Integral Formula of Product Functions

integral formula of product functions sin ax cos bx sin ax sin bx cos ax cos bx sin ax cos ax shahbaz ahmed June 2024 Introduction The formula involve sum and product of trigonometric functions sin (a + b)x = sin ax cos bx + cos ax sin bx............(1) sin (a − b)x = sin ax cos bx − cos ax sin bx..............(2) Adding equation (1) and equation (2) sin (a + b)x + sin (a − b)x = sin ax cos bx + cos ax sin bx + sin ax cos bx − cos ax sin bx sin (a + b)x + sin (a − b)x = sin ax cos bx + sin ax cos bx sin (a + b)x + sin (a − b)x = 2 sin ax cos bx Subtracting equation (2) from equation (1) sin (a + b)x − sin (a − b)x = sin ax cos bx + cos ax sin bx − [sin ax cos bx − cos ax sin bx] sin (a + b)x − sin (a − b)x = sin ax cos bx + cos ax sin bx − sin ax cos bx + cos ax sin bx sin (a + b)x − sin (a − b)x = 2 cos ax sin bx 1 Similarly cos (a + b)x = cos ax cos bx − sin ax sin bx............(3) cos (a − b)x = cos ax cos bx + sin ax sin bx..............(4) Adding equation (3) and equation (4) cos (a + b)x + cos (a − b)x = cos ax cos bx − sin ax sin bx + cos ax cos bx + sin ax sin bx cos (a + b)x + cos (a − b)x = 2 cos ax cos bx Subtracting equation (4) from equation (3) cos (a + b)x − cos (a − b)x = cos ax cos bx − sin ax sin bx − [cos ax cos bx + sin ax sin bx] cos (a + b)x − cos (a − b)x = cos ax cos bx − sin ax sin bx − cos ax cos bx − sin ax sin bx cos (a + b)x − cos (a − b)x = − sin ax sin bx − sin ax sin bx cos (a + b)x − cos (a − b)x = −2 sin ax sin bx ........................................................................... R sin ax cos bx dx Using the formula sin (a + b)x + sin (a − b)x = 2 sin ax cos bx 1 2 [sin (a + b)x + sin (a − b)x] = sin ax cos bx R sin ax cos bx dx = 12 [sin (a + b)x + sin (a − b)x] dx R R sin ax cos bx dx = 21 [sin (a + b)x + sin (a − b)x] dx R R R sin ax cos bx dx = 21 sin (a + b)x + 21 sin (a − b)x dx R Now d[cos (a + b)x] = − sin (a + b)x d[(a + b)x] d[cos (a + b)x] = −(a + b) sin (a + b)x dx 2 1 a+b d[cos (a + b)x] = − sin (a + b)x dx 1 d[cos (a + b)x] = sin (a + b)x dx − a+b Similarly d[cos (a − b)x] = − sin (a − b)x d[(a − b)x] d[cos (a − b)x] = −(a − b) sin (a − b)x dx 1 a−b d[cos (a − b)x] = − sin (a − b)x dx 1 − a−b d[cos (a − b)x] = sin (a − b)x dx Putting R sin ax R sin ax R sin ax in the above integral R R 1 1 cos bx dx = 21 [− a+b ]d[cos (a + b)x] + 21 [− a−b ]d[cos (a − b)x] R R 1 1 cos bx dx = 21 [− a+b ] d[cos (a + b)x] + 21 [− a−b ] d[cos (a − b)x] 1 1 cos bx dx = − 12 [ a+b ][cos (a + b)x] − 12 [ a−b ][cos (a − b)x] + C ............................................................ R sin ax sin bx dx Using the formula cos (a + b)x − cos (a − b)x = −2 sin ax sin bx − 12 [cos (a + b)x − cos (a − b)x] = sin ax sin bx Putting in the integral R R sin ax sin bx dx = − 12 [cos (a + b)x − cos (a − b)x]dx R R R sin ax sin bx dx = − 21 cos (a + b)x dx + 21 cos (a − b)x dx Now d[sin (a + b)x] = cos (a + b)x d(a + b)x d[sin (a + b)x] = (a + b) cos (a + b)x dx 1 a+b d[sin (a + b)x] = cos (a + b)x dx Similarly d[sin (a − b)x] = cos (a − b)x d(a − b)x d[sin (a − b)x] = (a − b) cos (a − b)x dx 3 1 a−b d[sin (a − b)x] = cos (a − b)x dx Putting in the above integral R R 1 R 1 sin ax sin bx dx = − 21 a+b d[sin (a + b)x] + 12 a−b d[sin (a − b)x] R R R 1 1 sin ax sin bx dx = − 21 a+b d[sin (a + b)x] + 21 a−b d[sin (a − b)x] R 1 1 1 1 sin ax sin bx dx = − 2 a+b [sin (a + b)x] + 2 a−b [sin (a − b)x] + C .......................................................................... R cos ax cos bx dx Using the formula cos (a + b)x + cos (a − b)x = 2 cos ax cos bx 1 2 cos (a + b)x + cos (a − b)x = cos ax cos bx Putting in the integral R R cos ax cos bx dx = 21 [cos (a + b)x + cos (a − b)x]dx R R R sin ax sin bx dx = 21 cos (a + b)x dx + 21 cos (a − b)x dx Now d[sin (a + b)x] = cos (a + b)x d(a + b)x d[sin (a + b)x] = (a + b) cos (a + b)x dx 1 a+b d[sin (a + b)x] = cos (a + b)x dx Similarly d[sin (a − b)x] = cos (a − b)x d(a − b)x d[sin (a − b)x] = (a − b) cos (a − b)x dx 1 a−b d[sin (a − b)x] = cos (a − b)x dx Putting in the R cos ax cos bx R cos ax cos bx R cos ax cos bx above integral R 1 d[sin (a + b)x] + dx = 21 a+b R 1 dx = 12 a+b d[sin (a + b)x] + dx = 1 1 2 a+b [sin (a + b)x] + R 1 a−b d[sin (a − b)x] 1 1 d[sin (a − b)x] 2 a−b 1 1 2 a−b [sin (a − b)x] + C .................................................................. 4 1 2 R Derive the formula R sin ax cos ax dx Putting a=b in the formula R 1 1 sin ax cos bx dx = − 12 [ a+b ][cos (a + b)x] − 12 [ a−b ][cos (a − b)x] + C (a−b)x Note that lima−→b cos a−b =0 R 1 ][cos (a + a)x] + C sin ax cos ax dx = − 12 [ a+a R 1 cos 2ax + C sin ax cos ax dx = − 4a ........................................................... Reference to the graph of the function y = sin 3x cos 3x, find the area of the curve for 0 ≤ x ≤ π 6 Since R 1 1 sin ax cos bx dx = − 12 [ a+b ][cos (a + b)x] − 12 [ a−b ][cos (a − b)x] + C Putting a = 2, b = 3 in the integral R 1 sin 2x cos 3x dx = − 21 [ 51 ][cos 5x] − 12 [ −1 ][cos (−x)] + C R 1 sin 2x cos 3x dx = − 10 [cos 5x] + 12 cos x + C Area of the curve for 0 ≤ x ≤ π6 π π Rπ 1 Area= 06 sin 2x cos 3x dx = − 10 [cos 5x]06 + 21 [cos x]06 Rπ 1 1 π Area= 06 sin 2x cos 3x dx = − 10 [cos 5π 6 − cos 0] + 2 [cos 6 − cos 0] 5 1 Area=− 10 [− √ 3 2 − 1] + 12 [ √ 3 2 − 1] √ √ Area=[ 2+20 3 ] + [ −2+4 3 ] √ 3 Area= −4+3 10 6