# Pulling dx out of cos(x+dx)

1. Nov 21, 2011

### Zula110100100

To get something into integral form the dx needs to be at the end right? How can I do this if what I have is rsin(x+dx)cos(x+dx)?

2. Nov 21, 2011

### dextercioby

It doesn't look ok. Can you perhaps write the whole thing/problem ?

3. Nov 21, 2011

### HallsofIvy

I am going to assume that your "dx" is not a differential but just some slight change in x. $\Delta x$ would be better notation.

The best you can do is use the sum identities:
sin(x+ dx)= sin(x)cos(dx)+ cos(x)sin(dx)
cos(x+ dx)= cos(x)cos(dx)- sin(x)sin(dx)

You could then argue that for dx sufficiently small sin(dx) is approximately equal to dx and cos(dx) is approximately equal to 1:
sin(x+ dx) is approximately sin(x)+ cos(x)dx
cos(x+ dx) is approximately cos(x)- sin(x)dx.

Now, r sin(x+ dx)cos(x+ dx) will be approximately
$sin(x)cos(x)+ cos^2(x)dx- sin^2(dx)- sin(x)cos(x)dx^2$

I suppose you could not argue that, for dx a "differential", you can drop $dx^2$ to get $sin(x)cos(x)+ cos^2(x)dx- sin^2(x)dx$

But those still will not give you something "times dx".

As dextercioby said, please show the entire problem.

4. Nov 21, 2011

### Zula110100100

The entire equation is : Lim $\Delta$x->0 $\Sigma$ (r$\Pi$$\Delta$x/90 - rsin(xi)cos(xi)+rsin(xi+$\Delta$x)cos(xi+$\Delta$x))

Thats ugly as hell, but I think you get the idea yes?