Partial integration of $\sin(x)\times 1$ results in an infinite series very similar to the Taylor series of cosine, but with some obvious differences.
$x\sin(x)-\frac{x^2}{2}\cos(x) -\frac{x^3}{3!}\sin(x)+\frac{x^4}{4!}\cos(x)+...+c$
What strikes me is the fact that this is almost identical to the Taylor series of cosine, with the $a$'s being replaced by $x$'s and the $--++$ series being shifted back by one, but this makes no sense as the indefinite integral of $\sin(x)$ is $-\cos(x)$, not $\cos(x)$.
Now, assuming I am wrong and this is indeed equal to $-\cos(x)$, I come on to my second problem: without replacing the $x$'s with constants, the series doesn't approach either, positive $\cos(x)$ or otherwise, and I can't see why not - it is equal to the integral of $\sin(x)$ after all. Isn't it?
I am sure this is a trivial result but I have struggled to find anything on it even in the darkest realms of triviality (i.e. Yahoo Answers), which is why I am having to ask this question.
If I am parsing your expression properly, your series would appear to be $$\sin (x)\times \left(x-\frac {x^3}{3!}+\frac {x^5}{5!}-\cdots\right) - \cos (x)\times \left(\frac {x^2}{2!}-\frac {x^4}{4!}-\cdots\right)$$ $$=\sin^2(x) -\cos (x)\times (1-\cos(x))=\sin^2(x)+\cos^2(x)-\cos(x)=1-\cos(x)$$
Which, up to a constant, is equal to $-\cos (x)$ as desired.