# 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.

• substitute $\theta = \frac{x}{2}$ – Peđa Terzić Feb 24 '12 at 13:12

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}
Consider $\theta = \frac{1}{2}x$. Then $$\cos\left(\frac{1}{2}x\right)\sin\left(\frac{1}{2}x\right) = \cos \theta \sin \theta$$ Notice that the double angle formula could be written: $$\sin \theta \cos \theta = \frac{1}{2}\sin(2\theta)$$ So the integrand is now: $$\cos\left(\frac{1}{2}x\right)\sin\left(\frac{1}{2}x\right) = \cos \theta \sin \theta = \frac{1}{2} \sin(2\theta) = \frac{1}{2}\sin\left(2\cdot \frac{1}{2}x\right) = \frac{1}{2}\sin x$$