Question #d6918

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Answer 1 · 2017-11-25T04:34:13

#e^4(-1/2xcos(2x)+1/4sin(2x))+C#

#inte(e^3xsin(2x)dx)=inte^4xsin(2x)dx#

#e^4# is a constant and can be pulled out:

#=e^4intxsin(2x)dx#

Let's use integration by parts since we have a product of two functions:

#u=x# and #dv=sin(2x)dx#. Then:

#du=dx# and #v=-1/2cos(2x)#

#intudv=uv-intvdu#

#intxsin(2x)dx=-1/2xcos(2x)-int-1/2cos(2x)dx=-1/2xcos(2x)+1/2intcos(2x)dx#

#=-1/2xcos(2x)+1/2*1/2sin(2x)+C#

Now we need to make sure we do not forget about the #e^4#:

#e^4intxsin(2x)dx=e^4(-1/2xcos(2x)+1/4sin(2x))+C#