Using double angle formulas in integration, trouble following an example. I have just started looking at integration and I am having trouble understanding what has been done in one of the examples in the book I am working through.  
It involves using the double angle formula for $\sin(2\theta)$ to provide a rearrangement for which an indefinite integral can then be found.
The double angle formula provided is $\sin(2\theta)=2\sin(\theta)\cos(\theta)$ and the example is as follows:
$$\int\cos\left(\frac{1}{2}x\right)\sin\left(\frac{1}{2}x\right)dx=\int\frac{1}{2}\sin\left(x\right)dx$$
$$=-\frac{1}{2}\cos\left(x\right)+c$$
The part of this example I am specifically stuck with is the first line where $\cos\left(\frac{1}{2}x\right)\sin\left(\frac{1}{2}x\right)$ is rewritten as $\frac{1}{2}\sin\left(x\right)$ using the previously stated double angle formula.

You know that $$\sin(2 \theta)=2\sin(\theta)\cdot\cos(\theta)$$ Now, put $\theta=\dfrac{1}{2}x$ to see that, $$\begin{align}\sin\left(2 \cdot\dfrac{1}{2}x\right)&=2\sin\left(\dfrac{1}{2}x\right)\cdot\cos\left(\dfrac{1}{2}x\right)\\\sin(x)&=2\sin\left(\dfrac{1}{2}x\right)\cdot\cos\left(\dfrac{1}{2}x\right)\\\sin\left(\dfrac{1}{2}x\right)\cdot\cos\left(\dfrac{1}{2}x\right)&=\dfrac{1}{2}\cdot\sin(x)\end{align}$$
