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 & Z = dom f & f = exp_R (#) (cos * exp_R) holds
integral (f,A) = ((sin * exp_R) . (upper_bound A)) - ((sin * exp_R) . (lower_bound A))
let f be PartFunc of REAL,REAL; :: thesis: for Z being open Subset of REAL st A c= Z & Z = dom f & f = exp_R (#) (cos * exp_R) holds
integral (f,A) = ((sin * exp_R) . (upper_bound A)) - ((sin * exp_R) . (lower_bound A))
let Z be open Subset of REAL; :: thesis: ( A c= Z & Z = dom f & f = exp_R (#) (cos * exp_R) implies integral (f,A) = ((sin * exp_R) . (upper_bound A)) - ((sin * exp_R) . (lower_bound A)) )
assume A1: ( A c= Z & Z = dom f & f = exp_R (#) (cos * exp_R) ) ; :: thesis: integral (f,A) = ((sin * exp_R) . (upper_bound A)) - ((sin * exp_R) . (lower_bound A))
then Z = (dom exp_R) /\ (dom (cos * exp_R)) by VALUED_1:def 4;
then A2: ( Z c= dom exp_R & Z c= dom (cos * exp_R) ) by XBOOLE_1:18;
for y being object st y in Z holds
y in dom (sin * exp_R)

proof

let y be object ; :: thesis: ( y in Z implies y in dom (sin * exp_R) )
assume y in Z ; :: thesis: y in dom (sin * exp_R)
then ( y in dom exp_R & exp_R . y in dom sin ) by A2, SIN_COS:24, XREAL_0:def 1;
hence y in dom (sin * exp_R) by FUNCT_1:11; :: thesis: verum

end;

then A3: Z c= dom (sin * exp_R) ;
A4: cos * exp_R is_differentiable_on Z by A2, FDIFF_7:35;
exp_R is_differentiable_on Z by FDIFF_1:26, TAYLOR_1:16;
then f | Z is continuous by A1, A4, FDIFF_1:21, FDIFF_1:25;
then f | A is continuous by A1, FCONT_1:16;
then A5: ( f is_integrable_on A & f | A is bounded ) by A1, INTEGRA5:10, INTEGRA5:11;
A6: sin * exp_R is_differentiable_on Z by A3, FDIFF_7:34;
A7: for x being Real st x in Z holds
f . x = (exp_R . x) * (cos . (exp_R . x))

proof

let x be Real; :: thesis: ( x in Z implies f . x = (exp_R . x) * (cos . (exp_R . x)) )
assume A8: x in Z ; :: thesis: f . x = (exp_R . x) * (cos . (exp_R . x))
then (exp_R (#) (cos * exp_R)) . x = (exp_R . x) * ((cos * exp_R) . x) by A1, VALUED_1:def 4
.= (exp_R . x) * (cos . (exp_R . x)) by A2, A8, FUNCT_1:12 ;
hence f . x = (exp_R . x) * (cos . (exp_R . x)) by A1; :: thesis: verum

end;

A9: for x being Element of REAL st x in dom ((sin * exp_R) `| Z) holds
((sin * exp_R) `| Z) . x = f . x

proof

let x be Element of REAL ; :: thesis: ( x in dom ((sin * exp_R) `| Z) implies ((sin * exp_R) `| Z) . x = f . x )
assume x in dom ((sin * exp_R) `| Z) ; :: thesis: ((sin * exp_R) `| Z) . x = f . x
then A10: x in Z by A6, FDIFF_1:def 7;
then ((sin * exp_R) `| Z) . x = (exp_R . x) * (cos . (exp_R . x)) by A3, FDIFF_7:34
.= f . x by A10, A7 ;
hence ((sin * exp_R) `| Z) . x = f . x ; :: thesis: verum

end;

dom ((sin * exp_R) `| Z) = dom f by A1, A6, FDIFF_1:def 7;
then (sin * exp_R) `| Z = f by A9, PARTFUN1:5;
hence integral (f,A) = ((sin * exp_R) . (upper_bound A)) - ((sin * exp_R) . (lower_bound A)) by A1, A5, A3, FDIFF_7:34, INTEGRA5:13; :: thesis: verum