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How do you integrate by parts #[xcos2x]#?

Answer 1

Isabella Ackerman

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Answer 2

Charles Arocha

To integrate (x \cos(2x)) by parts, follow these steps:

Choose (u) and (dv).
Find (du) and (v).
Apply the integration by parts formula: (\int u , dv = uv - \int v , du).
Evaluate the integral.

Let's proceed with these steps:

Choose (u = x) and (dv = \cos(2x) , dx).
Find (du) and (v): [du = dx] [v = \frac{1}{2}\sin(2x)]
Apply the integration by parts formula: [\int x \cos(2x) , dx = uv - \int v , du] [= x \cdot \frac{1}{2}\sin(2x) - \int \frac{1}{2}\sin(2x) , dx]
Evaluate the integral: [= \frac{x}{2}\sin(2x) + \frac{1}{4}\cos(2x) + C]

Therefore, the integral of (x \cos(2x)) with respect to (x) is (\frac{x}{2}\sin(2x) + \frac{1}{4}\cos(2x) + C), where (C) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.