let n be Element of NAT ; :: thesis: for A being closed-interval Subset of REAL st A = [.0 ,(2 * PI ).] holds
integral (((#Z n) * cos ) (#) sin ),A = 0
let A be closed-interval Subset of REAL ; :: thesis: ( A = [.0 ,(2 * PI ).] implies integral (((#Z n) * cos ) (#) sin ),A = 0 )
assume A = [.0 ,(2 * PI ).] ; :: thesis: integral (((#Z n) * cos ) (#) sin ),A = 0
then ( upper_bound A = 2 * PI & lower_bound A = 0 ) by INTEGRA8:37;
then integral (((#Z n) * cos ) (#) sin ),A = (((- (1 / (n + 1))) (#) ((#Z (n + 1)) * cos )) . (2 * PI )) - (((- (1 / (n + 1))) (#) ((#Z (n + 1)) * cos )) . 0 ) by Th22
.= ((- (1 / (n + 1))) * (((#Z (n + 1)) * cos ) . (2 * PI ))) - (((- (1 / (n + 1))) (#) ((#Z (n + 1)) * cos )) . 0 ) by VALUED_1:6
.= ((- (1 / (n + 1))) * (((#Z (n + 1)) * cos ) . (2 * PI ))) - ((- (1 / (n + 1))) * (((#Z (n + 1)) * cos ) . 0 )) by VALUED_1:6
.= ((- (1 / (n + 1))) * ((#Z (n + 1)) . (cos . (2 * PI )))) - ((- (1 / (n + 1))) * (((#Z (n + 1)) * cos ) . 0 )) by FUNCT_1:23, SIN_COS:27
.= ((- (1 / (n + 1))) * ((#Z (n + 1)) . (cos . (2 * PI )))) - ((- (1 / (n + 1))) * ((#Z (n + 1)) . (cos . 0 ))) by FUNCT_1:23, SIN_COS:27
.= 0 by SIN_COS:33, SIN_COS:81 ;
hence integral (((#Z n) * cos ) (#) sin ),A = 0 ; :: thesis: verum