How do you integrate #[(e^(2x))sinx]dx#?
Integrate by parts twice using
Following the second integration in sections, you will
You are free to rewrite, simplify, or factor as necessary.
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To integrate (\int e^{2x} \sin(x) , dx), you can use integration by parts. Let (u = e^{2x}) and (dv = \sin(x) , dx). Then, (du = 2e^{2x} , dx) and (v = -\cos(x)).
Using the integration by parts formula:
[ \int u , dv = uv - \int v , du ]
we have:
[ \int e^{2x} \sin(x) , dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot (2e^{2x}) , dx ]
which simplifies to:
[ \int e^{2x} \sin(x) , dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) , dx ]
Now, we integrate (\int e^{2x} \cos(x) , dx) using integration by parts again. Let (u = e^{2x}) and (dv = \cos(x) , dx). Then, (du = 2e^{2x} , dx) and (v = \sin(x)).
Applying the integration by parts formula again, we get:
[ \int e^{2x} \cos(x) , dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} , dx ]
This simplifies to:
[ \int e^{2x} \cos(x) , dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) , dx ]
We can now substitute this result back into our original integral to obtain:
[ \int e^{2x} \sin(x) , dx = -e^{2x} \cos(x) + 2 \left(e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) , dx\right) ]
Solving for the integral term on the right-hand side and simplifying, we get:
[ \int e^{2x} \sin(x) , dx = -e^{2x} \cos(x) + 2e^{2x} \sin(x) - 4 \int e^{2x} \sin(x) , dx ]
Rearranging terms, we find:
[ 5 \int e^{2x} \sin(x) , dx = -e^{2x} \cos(x) + 2e^{2x} \sin(x) ]
And finally:
[ \int e^{2x} \sin(x) , dx = \frac{-e^{2x} \cos(x) + 2e^{2x} \sin(x)}{5} + 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.
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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