How do you use integration by parts to evaluate the integral #2xsin(x)dx#?

3 Answers
Mar 3, 2018

I tried this:

Explanation:

Have a look:

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Mar 3, 2018

#sin(x)-xcos(x)+C#

Explanation:

We have #int2xsin(x)dx#

Take the constant out:

#2intxsin(x)dx#

According to integration by parts, where #u# and #v# are functions,

#intuvdx=uintvdx-intu'(intvdx)dx#

Here, #u=x# and #v=sin(x)#

So we have:

#x intsin(x)dx-int(x)'(intsin(x)dx)dx#

#x(-cos(x))-int(-cos(x))dx#

#-xcos(x)+intcos(x)dx#

#-xcos(x)+sin(x)#

Add the constant of integration:

#sin(x)-xcos(x)+C#

The answer.

Mar 3, 2018

#I=2(sinx-xcosx)+C#

Explanation:

Method of Integration by Parts:
#intu*vdx=uintvdx-int((du)/(dx)intvdx)dx#
Let,#u=2xandv=sinx#
So,#(du)/(dx)=2andintvdx=intsinxdx=-cosx#
#I=int2xsinxdx=2x(-cosx)-int(2)(-cosx)dx#
#I=-2xcosx+2intcosxdx#
#I=-2xcosx+2sinx+C#
#I=2(sinx-xcosx)+C#