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 = (sin (#) cos) ^ & Z c= dom (ln * tan) & Z = dom f & f | A is continuous holds
integral (f,A) = ((ln * tan) . (upper_bound A)) - ((ln * tan) . (lower_bound A))
let f be PartFunc of REAL,REAL; :: thesis: for Z being open Subset of REAL st A c= Z & f = (sin (#) cos) ^ & Z c= dom (ln * tan) & Z = dom f & f | A is continuous holds
integral (f,A) = ((ln * tan) . (upper_bound A)) - ((ln * tan) . (lower_bound A))
let Z be open Subset of REAL; :: thesis: ( A c= Z & f = (sin (#) cos) ^ & Z c= dom (ln * tan) & Z = dom f & f | A is continuous implies integral (f,A) = ((ln * tan) . (upper_bound A)) - ((ln * tan) . (lower_bound A)) )
assume A1: ( A c= Z & f = (sin (#) cos) ^ & Z c= dom (ln * tan) & Z = dom f & f | A is continuous ) ; :: thesis: integral (f,A) = ((ln * tan) . (upper_bound A)) - ((ln * tan) . (lower_bound A))
then A2: ( f is_integrable_on A & f | A is bounded ) by INTEGRA5:10, INTEGRA5:11;
A3: ln * tan is_differentiable_on Z by A1, FDIFF_8:18;
A4: for x being Real st x in Z holds
f . x = 1 / ((sin . x) * (cos . x))

proof

let x be Real; :: thesis: ( x in Z implies f . x = 1 / ((sin . x) * (cos . x)) )
assume x in Z ; :: thesis: f . x = 1 / ((sin . x) * (cos . x))
then ((sin (#) cos) ^) . x = 1 / ((sin (#) cos) . x) by A1, RFUNCT_1:def 2
.= 1 / ((sin . x) * (cos . x)) by VALUED_1:5 ;
hence f . x = 1 / ((sin . x) * (cos . x)) by A1; :: thesis: verum

end;

A5: for x being Element of REAL st x in dom ((ln * tan) `| Z) holds
((ln * tan) `| Z) . x = f . x

proof

let x be Element of REAL ; :: thesis: ( x in dom ((ln * tan) `| Z) implies ((ln * tan) `| Z) . x = f . x )
assume x in dom ((ln * tan) `| Z) ; :: thesis: ((ln * tan) `| Z) . x = f . x
then A6: x in Z by A3, FDIFF_1:def 7;
then ((ln * tan) `| Z) . x = 1 / ((sin . x) * (cos . x)) by A1, FDIFF_8:18
.= f . x by A4, A6 ;
hence ((ln * tan) `| Z) . x = f . x ; :: thesis: verum

end;

dom ((ln * tan) `| Z) = dom f by A1, A3, FDIFF_1:def 7;
then (ln * tan) `| Z = f by A5, PARTFUN1:5;
hence integral (f,A) = ((ln * tan) . (upper_bound A)) - ((ln * tan) . (lower_bound A)) by A1, A2, FDIFF_8:18, INTEGRA5:13; :: thesis: verum