# Evaluate the integral? : # int \ 2x \ sin3x \ dx #

# int \ 2x \ sin3x \ dx = -2/3xcos3x +2/9 sin3x + C #

We want to evaluate:

We can use Integration By Parts (IBP). Essentially we would like to identify one function that simplifies when differentiated, and identify one that simplifies when integrated (or is at least is integrable).

Then plugging into the IBP formula:

Hence:

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To evaluate the integral ( \int 2x \sin(3x) , dx ), we can use integration by parts, which is based on the formula: [ \int u , dv = uv - \int v , du ] Here, we choose:

- ( u = 2x ) (which implies ( du = 2 , dx ))
- ( dv = \sin(3x) , dx ) (which implies ( v = -\frac{1}{3}\cos(3x) ), because the derivative of (\cos(3x)) is (-3\sin(3x)), and integrating ( dv ) gives ( v ))

Now, substitute ( u ), ( du ), ( v ), and ( dv ) into the integration by parts formula: [ \int 2x \sin(3x) , dx = uv - \int v , du = -\frac{2x}{3}\cos(3x) - \int -\frac{1}{3}\cos(3x) \cdot 2 , dx ] [ = -\frac{2x}{3}\cos(3x) + \frac{2}{3} \int \cos(3x) , dx ] To integrate ( \cos(3x) ), we have: [ \int \cos(3x) , dx = \frac{1}{3}\sin(3x) + C ] Therefore, the original integral becomes: [ -\frac{2x}{3}\cos(3x) + \frac{2}{3} \cdot \frac{1}{3}\sin(3x) + C ] [ = -\frac{2x}{3}\cos(3x) + \frac{2}{9}\sin(3x) + C ] So, the evaluated integral is: [ \int 2x \sin(3x) , dx = -\frac{2x}{3}\cos(3x) + \frac{2}{9}\sin(3x) + C ]

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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.

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.

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.

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