How do you find the antiderivative of #(e^(2x))sin3x#?
1 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
# (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 #