How do you find the indefinite integral of xsinx

1. Mar 3, 2005

laker88116

I am just curious I was thinking about this and if anyone could explain I would appreciate it. I am curious to know how to find the indefinite integral of xsinx with respect to x.

Thanks, David

2. Mar 3, 2005

dextercioby

Are u familiar to the part integration method...?If so,then do it...

Daniel.

3. Mar 3, 2005

laker88116

im not, i did a lot bit of research however i am having difficulty comprehending it, im in calculus ab now, we have yet to reach that if it is in our agenda

4. Mar 3, 2005

dextercioby

Since part integration is the first method of integration one learns,i suspect you haven't reached indefinite integration at all.In which case,why bother...?Is it curiosity,or what...?

Daniel.

5. Mar 3, 2005

James R

The integration by parts formula is:

$$\int u \frac{dv}{dx} dx = uv - \int \frac{du}{dx} v dx$$.

In your problem, put

$$u = x, \qquad \frac{dv}{dx} = \sin x$$.

Now find $du/dx$ and $v$, then plug them into the formula above. Do the second integration, and you're done.

6. Mar 3, 2005

laker88116

ive reached integration, i can integrate like 2sin2x when using substitution as long as the constant cancels out from du, in my book, thomas/finney 9th edition calculus, it isn't introduced until the latter part of the book, and since my class is only for the ap test, its not like we will get that far anyway i dont think and im just curious, i looked in the book but its past me

7. Mar 3, 2005

laker88116

ok that helps i get it that way, thanks

8. Mar 3, 2005

dextercioby

Part integration uses product rule of differentiation for proving.I think that's a little easier than substitution,which would require chain rule for proving...

Daniel.

P.S.I think someone else offered the solution.

9. Mar 3, 2005

laker88116

thats it, it slipped my mind i couldnt remember what its called, but yeah, chain rule is what i was getting at, we just havent done integration by parts yet

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