# How do you find the antiderivative of #(e^(2x))sin3x#?

# int \ e^(2x)sin3x \ dx = e^(2x)/13(2sin3x -3cos3x ) + C #

Here's a slightly different approach to the classic double application of Integration By Parts that we could employ.

Let:

Now combine the previously mentioned outcomes:

3Eq [A] plus 2Eq [B}:

Also, from [A] we obtain:

This yields the following two outcomes:

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To find the antiderivative of ( e^{2x}\sin(3x) ), you can use integration by parts. Let ( u = e^{2x} ) and ( dv = \sin(3x)dx ). Then, ( du = 2e^{2x}dx ) and ( v = -\frac{1}{3}\cos(3x) ).

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

( \int e^{2x}\sin(3x)dx = -\frac{1}{3}e^{2x}\cos(3x) - \int -\frac{2}{3}e^{2x}\cos(3x)dx )

This integral on the right-hand side can be further integrated using integration by parts or by recognizing it as a multiple of the original integral. Repeat the process until the integral is expressed solely in terms of known functions.

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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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