let A be non empty closed_interval Subset of REAL; :: thesis: for f being PartFunc of REAL,REAL
for Z being open Subset of REAL st A c= Z & f = 2 (#) (exp_R (#) cos) & Z c= dom (exp_R (#) (sin + cos)) & Z = dom f & f | A is continuous holds
integral (f,A) = ((exp_R (#) (sin + cos)) . (upper_bound A)) - ((exp_R (#) (sin + cos)) . (lower_bound A))
let f be PartFunc of REAL,REAL; :: thesis: for Z being open Subset of REAL st A c= Z & f = 2 (#) (exp_R (#) cos) & Z c= dom (exp_R (#) (sin + cos)) & Z = dom f & f | A is continuous holds
integral (f,A) = ((exp_R (#) (sin + cos)) . (upper_bound A)) - ((exp_R (#) (sin + cos)) . (lower_bound A))
let Z be open Subset of REAL; :: thesis: ( A c= Z & f = 2 (#) (exp_R (#) cos) & Z c= dom (exp_R (#) (sin + cos)) & Z = dom f & f | A is continuous implies integral (f,A) = ((exp_R (#) (sin + cos)) . (upper_bound A)) - ((exp_R (#) (sin + cos)) . (lower_bound A)) )
assume A1: ( A c= Z & f = 2 (#) (exp_R (#) cos) & Z c= dom (exp_R (#) (sin + cos)) & Z = dom f & f | A is continuous ) ; :: thesis: integral (f,A) = ((exp_R (#) (sin + cos)) . (upper_bound A)) - ((exp_R (#) (sin + cos)) . (lower_bound A))
then A2: ( f is_integrable_on A & f | A is bounded ) by INTEGRA5:10, INTEGRA5:11;
A3: exp_R (#) (sin + cos) is_differentiable_on Z by A1, FDIFF_7:41;
A4: for x being Real st x in Z holds
f . x = (2 * (exp_R . x)) * (cos . x)

proof

let x be Real; :: thesis: ( x in Z implies f . x = (2 * (exp_R . x)) * (cos . x) )
assume x in Z ; :: thesis: f . x = (2 * (exp_R . x)) * (cos . x)
(2 (#) (exp_R (#) cos)) . x = 2 * ((exp_R (#) cos) . x) by VALUED_1:6
.= 2 * ((exp_R . x) * (cos . x)) by VALUED_1:5 ;
hence f . x = (2 * (exp_R . x)) * (cos . x) by A1; :: thesis: verum

end;

A5: for x being Element of REAL st x in dom ((exp_R (#) (sin + cos)) `| Z) holds
((exp_R (#) (sin + cos)) `| Z) . x = f . x

proof

let x be Element of REAL ; :: thesis: ( x in dom ((exp_R (#) (sin + cos)) `| Z) implies ((exp_R (#) (sin + cos)) `| Z) . x = f . x )
assume x in dom ((exp_R (#) (sin + cos)) `| Z) ; :: thesis: ((exp_R (#) (sin + cos)) `| Z) . x = f . x
then A6: x in Z by A3, FDIFF_1:def 7;
then ((exp_R (#) (sin + cos)) `| Z) . x = (2 * (exp_R . x)) * (cos . x) by A1, FDIFF_7:41
.= f . x by A6, A4 ;
hence ((exp_R (#) (sin + cos)) `| Z) . x = f . x ; :: thesis: verum

end;

dom ((exp_R (#) (sin + cos)) `| Z) = dom f by A1, A3, FDIFF_1:def 7;
then (exp_R (#) (sin + cos)) `| Z = f by A5, PARTFUN1:5;
hence integral (f,A) = ((exp_R (#) (sin + cos)) . (upper_bound A)) - ((exp_R (#) (sin + cos)) . (lower_bound A)) by A1, A2, FDIFF_7:41, INTEGRA5:13; :: thesis: verum