# How do you find the integral of #e^(2x) cos4x dx#?

Integrate by parts twice using

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To find the integral of ( e^{2x} \cos(4x) , dx ), you can use integration by parts. Let ( u = e^{2x} ) and ( dv = \cos(4x) , dx ). Then, ( du = 2e^{2x} , dx ) and ( v = \frac{1}{4} \sin(4x) ).

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

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

Now, we have another integral of the form ( e^{2x} \sin(4x) ). To solve this, we can use integration by parts again. Let ( u = e^{2x} ) and ( dv = \sin(4x) , dx ). Then, ( du = 2e^{2x} , dx ) and ( v = -\frac{1}{4} \cos(4x) ).

Applying the integration by parts formula again, we get:

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

Substituting this result back into our original integral, we have:

[ \int e^{2x} \cos(4x) , dx = \frac{1}{4} e^{2x} \sin(4x) + \frac{1}{8} e^{2x} \cos(4x) - \frac{1}{4} \int e^{2x} \cos(4x) , dx ]

Now, we can solve for the integral by isolating it on one side:

[ \frac{5}{4} \int e^{2x} \cos(4x) , dx = \frac{1}{4} e^{2x} \sin(4x) + \frac{1}{8} e^{2x} \cos(4x) ]

[ \int e^{2x} \cos(4x) , dx = \frac{1}{5} e^{2x} \sin(4x) + \frac{1}{10} e^{2x} \cos(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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