What is the integral of #int (e^x * cos 2x) dx#?

1 Answer
Aug 16, 2017

# int \ e^(x)cos2x \ dx = e^(x)/5(2sin2x+cos2x) + C#

Explanation:

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

Let:

# s = e^(x)sin2x \ \ \ \ # and #I_s = int e^(x)sin2x#
# c = e^(x)cos2x \ \ \ \ # and #I_c = int e^(x)cos2x#

Differentiating wrt #x# we get:

# (ds)/dx = e^(x)(d/dx sin2x) + (d/dx e^(x))sin2x #
# \ \ \ \ \ \ = 2e^(x)cos2x + e^(x)sin2x #

# (dc)/dx = e^(x)(d/dx cos2x) + (d/dx e^(x))cos2x #
# \ \ \ \ \ \ = -2e^(x)sin2x + e^(x)cos2x #

Now integrate the above results:

# int \ (ds)/dx \ dx = int \ 2e^(x)cos2x + e^(x)sin2x \ dx#
# => s= 2I_c + I_s # ... [A]

# int \ (dc)/dx \ dx = int \ -2e^(x)sin2x + e^(x)cos2x \ dx#
# => c = -2I_s + I_c # ... [B]

2Eq [A] + Eq [B}:

# 2s+c=5I_c #
# :. I_c = 1/5(2s+c)#

From [B] we also get:

# c = -2I_s+1/5(2s+c) #
# :. c = -2I_s+2/5s+1/5c #
# :. I_s = 1/5(s-2c) #

Hence we get the two results:

# int \ e^(x)cos2x \ dx = 1/5(2s+c) + C#
# " " = e^(x)/5(2sin2x+cos2x) + C#

# int \ e^(x)sin2x \ dx = 1/5(s-2c) + C #
# " " = e^(x)/5(sin2x -2cos2x ) + C #