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

For this one, Integration by Parts will work just fine.

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

is the integral you are given. The hard part in using this method is deciding which piece of the original is the ##u## and which piece is the ##dv##.

One way to decide is to think about the original in two separate pieces (i.e. ##xsin2x## as ##x## and ##sin2x## – note, the ##dx## will go with whichever piece you select as the ##dv##).

When selecting ##dv##, which of the two pieces is the most difficult part you know how to integrate – in this case you should select ##dv=sin2xdx## – notice the ##dx## is now associated with this piece. Thus, we let ##u=x##, the remaining part of the integrand.

Notice in the end formula (the right side of the parts equation, there is a ##du## and a ##v##. So, we need to integrate our ##dv## to get ##v##, and differentiate 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 pieces we just chose and created as seen below.

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

by integrating the integrand leftover on the right side of the equation.

By simplification:

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

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