let A be closed-interval Subset of REAL ; :: thesis: for Z being open Subset of REAL st A c= Z & ( for x being Real st x in Z holds
sin . x > 0 ) & Z c= dom (ln * sin ) & Z = dom cot & cot | A is continuous holds
integral cot ,A = ((ln * sin ) . (sup A)) - ((ln * sin ) . (inf A))
let Z be open Subset of REAL ; :: thesis: ( A c= Z & ( for x being Real st x in Z holds
sin . x > 0 ) & Z c= dom (ln * sin ) & Z = dom cot & cot | A is continuous implies integral cot ,A = ((ln * sin ) . (sup A)) - ((ln * sin ) . (inf A)) )
set f = cot ;
assume A1: ( A c= Z & ( for x being Real st x in Z holds
sin . x > 0 ) & Z c= dom (ln * sin ) & Z = dom cot & cot | A is continuous ) ; :: thesis: integral cot ,A = ((ln * sin ) . (sup A)) - ((ln * sin ) . (inf A))
then A2: ( cot is_integrable_on A & cot | A is bounded ) by INTEGRA5:10, INTEGRA5:11;
A3: ln * sin is_differentiable_on Z by A1, FDIFF_4:43;
A4: for x being Real st x in dom ((ln * sin ) `| Z) holds
((ln * sin ) `| Z) . x = cot . x

proof

let x be Real; :: thesis: ( x in dom ((ln * sin ) `| Z) implies ((ln * sin ) `| Z) . x = cot . x )
assume x in dom ((ln * sin ) `| Z) ; :: thesis: ((ln * sin ) `| Z) . x = cot . x
then A5: x in Z by A3, FDIFF_1:def 8;
then AA: ( x is Real & sin . x <> 0 ) by A1;
((ln * sin ) `| Z) . x = cot x by A5, A1, FDIFF_4:43
.= cot . x by SIN_COS9:16, AA ;
hence ((ln * sin ) `| Z) . x = cot . x ; :: thesis: verum

end;

dom ((ln * sin ) `| Z) = dom cot by A1, A3, FDIFF_1:def 8;
then (ln * sin ) `| Z = cot by A4, PARTFUN1:34;
hence integral cot ,A = ((ln * sin ) . (sup A)) - ((ln * sin ) . (inf A)) by A1, A2, FDIFF_4:43, INTEGRA5:13; :: thesis: verum