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Evaluate the integral 4sin5xcos8xdx?
Once again, the integral I is 4sin5xcos8xdx
Thanks for the help!
2 Answers
- MechEng2030Lv 78 years agoFavorite Answer
∫4sin(5x)cos(8x) dx
sin(3x) = sin(8x - 5x) = sin(8x)cos(5x) - cos(8x)sin(5x)
sin(13x) = sin(8x + 5x) = sin(8x)cos(5x) + cos(8x)sin(5x)
Subtracting the two:
sin(13x) - sin(3x) = 2cos(8x)sin(5x) => 2(sin(13x) - sin(3x)) = 4sin(5x)cos(8x)
∫2(sin(13x) - sin(3x)) dx
= -2/13*cos(13x) + 2/3*cos(3x) + C
- 8 years ago
4 * sin(5x) * cos(8x) * dx
sin(5x)cos(8x) =>
(1/2) * (2sin(5x)cos(8x)) =>
(1/2) * (sin(5x)cos(8x) + sin(5x)cos(8x)) =>
(1/2) * (sin(5x)cos(8x) - sin(8x)cos(5x) + sin(8x)cos(5x) + sin(5x)cos(8x)) =>
(1/2) * (sin(5x - 8x) + sin(8x + 5x)) =>
(1/2) * (sin(-3x) + sin(13x)) =>
(1/2) * (sin(13x) - sin(3x))
4 * (1/2) * (sin(13x) - sin(3x)) * dx =>
2 * sin(13x) * dx - 2 * sin(3x) * dx
Integrate
(-2/13) * cos(13x) + (2/3) * cos(3x) + C =>
(2/3) * (1/13) * (13 * cos(3x) - 3 * cos(13x)) + C =>
(2/39) * (13cos(3x) - 3cos(13x)) + C
