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

3 Answers
Gió
Mar 3, 2018

I tried this:

Explanation:

Have a look:

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Surya K.
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.

maganbhai P.
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

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