How do you integrate #int xsin(4x)# by integration by parts method?

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Answer 1 · 2016-07-25T14:21:16

#= - 1/4x cos(4x) + 1/16 sin(4x) + C#

#int xsin(4x) \ dx#

#= int x d/dx ( - 1/4 cos(4x)) \ dx#

which by IBP:
#= - 1/4x cos(4x) + 1/4 int d/dx( x) * cos(4x) \ dx#

#= - 1/4x cos(4x) + 1/4 int cos(4x) \ dx#

#= - 1/4x cos(4x) + 1/4 * 1/4 sin(4x) + C#

#= - 1/4x cos(4x) + 1/16 sin(4x) + C#

Answer 2 · 2016-07-25T14:29:00

#-1/4xcos(4x)+1/16sin(4x)+C#

Before integrating by parts, we can first use substitution to get rid of the #4x# in the sine function.

Let #color(purple)(t=4x#. This implies that #color(green)(dt=4dx# and that #color(blue)(x=t/4#. Thus:

#ttintxsin(4x)dx=1/4intcolor(green)4color(blue)xsin(color(purple)(4x))color(green)(dx)=1/4intt/4sin(t)dt=1/16inttsin(t)dt#

Now is a better time to use integration by parts, which takes the form:

#intudv=uv-intvdu#

So, for #intcolor(red)tcolor(brown)(sin(t)dt#, let:

#color(red)(u=t)" "=>" "du=dt#

#color(brown)(dv=sin(t)dt)" "=>" "intdv=intsin(t)dt" "=>" "v=-cos(t)#

This gives us, following #intudv=uv-intvdu#:

#inttsin(t)dt=-tcos(t)-int(-cos(t))dt#

#=-tcos(t)+intcos(t)dt#

#=-tcos(t)+sin(t)#

Since #t=4x#:

#inttsin(t)dt=-4xcos(4x)+sin(4x)#

And since #intxsin(4x)dx=1/16inttsin(t)dt#, divide the expression by #16#:

#intxsin(4x)dx=-1/4xcos(4x)+1/16sin(4x)+C#