Can we find the area between $\sin x$ and $\sin 2x$?

Can we find the area between $\sin x$ and $\sin 2x$?

Question

Using the same axes, sketch the curves $y = \sin x$ and $y = \sin 2x$ from $x = 0$ to $\tfrac{1}{2}\pi$, where $x$ is in radians.

Prove that the curves intersect at the points $(0,0) \quad\text{and}\quad \left(\frac{\pi}{3}, \frac{\sqrt{3}}{2} \right).$

Calculate the area of the region bounded by the parts of the curves between these two points.

Solution

The curve $y = \sin 2x$ has the same shape as $y = \sin x$, but twice the frequency; we thus have the following sketch.

The curves intersect exactly when $\sin x = \sin 2x$, or $\sin x = 2 \sin x \cos x$.

This gives us that either $\sin x = 0$ or $\cos x = \dfrac{1}{2}$. Within our region this means $x = 0$ or $x=\dfrac{\pi}{3}$.

So the curves intersect at $(0,0)$ and at $\left(\dfrac{\pi}{3}, \dfrac{\sqrt{3}}{2}\right)$.

The area we are to calculate is the yellow area indicated in the following diagram.

Thinking about scaling

We are looking for the area

$$\begin{align*} A &= \int_0^{\dfrac{\pi}{3}} \sin 2x\:dx - \int_0^{\dfrac{\pi}{3}}\sin x\:dx\\ &= \frac{1}{2}\int_0^{\dfrac{2\pi}{3}} \sin x\:dx - \int_0^{\dfrac{\pi}{3}}\sin x\:dx\\ &= \frac{1}{2}\biggl[-\cos x\biggr]_0^{\dfrac{2\pi}{3}} - \biggl[-\cos x\biggr]_0^{\dfrac{\pi}{3}}\\ &= \frac{1}{2}\left(\frac{1}{2}+1\right) - \left(-\frac{1}{2}+1\right)\\ &= \frac{1}{4}. \end{align*}$$

Using an implicit substitution

We are looking for the area

$$\begin{align*} A &= \int_0^{\dfrac{\pi}{3}} \sin 2x -\sin x\:dx \\ &= \biggl[ -\frac{\cos 2x}{2}+\cos x \biggr]_0^{\dfrac{\pi}{3}} \\ &= \frac{1}{4} + \frac{1}{2} + \frac{1}{2}-1\\ &= \frac{1}{4}. \end{align*}$$