let A be non empty closed_interval Subset of REAL; :: thesis: for x being Real
for n being Element of NAT st A = [.(x + ((2 * n) * PI)),(x + (((2 * n) + 1) * PI)).] holds
integral ((sin (#) cos),A) = 0
let x be Real; :: thesis: for n being Element of NAT st A = [.(x + ((2 * n) * PI)),(x + (((2 * n) + 1) * PI)).] holds
integral ((sin (#) cos),A) = 0
let n be Element of NAT ; :: thesis: ( A = [.(x + ((2 * n) * PI)),(x + (((2 * n) + 1) * PI)).] implies integral ((sin (#) cos),A) = 0 )
assume A = [.(x + ((2 * n) * PI)),(x + (((2 * n) + 1) * PI)).] ; :: thesis: integral ((sin (#) cos),A) = 0
then ( upper_bound A = x + (((2 * n) + 1) * PI) & lower_bound A = x + ((2 * n) * PI) ) by Th37;
then integral ((sin (#) cos),A) = (1 / 2) * (((cos . (x + ((2 * n) * PI))) * (cos . (x + ((2 * n) * PI)))) - ((cos . (x + (((2 * n) + 1) * PI))) * (cos . (x + (((2 * n) + 1) * PI))))) by Th90
.= (1 / 2) * (((cos x) * (cos (x + ((2 * n) * PI)))) - ((cos (x + (((2 * n) + 1) * PI))) * (cos (x + (((2 * n) + 1) * PI))))) by Th3
.= (1 / 2) * (((cos x) * (cos x)) - ((cos (x + (((2 * n) + 1) * PI))) * (cos (x + (((2 * n) + 1) * PI))))) by Th3
.= (1 / 2) * (((cos x) * (cos x)) - ((- (cos x)) * (cos (x + (((2 * n) + 1) * PI))))) by Th4
.= (1 / 2) * (((cos x) * (cos x)) - ((- (cos x)) * (- (cos x)))) by Th4
.= 0 ;
hence integral ((sin (#) cos),A) = 0 ; :: thesis: verum