# Two (equivalent?) integrals

1. Oct 7, 2008

### musemonkey

1. This is part of a larger calculation. I do it one way and get the right answer and do it a seemingly equivalent way and get the wrong answer. The question is just about reconciling the two integrals below.

2.

$$I_1 = \int \frac{sin(2x)}{2} dx = -\frac{cos(2x)}{4}$$

$$\frac{sin(2x)}{2} = sin(x)cos(x)$$

$$I_2 = \int sin(x)cos(x) dx = - \frac{cos^2(x)}{2}$$

For $$x= 0$$, $$I_1 = -1/4 \neq I_2 = -1/2$$.

How can this be?

2. Oct 7, 2008

### Dick

cos(2x)=cos(x+x)=cos(x)^2-sin(x)^2=2*cos(x)^2-1. Your two expressions differ by a constant. Don't forget the +C on an indefinite integral.

3. Oct 7, 2008

Thanks!