R ) & Z = dom f & f | A is continuous holds
integral f,A = ((sec * exp_R ) . (upper_bound A)) - ((sec * exp_R ) . (lower_bound A))
let f be PartFunc of REAL ,REAL ; :: thesis: for Z being open Subset of REAL st A c= Z & ( for x being Real st x in Z holds
f . x = ((exp_R . x) * (sin . (exp_R . x))) / ((cos . (exp_R . x)) ^2 ) ) & Z c= dom (sec * exp_R ) & Z = dom f & f | A is continuous holds
integral f,A = ((sec * exp_R ) . (upper_bound A)) - ((sec * exp_R ) . (lower_bound A))
let Z be open Subset of REAL ; :: thesis: ( A c= Z & ( for x being Real st x in Z holds
f . x = ((exp_R . x) * (sin . (exp_R . x))) / ((cos . (exp_R . x)) ^2 ) ) & Z c= dom (sec * exp_R ) & Z = dom f & f | A is continuous implies integral f,A = ((sec * exp_R ) . (upper_bound A)) - ((sec * exp_R ) . (lower_bound A)) )
assume A1: ( A c= Z & ( for x being Real st x in Z holds
f . x = ((exp_R . x) * (sin . (exp_R . x))) / ((cos . (exp_R . x)) ^2 ) ) & Z c= dom (sec * exp_R ) & Z = dom f & f | A is continuous ) ; :: thesis: integral f,A = ((sec * exp_R ) . (upper_bound A)) - ((sec * exp_R ) . (lower_bound A))
then A2: ( f is_integrable_on A & f | A is bounded ) by INTEGRA5:10, INTEGRA5:11;
A3: sec * exp_R is_differentiable_on Z by A1, FDIFF_9:12;
A4: for x being Real st x in dom ((sec * exp_R ) `| Z) holds
((sec * exp_R ) `| Z) . x = f . x

proof

let x be Real; :: thesis: ( x in dom ((sec * exp_R ) `| Z) implies ((sec * exp_R ) `| Z) . x = f . x )
assume x in dom ((sec * exp_R ) `| Z) ; :: thesis: ((sec * exp_R ) `| Z) . x = f . x
then A5: x in Z by A3, FDIFF_1:def 8;
then ((sec * exp_R ) `| Z) . x = ((exp_R . x) * (sin . (exp_R . x))) / ((cos . (exp_R . x)) ^2 ) by A1, FDIFF_9:12
.= f . x by A1, A5 ;
hence ((sec * exp_R ) `| Z) . x = f . x ; :: thesis: verum

end;

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