Evaluate the integral. int-3x sin(-2x)dx Answer:

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Evaluate the integral.  int-3x sin(-2x)dx Answer:

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Identify integral: Identify the integral to be solved. We need to evaluate the integral of the function -3x sin(-2x) with respect to x.Apply integration by parts: Apply the integration by parts formula. Integration by parts states that ∫u dv = uv - ∫v du, where u and dv are differentiable functions of x. We choose u = -3x and dv = sin(-2x)dx. We need to find du and v.Differentiate u and integrate dv: Differentiate u and integrate dv. Differentiating u with respect to x gives us du = -3 dx. Integrating dv, we have v = ∫sin(-2x)dx. To integrate sin(-2x), we use the substitution method. Let w = -2x, then dw = -2 dx, and dx = -dw/2. Now we integrate sin(w) with respect to w and then substitute back. v = ∫sin(w) * (-1/2) dw = (-1/2) * (-cos(w)) = (1/2)cos(-2x).Apply integration by parts: Apply the integration by parts formula. Now we have u = -3x, du = -3 dx, and v = (1/2)cos(-2x). Plugging these into the integration by parts formula gives us: ∫-3x sin(-2x)dx = uv - ∫v du = (-3x) * (1/2)cos(-2x) - ∫(1/2)cos(-2x) * (-3) dx = (-3/2)x cos(-2x) + (3/2) ∫cos(-2x) dxIntegrate remaining integral: Integrate the remaining integral. We need to integrate (3/2)cos(-2x) with respect to x. Using the substitution method again, let w = -2x, then dw = -2 dx, and dx = -dw/2. Now we integrate cos(w) with respect to w and then substitute back. ∫cos(-2x) dx = ∫cos(w) * (-1/2) dw = (-1/2) * sin(w) = (-1/2)sin(-2x). So, (3/2) ∫cos(-2x) dx = (3/2) * (-1/2)sin(-2x) = (-3/4)sin(-2x).Combine and add constant: Combine the results and add the constant of integration. The integral of -3x sin(-2x) with respect to x is: (-3/2)x cos(-2x) - (3/4)sin(-2x) + C, where C is the constant of integration.

Evaluate the integral. $\newline$ $\int-3 x \sin (-2 x) d x$ $\newline$ Answer:

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