Prove the formula, where m and n are positive integers. sin ( mx ) sin ( n x ) dx = 0 if m # n if m = n Let m and n be positive integers. First consider the case where m # n. By the product identity sin (x) sin(y) = = cos(x - y) - cos(x + y) , the integral can be rewritten as follows, [, sin(mx) sin(ox) ox - , = [cos( - cos ( mx + nx ) LE [ cos (( [m = " )x - cos ( ( m + n ) x ) 0 - 0 x =0 Now, consider the case where m = n. Note that the integral can be rewritten as follows. sin ( mx ) sin ( nx ) dx = sin ( mx ) sin ( mx ) dx "sin ? ( mix ) dx By the half angle formula sin(x) = 1 - cos(2x) 2 [ sin ?(mxax - L =(1 -([cos ( 2x) = 7 Thus, for positive integers m and n, sin(mx) sin(nx) dx _ JO if m # n x if m = n