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Integrate by parts:

∫ uv dx = u ∫ v dx - ∫ (u' ∫ v dx) dx

Let u = -2x

Let v = cos 3x

→ u' = d/dx -2x = -2

→ ∫ -2x cos 3x dx = -2x ∫ cos 3x dx - ∫ (-2 ∫ cos 3x dx) dx

= -2x/3 sin 3x - ∫ -2/3 sin 3x dx

= -2x/3 sin 3x - 2/9 cos 3x + c

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8y ago
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8y ago

Let u=-2x and let dv/dx = -cos3xthen du/dx = 2 and v = (sin3x)/3

So, I = uv - int[du/dx*v]dx = 2x*sin(3x)/3 - int[2*sin(3x)/3]dx

= 2x/3*sin(3x) - 2/9*cos(3x) + C

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Q: How would you evaluate the indefinite integral -2xcos3xdx?
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