Pulling dx out of cos(x+dx)

  • #1
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)?
 

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  • #2
dextercioby
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It doesn't look ok. Can you perhaps write the whole thing/problem ?
 
  • #3
HallsofIvy
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I am going to assume that your "dx" is not a differential but just some slight change in x. [itex]\Delta x[/itex] 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
[itex]sin(x)cos(x)+ cos^2(x)dx- sin^2(dx)- sin(x)cos(x)dx^2[/itex]

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

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


As dextercioby said, please show the entire problem.
 
  • #4
The entire equation is : Lim [itex]\Delta[/itex]x->0 [itex]\Sigma[/itex] (r[itex]\Pi[/itex][itex]\Delta[/itex]x/90 - rsin(xi)cos(xi)+rsin(xi+[itex]\Delta[/itex]x)cos(xi+[itex]\Delta[/itex]x))

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

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