# Integrating infinite sums

1. Nov 27, 2007

### rock.freak667

1. The problem statement, all variables and given/known data
Using the macluarin's expansion for sinx show that $\int sinx dx=-cosx+c$

2. Relevant equations

$$sinx=\sum_{n=0} ^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}$$

3. The attempt at a solution
Well I can easily write out some of the series and just show that it is equal to -cosx

but if I integrate the representation for the infinite series i get

$$\int sinx dx= \sum_{n=0} ^\infty \frac{(-1)^nx^{2n+2}}{(2n+1)!}$$
shouldn't -cosx be:
$$\int sinx dx= \sum_{n=0} ^\infty \frac{(-1)^{n+1}x^{2n+2}}{(2n+1)!}$$

and also I am supposed to get $x^{2n}$ not what I got

2. Nov 27, 2007

### Avodyne

Your denominators should be (2n+2)! in the last two lines.

With this correction, your last series is cos(x)-1; write out the first few terms to see.

3. Nov 27, 2007

### rock.freak667

ah yes I made a typo...but even if I change it
how do I manipulate

$$\sum_{n=0} ^\infty \frac{(-1)^nx^{2n+2}}{(2n+2)!}$$ (what I integrated and got)

Into

$$-\sum_{n=0} ^\infty \frac{(-1)^nx^{2n}}{(2n)!}$$ (What I am supposed to get)

4. Nov 28, 2007

### HallsofIvy

Staff Emeritus
Let j= n+1 and see what you get with j as the index.