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 & f = 2 (#) (exp_R (#) sin ) & 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 (#) sin ) & 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 (#) sin ) & 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 (#) sin ) & 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_7:40;
B1: for x being Real st x in Z holds
f . x = (2 * (exp_R . x)) * (sin . x)

proof

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

end;

A4: for x being Real st x in dom ((exp_R (#) (sin - cos )) `| Z) holds
((exp_R (#) (sin - cos )) `| Z) . x = f . x

proof

let x be 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 A5: x in Z by A3, FDIFF_1:def 8;
then ((exp_R (#) (sin - cos )) `| Z) . x = (2 * (exp_R . x)) * (sin . x) by A1, FDIFF_7:40
.= f . x by B1, A5 ;
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 8;
then (exp_R (#) (sin - cos )) `| Z = f by A4, PARTFUN1:34;
hence integral f,A = ((exp_R (#) (sin - cos )) . (upper_bound A)) - ((exp_R (#) (sin - cos )) . (lower_bound A)) by A1, A2, FDIFF_7:40, INTEGRA5:13; :: thesis: verum