What is #int \ e^(5x+1) sin(2x+3) \ dx #?

We seek:

# I = int \ e^(5x+1) sin(2x+3) \ dx #

Utilising the given result:

# int \ e^(ax) \ cosbx \ dx = (e^(ax)(asinbx-bcosbx)) / (a^2+b^2) \ dx # ..... [A}

We now perform a substitution:

Let #u=2x+3 => (du)/dx = 2 #; and #x=1/2(u-3) #

Substituting into the integral we get:

# I = int \ e^(5/2(u-3)+1) sinu \ (1/2) \ du #
# \ \ = 1/2 \ int \ e^(5/2u-15/2+1) sinu \ du #
# \ \ = 1/2 \ int \ e^(5/2u)e^(-13/2) sinu \ du #
# \ \ = 1/2 e^(-13/2) int \ e^(5/2u) sinu \ du #

We can now use the given result [A] with:

# a = 5/2 #
# b = 1 #

Giving:

# I = 1/2 e^(-13/2){(e^(5/2u)(5/2sin1u-1cos1u)) / ((5/2)^2+1^2)} + C#
# \ \ = 1/2 e^(-13/2){ (e^(5/2u) \ 1/2 \ (5sinu-2cosu) ) / (25/4+1)} + C#
# \ \ = 1/4 e^(-13/2){ (e^(5/2u) \ (5sinu-2cosu) ) / (29/4)} + C#
# \ \ = e^(-13/2) * (e^(5/2u) \ (5sinu-2cosu) ) / 29 + C#
# \ \ = (e^(5/2u-13/2) \ (5sinu-2cosu) ) / 29 + C#

Now if we restore the earlier substitution:

# I = (e^(5/2(2x+3)-13/2) \ (5sin(2x+3)-2cos(2x+3)) ) / 29 + C#
# \ \ = (e^(5x+15/2-13/2) \ (5sin(2x+3)-2cos(2x+3)) ) / 29 + C#
# \ \ = (e^(5x+1) \ (5sin(2x+3)-2cos(2x+3)) ) / 29 + C#