let A be non empty closed_interval Subset of REAL; 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 = (sin . x) + ((sin . x) / ((cos . x) ^2)) ) & Z c= dom (sin (#) tan) & Z = dom f & f | A is continuous holds
integral (f,A) = ((sin (#) tan) . (upper_bound A)) - ((sin (#) tan) . (lower_bound A))
let f be 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 = (sin . x) + ((sin . x) / ((cos . x) ^2)) ) & Z c= dom (sin (#) tan) & Z = dom f & f | A is continuous holds
integral (f,A) = ((sin (#) tan) . (upper_bound A)) - ((sin (#) tan) . (lower_bound A))
let Z be open Subset of REAL; ( A c= Z & ( for x being Real st x in Z holds
f . x = (sin . x) + ((sin . x) / ((cos . x) ^2)) ) & Z c= dom (sin (#) tan) & Z = dom f & f | A is continuous implies integral (f,A) = ((sin (#) tan) . (upper_bound A)) - ((sin (#) tan) . (lower_bound A)) )
assume A1:
( A c= Z & ( for x being Real st x in Z holds
f . x = (sin . x) + ((sin . x) / ((cos . x) ^2)) ) & Z c= dom (sin (#) tan) & Z = dom f & f | A is continuous )
; integral (f,A) = ((sin (#) tan) . (upper_bound A)) - ((sin (#) tan) . (lower_bound A))
then A2:
( f is_integrable_on A & f | A is bounded )
by INTEGRA5:10, INTEGRA5:11;
A3:
sin (#) tan is_differentiable_on Z
by A1, FDIFF_10:12;
A4:
for x being Element of REAL st x in dom ((sin (#) tan) `| Z) holds
((sin (#) tan) `| Z) . x = f . x
dom ((sin (#) tan) `| Z) = dom f
by A1, A3, FDIFF_1:def 7;
then
(sin (#) tan) `| Z = f
by A4, PARTFUN1:5;
hence
integral (f,A) = ((sin (#) tan) . (upper_bound A)) - ((sin (#) tan) . (lower_bound A))
by A1, A2, A3, INTEGRA5:13; verum