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 & dom sec = Z & ( for x being Real st x in Z holds
f . x = (sin . x) / ((cos . x) ^2 ) ) & Z = dom f & f | A is continuous holds
integral f,A = (sec . (upper_bound A)) - (sec . (lower_bound A))
let f be PartFunc of REAL ,REAL ; :: thesis: for Z being open Subset of REAL st A c= Z & dom sec = Z & ( for x being Real st x in Z holds
f . x = (sin . x) / ((cos . x) ^2 ) ) & Z = dom f & f | A is continuous holds
integral f,A = (sec . (upper_bound A)) - (sec . (lower_bound A))
let Z be open Subset of REAL ; :: thesis: ( A c= Z & dom sec = Z & ( for x being Real st x in Z holds
f . x = (sin . x) / ((cos . x) ^2 ) ) & Z = dom f & f | A is continuous implies integral f,A = (sec . (upper_bound A)) - (sec . (lower_bound A)) )
assume A1: ( A c= Z & dom sec = Z & ( for x being Real st x in Z holds
f . x = (sin . x) / ((cos . x) ^2 ) ) & Z = dom f & f | A is continuous ) ; :: thesis: integral f,A = (sec . (upper_bound A)) - (sec . (lower_bound A))
then A2: ( f is_integrable_on A & f | A is bounded ) by INTEGRA5:10, INTEGRA5:11;
A3: sec is_differentiable_on Z by A1, FDIFF_9:4;
A4: for x being Real st x in dom (sec `| Z) holds
(sec `| Z) . x = f . x

proof

let x be Real; :: thesis: ( x in dom (sec `| Z) implies (sec `| Z) . x = f . x )
assume x in dom (sec `| Z) ; :: thesis: (sec `| Z) . x = f . x
then A5: x in Z by A3, FDIFF_1:def 8;
then (sec `| Z) . x = (sin . x) / ((cos . x) ^2 ) by A1, FDIFF_9:4
.= f . x by A1, A5 ;
hence (sec `| Z) . x = f . x ; :: thesis: verum

end;

dom (sec `| Z) = dom f by A1, A3, FDIFF_1:def 8;
then sec `| Z = f by A4, PARTFUN1:34;
hence integral f,A = (sec . (upper_bound A)) - (sec . (lower_bound A)) by A1, A2, FDIFF_9:4, INTEGRA5:13; :: thesis: verum