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 & ( for x being Real st x in Z holds
f . x = (cos . (sin . x)) * (cos . x) ) & Z = dom f & f | A is continuous holds
integral (f,A) = ((sin * sin) . (upper_bound A)) - ((sin * sin) . (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 = (cos . (sin . x)) * (cos . x) ) & Z = dom f & f | A is continuous holds
integral (f,A) = ((sin * sin) . (upper_bound A)) - ((sin * sin) . (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 = (cos . (sin . x)) * (cos . x) ) & Z = dom f & f | A is continuous implies integral (f,A) = ((sin * sin) . (upper_bound A)) - ((sin * sin) . (lower_bound A)) )
assume A1: ( A c= Z & ( for x being Real st x in Z holds
f . x = (cos . (sin . x)) * (cos . x) ) & Z = dom f & f | A is continuous ) ; :: thesis: integral (f,A) = ((sin * sin) . (upper_bound A)) - ((sin * sin) . (lower_bound A))
then A2: ( f is_integrable_on A & f | A is bounded ) by INTEGRA5:10, INTEGRA5:11;
A3: sin * sin is_differentiable_on Z by FDIFF_10:7;
A4: for x being Element of REAL st x in dom ((sin * sin) `| Z) holds
((sin * sin) `| Z) . x = f . x

proof

let x be Element of REAL ; :: thesis: ( x in dom ((sin * sin) `| Z) implies ((sin * sin) `| Z) . x = f . x )
assume x in dom ((sin * sin) `| Z) ; :: thesis: ((sin * sin) `| Z) . x = f . x
then A5: x in Z by A3, FDIFF_1:def 7;
then ((sin * sin) `| Z) . x = (cos . (sin . x)) * (cos . x) by FDIFF_10:7
.= f . x by A1, A5 ;
hence ((sin * sin) `| Z) . x = f . x ; :: thesis: verum

end;

dom ((sin * sin) `| Z) = dom f by A1, A3, FDIFF_1:def 7;
then (sin * sin) `| Z = f by A4, PARTFUN1:5;
hence integral (f,A) = ((sin * sin) . (upper_bound A)) - ((sin * sin) . (lower_bound A)) by A1, A2, A3, INTEGRA5:13; :: thesis: verum