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

1 Answer
Aug 14, 2017

Answer:

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

Explanation:

We could use a traditional double application of Integration By Parts. Here is a slightly different approach.

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 integrate the above results:

# 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] + 2Eq [B}:

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

From [A] we also get:

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

Hence we get the two results:

# 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 #