# How do you integrate #int 2x sin 4x dx#?

Use integration by parts to find that

Using integration by parts:

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To integrate ( \int 2x \sin(4x) , dx ), you can use integration by parts method. Let ( u = 2x ) and ( dv = \sin(4x) , dx ). Then, ( du = 2 , dx ) and ( v = -\frac{1}{4} \cos(4x) ).

Applying the integration by parts formula ( \int u , dv = uv - \int v , du ), we get:

[ \int 2x \sin(4x) , dx = -\frac{1}{2}x \cos(4x) - \int -\frac{1}{4} \cos(4x) \cdot 2 , dx ]

[ = -\frac{1}{2}x \cos(4x) + \frac{1}{2} \int \cos(4x) , dx ]

Integrating ( \int \cos(4x) , dx ) gives:

[ = -\frac{1}{2}x \cos(4x) + \frac{1}{8} \sin(4x) + C ]

where ( C ) is the constant of integration.

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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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