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

Answer 1

# 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:

# s = e^(2x)sin3x \ \ \ \ # and #I_s = int e^(2x)sin3x# # c = e^(2x)cos3x \ \ \ \ # and #I_c = int e^(2x)cos3x#
Differentiating wrt #x# we get:
# (ds)/dx = e^(2x)(d/dx sin3x) + (d/dx e^(2x))sin3x # # \ \ \ \ \ \ = 3e^(2x)cos3x + 2e^(2x)sin3x #
# (dc)/dx = e^(2x)(d/dx cos3x) + (d/dx e^(2x))cos3x # # \ \ \ \ \ \ = -3e^(2x)sin3x + 2e^(2x)cos3x #

Now combine the previously mentioned outcomes:

# int \ (ds)/dx \ dx = int \ 3e^(2x)cos3x + 2e^(2x)sin3x \ dx# # => s= 3I_c + 2I_s # ... [A]
# int \ (dc)/dx \ dx = int \ -3e^(2x)sin3x + 2e^(2x)cos3x \ dx# # => c = -3I_s + 2I_c # ... [B]

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

# 3s+2c = 9I_c + 6I_s -6I_s + 4I_c # # :. 3s+2c = 13I_c # # :. I_c = 1/13(3s+2c)#

Also, from [A] we obtain:

# s = 3/13(3s+2c) + 2I_s # # :. s = 9/13s+6/13c + 2I_s # # :. I_s = 1/13(2s -3c) #

This yields the following two outcomes:

# int \ e^(2x)cos3x \ dx = 1/13(3e^(2x)sin3x+2e^(2x)cos3x) + C# # " " = e^(2x)/13(3sin3x+2cos3x) + C#
# int \ e^(2x)sin3x \ dx = 1/13(2e^(2x)sin3x -3e^(2x)cos3x ) + C # # " " = e^(2x)/13(2sin3x -3cos3x ) + C #
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Answer 2

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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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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