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How would you evaluate the indefinite integral -2xcos3xdx?
Dirtyarmor
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
Wiki User
∙ 7y ago
Wiki User
∙ 7y agoLet 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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amphipathic
