# Integrate cos(x)-cos(x-c) from 0 to c/2

1. Oct 16, 2012

### sugarxsweet

1. The problem statement, all variables and given/known data
Integrate cos(x)-cos(x-c) from 0 to c/2

2. Relevant equations
sin(x-c)=sinxcosc-cosxsinc

3. The attempt at a solution
sin(x)-sin(x-c) from 0 to c/2
=sin(c/2)-sin(c/2)cos(c)+cos(c/2)sin(c)-sin(0)+sin(-c)
=sin(c/2)-sin(c/2)cos(c)+cos(c/2)sin(c)-0-sin(c)

Correct response:2sin(c/2)-sin(c)

What am I doing wrong?

2. Oct 16, 2012

### SammyS

Staff Emeritus
You're doing nothing wrong. They are equivalent.

To get the answer in a more straight forward manner, simply plug the limits of integration into the anti-derivative, sin(x)-sin(x-c) .

3. Oct 16, 2012

### sugarxsweet

Sorry, I guess this is a question of trig then - how do I simplify my answer to the correct answer? It's probably something stupid but I'm having trouble figuring it out!

4. Oct 16, 2012

### BloodyFrozen

$$sin(\frac{c}{2})-sin(\frac{c}{2})cos(c)+cos(\frac{c}{2})sin(c)-sin(c)$$
$$sin(\frac{c}{2})-sin(c)+cos(\frac{c}{2})sin(c)-sin(\frac{c}{2})cos(c)$$

You have to realize (from trig addition identity) that $cos(\frac{c}{2})sin(c)-sin(\frac{c}{2})cos(c) = sin(c-\frac{c}{2}) = sin(\frac{c}{2})$

Using the above, we have:

$$sin(\frac{c}{2})-sin(c)+sin(\frac{c}{2})$$
$$2sin(\frac{c}{2})-sin(c)$$

5. Oct 16, 2012

Thanks!