题目

Problem

Figure 4 shows a sketch of part of the curve with equation

y=(1+2\cos 2x)^2

(a) Show that

(1+2\cos 2x)^2\equiv p+q\cos 2x+r\cos 4x

where $p$ , $q$ and $r$ are constants to be found.

(2)

The curve touches the positive $x$ -axis for the second time when $x=a$ , as shown in Figure 4.

The regions bounded by the curve, the $y$ -axis and the $x$ -axis up to $x=a$ are shown shaded in Figure 4.

(b) Find, using algebraic integration and making your method clear, the exact total area of the shaded regions. Write your answer in simplest form.

(5)

题目中文翻译

图 4 给出了部分曲线的示意图，其方程为

y=(1+2\cos 2x)^2

(a) 证明

(1+2\cos 2x)^2\equiv p+q\cos 2x+r\cos 4x

其中 $p$ 、 $q$ 和 $r$ 是需要求出的常数。

如图 4 所示，当 $x=a$ 时，曲线第二次与正 $x$ 轴相切。

图中从 $x=0$ 到 $x=a$ ，由曲线、 $y$ 轴和 $x$ 轴围成的区域已用阴影标出。

(b) 使用代数积分并清楚写出方法，求阴影区域的精确总面积，并把答案写成最简形式。

解答

We have

(1+2\cos 2x)^2 =1+4\cos 2x+4\cos^2 2x

Using

\cos 4x=2\cos^2 2x-1

we get

4\cos^2 2x=2(2\cos^2 2x)=2(\cos 4x+1)

Therefore

(1+2\cos 2x)^2 =1+4\cos 2x+2(\cos 4x+1)

(1+2\cos 2x)^2 \equiv 3+4\cos 2x+2\cos 4x

Hence

p=3,\qquad q=4,\qquad r=2

The curve touches the $x$ -axis when

(1+2\cos 2x)^2=0

1+2\cos 2x=0

and hence

\cos 2x=-\frac12

The second positive value of $x$ is

a=\frac{2\pi}{3}

因为曲线方程是一个平方，所以曲线在 $x$ 轴上方或与 $x$ 轴相切，面积可以直接积分。

Using the result from part (a),

\text{Area} =\int_0^{2\pi/3}(3+4\cos 2x+2\cos 4x)\,\mathrm{d}x

Therefore

\begin{aligned} \text{Area} &=\left[3x+2\sin 2x+\frac12\sin 4x\right]_0^{2\pi/3} \\ &=2\pi+2\sin\frac{4\pi}{3} +\frac12\sin\frac{8\pi}{3} \end{aligned}

Now

\sin\frac{4\pi}{3}=-\frac{\sqrt3}{2}, \qquad \sin\frac{8\pi}{3}=\sin\frac{2\pi}{3}=\frac{\sqrt3}{2}

Hence

\text{Area} =2\pi-\sqrt3+\frac{\sqrt3}{4}

So the exact total area is

\boxed{2\pi-\frac{3\sqrt3}{4}}