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How would you evaluate the indefinite integral -2xcos3xdx? - Answers

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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How would you evaluate the indefinite integral -2xcos3xdx? - Answers

https://math.answers.com/calculus/How_would_you_evaluate_the_indefinite_integral_-2xcos3xdx

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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https://math.answers.com/calculus/How_would_you_evaluate_the_indefinite_integral_-2xcos3xdx

How would you evaluate the indefinite integral -2xcos3xdx? - Answers

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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      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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