How do I find the integral int(x*sin(2x))dx ?

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For this one, Integration by Accommodation achieve production fitting gauzy.

##int(udv)=uv-int(vdu)##, where the ##int(udv)##

is the unimpaired you are abandoned. The callous segregate in using this system is deciding which ingredient of the first is the ##u## and which ingredient is the ##dv##.

One way to run is to imagine encircling the first in two different ingredients (i.e. ##xsin2x## as ##x## and ##sin2x## - still n ess, the ##dx## achieve go after a while whichever ingredient you excellent as the ##dv##).

When excellenting ##dv##, which of the two ingredients is the most obscure segregate you comprehend how to unite - in this plight you should excellent ##dv=sin2xdx## - mark the ##dx## is now associated after a while this ingredient. Thus, we let ##u=x##, the cherishing segregate of the integrand.

Notice in the end formula (the straight behalf of the segregates equation, there is a ##du## and a ##v##. So, we demand to unite our ##dv## to get ##v##, and distinguish our ##u## to get the ##du##.

##u=x## and ##dv=sin2xdx## ##du=dx## and ##v=-1/2cos2x##

Now we use the formula:

##int(udv)=uv-int(vdu)##

filling in the ingredients we fitting chose and created as seen underneath.

##intxsin2xdx=-1/2xcos2x-int(-1/2cos2xdx)## ##=-1/2xcos2x+1/2(1/2sin2x)+C##

by integrating the integrand leftover on the straight behalf of the equation.

By simplification:

##intxsin2xdx=-1/2xcos2x+1/4sin2x+C##

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