How do you integrate #[(e^(2x))sinx]dx#?

Answer 1

Integrate by parts twice using #u = e^(2x)# both times.

Following the second integration in sections, you will

#int e^(2x)sinx dx = -e^(2x)cosx + 2e^(2x)sinx - 4 int e^(2x)sinx dx#
Note that the last integral is the same as the one we want. Call it #I# for now.
#I = -e^(2x)cosx + 2e^(2x)sinx - 4 I#
So #I = 1/5[-e^(2x)cosx + 2e^(2x)sinx] +C#

You are free to rewrite, simplify, or factor as necessary.

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Answer 2

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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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.

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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