e 3t sin4t d t Solution Well, you gotta do with integration by parts. Let u = si

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Well, you gotta do with integration by parts. Let u = sin(4x) ==> du = 4cos(4x) dx dv = e^(3x) dx ==> v = (1/3)e^(3x). It will make ? e^(3x)sin(4x) dx ==> (1/3)e^(3x)sin(4x) - ? (4/3)e^(3x)cos(4x) dx ==> (1/3)e^(3x)sin(4x) - 4/3 ? e^(3x)cos(4x) dx. Then, integrate ? e^(3x)cos(4x) dx by parts with: u = cos(4x) ==> du = -4sin(4x) dx dv = e^(3x) dx ==> v = (1/3)e^(3x). This gives: ? e^(3x)sin(4x) dx = (1/3)e^(3x)sin(4x) - 4/3 ? e^(3x)cos(4x) dx = (1/3)e^(3x)sin(4x) - (4/3)(uv - ? v du) = (1/3)e^(3x)sin(4x) - (4/3)[(1/3)e^(3x)cos(4x) - ? (-4/3)e^(3x)sin(4x) dx] = (1/3)e^(3x)sin(4x) - (4/9)e^(3x)cos(4x) - 16/9 ? e^(3x)sin(4x) dx. This implies that: ? e^(3x)sin(4x) dx = (1/3)e^(3x)sin(4x) - (4/9)e^(3x)cos(4x) - 16/9 ? e^(3x)sin(4x) dx. Adding 16/9 ? e^(3x)sin(4x) dx to both sides: 25/9 ? e^(3x)sin(4x) dx = (1/3)e^(3x)sin(4x) - (4/9)e^(3x)cos(4x). Multiplying both sides by 9/25 and tacking on the integration constant: ? e^(3x)sin(4x) dx = 9/25 * [(1/3)e^(3x)sin(4x) - (4/9)e^(3x)cos(4x)] + C ==> ? e^(3x)sin(4x) dx = e^(3x) * [3sin(4x) - 4cos(4x)]/25 + C. is an answer!

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