Newbie here: antiderivative for a trig function

[paretoptimal]

Greetings,

Sure wish I found this forum like two years ago! Anyways, as my cherry-popping post I shall seek knowledge for the fore-going:


The anti-derivative for xcos(x)?

I was able to do (2x)cos(x^2) = sin(x^2), but xcos(x) has boggled me. I am not seeing it.
Can anyone give me the answer and a good strategy/rule of thumb for tackling similar anti-derivatives?

Thanks a million!

 

[asd1249jf]

Welcome to PF, paretoptimal!

There is a specific method to take the antiderivative of the function you've shown. Have you studied integration by parts?

If no, I suggest you first read the following article to understand what it is.

http://en.wikipedia.org/wiki/Integration_by_parts

If yes, set u equal to the diminishing term and dv equal to the non-diminishing term. This should be pretty obvious to do.

 

[paretoptimal]

Welcome to PF, paretoptimal!

There is a specific method to take the antiderivative of the function you've shown. Have you studied integration by parts?

If no, I suggest you first read the following article to understand what it is.

http://en.wikipedia.org/wiki/Integration_by_parts

If yes, set u equal to the diminishing term and dv equal to the non-diminishing term. This should be pretty obvious to do.

Thanks for the info, l46kok! We haven't gone over integration by parts in my calc class. I should probably re-clarify my post. The whole equation was: sinx + xcosx. I anti-derived the first term as (-)cosx. It is the second which is giving me problems. Perhaps the whole equation can be anti-derived as a single term? I am not sure. But I don't think I should attempt solving this by using tools I haven't been endowed with yet.

Thanks again!

 

[CompuChip]

Actually, the longer expression makes it easier to solve
For, what is the derivative of (x sin x) ?

As a small aside, which is not really relevant for the question but since the term partial integration has already been dropped, it might interest you: Observe that the key here lies in the product rule. In general, if you have an integrand of the form f'(x) g(x) + g(x) f'(x) you can write this as the derivative of f(x) g(x), and the idea of partial integration is based on this.

(By the way, before you ask me how the hell I ever thought of this -- if you haven't learned partial integration yet, I wouldn't really know how to solve this formally. It's just a lot of training and a little luck that allowed me to recognize sin(x) + x cos(x) as a derivative)

 

[paretoptimal]

Actually, the longer expression makes it easier to solve
For, what is the derivative of (x sin x) ?

As a small aside, which is not really relevant for the question but since the term partial integration has already been dropped, it might interest you: Observe that the key here lies in the product rule. In general, if you have an integrand of the form f'(x) g(x) + g(x) f'(x) you can write this as the derivative of f(x) g(x), and the idea of partial integration is based on this.

(By the way, before you ask me how the hell I ever thought of this -- if you haven't learned partial integration yet, I wouldn't really know how to solve this formally. It's just a lot of training and a little luck that allowed me to recognize sin(x) + x cos(x) as a derivative)

Beautiful, CompuChip, I should have used reverse psychology on that equation. If I see an equation like that again, I will think of the "reverse product rule". I totally see it!