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P4 2021 Oct Q6

A Level / Edexcel / P4

IAL 2021 Oct Paper · Question 6

题目

Problem

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

Figure 2 shows a sketch of the curve with equation

y= 16sin⁡2x(3+4sin⁡x)2 0≤x≤π2\begin{align*} y=&\,\frac{16\sin 2x}{(3+4\sin x)^2} \\ &\,0\leq x\leq \frac{\pi}{2} \end{align*}

The region RR, shown shaded in Figure 2, is bounded by the curve, the xx-axis and the line with equation x=π6x=\dfrac{\pi}{6}

Using the substitution u=3+4sin⁡xu=3+4\sin x, show that the area of RR can be written in the form a+ln⁡ba+\ln b, where aa and bb are rational constants to be found.

(7)
题目中文翻译

本题你必须写出所有计算步骤。

不接受依赖计算器技术的解法。

图 2 给出了曲线的草图，其方程为

y= 16sin⁡2x(3+4sin⁡x)2 0≤x≤π2\begin{align*} y=&\,\frac{16\sin 2x}{(3+4\sin x)^2} \\ &\,0\leq x\leq \frac{\pi}{2} \end{align*}

图 2 中阴影部分所示区域 RR 由该曲线、xx 轴以及直线 x=π6x=\dfrac{\pi}{6} 围成。

使用代换 u=3+4sin⁡xu=3+4\sin x，证明区域 RR 的面积可写成 a+ln⁡ba+\ln b 的形式，其中 a,ba,b 为待求有理常数。

(7)

解答

解法一

思路

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区域位于 x=π6x=\frac{\pi}{6} 与曲线在 x=π2x=\frac{\pi}{2} 的零点之间。先用 sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x 整理被积函数，再作题目指定的代换 u=3+4sin⁡xu=3+4\sin x，同时把上下限由 xx 转换为 uu。

答题过程

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The required area is

 A= ∫π/6π/216sin⁡2x(3+4sin⁡x)2 dx.\begin{align*} &\,A=\\ &\,\int_{\pi/6}^{\pi/2} \frac{16\sin2x}{(3+4\sin x)^2} \,\mathrm{d}x. \end{align*}

Using sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x,

 A= ∫π/6π/232sin⁡xcos⁡x(3+4sin⁡x)2 dx.\begin{align*} &\,A=\\ &\,\int_{\pi/6}^{\pi/2} \frac{32\sin x\cos x}{(3+4\sin x)^2} \,\mathrm{d}x. \end{align*}

Let

u=3+4sin⁡x.u=3+4\sin x.

Then

du=4cos⁡x dx\mathrm{d}u=4\cos x\,\mathrm{d}x

and

sin⁡x=u−34.\sin x=\frac{u-3}{4}.

The limits become

xu=3+4sin⁡xπ65π27\begin{array}{c|c} x & u=3+4\sin x\\ \hline \frac{\pi}{6} & 5\\ \frac{\pi}{2} & 7 \end{array}

Therefore,

 A= ∫5732(u−34)u2 ⋅14 du= ∫572(u−3)u2 du= ∫57(2u−6u2) du= [2ln⁡u+6u]57= 2ln⁡7+67 −2ln⁡5−65= 2ln⁡(75)−1235= −1235+ln⁡(4925).\begin{align*} &\,A\\ =&\,\int_5^7 \frac{32\big(\frac{u-3}{4}\big)}{u^2} \\ &\,\hspace{2pt}\cdot\frac14\,\mathrm{d}u\\ =&\,\int_5^7\frac{2(u-3)}{u^2}\,\mathrm{d}u\\ =&\,\int_5^7 \bigg(\frac2u-\frac6{u^2}\bigg)\,\mathrm{d}u\\ =&\,\bigg[2\ln u+\frac6u\bigg]_5^7\\ =&\,2\ln7+\frac67\\ &\,\hspace{2pt}-2\ln5-\frac65\\ =&\,2\ln\bigg(\frac75\bigg)-\frac{12}{35}\\ =&\,\boxed{ -\frac{12}{35}+\ln\bigg(\frac{49}{25}\bigg) }. \end{align*}

Thus the area has the required form a+ln⁡ba+\ln b, where

a=−1235,b=4925.\boxed{ a=-\frac{12}{35}, \qquad b=\frac{49}{25} }.

也可以把原函数换回 xx，再使用原来的上下限。

Writing the antiderivative in terms of xx gives

F(x)= 2ln⁡(3+4sin⁡x) +63+4sin⁡x.\begin{align*} F(x)=&\,2\ln(3+4\sin x) \\ &\,\hspace{2pt}+\frac{6}{3+4\sin x}. \end{align*}

Hence,

 A=F(π2)−F(π6)= 2ln⁡7+67 −(2ln⁡5+65)= −1235+ln⁡(4925).\begin{align*} &\,A=F\bigg(\frac{\pi}{2}\bigg)-F\bigg(\frac{\pi}{6}\bigg) \\ =&\,2\ln7+\frac67 \\ &\,\hspace{2pt}-\bigg(2\ln5+\frac65\bigg) \\ =&\,-\frac{12}{35}+\ln\bigg(\frac{49}{25}\bigg). \end{align*}

解法二

思路

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仍先作题目指定的换元。得到 ∫57(2u−6)u−2 du\int_5^7(2u-6)u^{-2}\,\mathrm{d}u 后，不拆分被积函数，而是用分部积分：对 2u−62u-6 求导、对 u−2u^{-2} 积分，剩下的积分就是对数。

答题过程

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Using u=3+4sin⁡xu=3+4\sin x as in the first solution gives

A=∫57(2u−6)u−2 du.A=\int_5^7(2u-6)u^{-2}\,\mathrm{d}u.

For integration by parts, take

p= 2u−6,dv= u−2 du,dpdu= 2,v= −1u.\begin{align*} p=&\,2u-6, \\ \mathrm{d}v=&\,u^{-2}\,\mathrm{d}u, \\ \frac{\mathrm{d}p}{\mathrm{d}u}=&\,2, \\ v=&\,-\frac1u. \end{align*}

Then,

 A=[−2u−6u]57 −∫57(−1u)2 du= [−2u−6u+2ln⁡u]57= −87+45 +2ln⁡7−2ln⁡5= −1235+ln⁡(4925).\begin{align*} &\,A=\bigg[-\frac{2u-6}{u}\bigg]_5^7 \\ &\,\hspace{2pt}-\int_5^7\bigg(-\frac1u\bigg)2\,\mathrm{d}u \\ =&\,\bigg[-\frac{2u-6}{u}+2\ln u\bigg]_5^7 \\ =&\,-\frac87+\frac45 \\ &\,\hspace{2pt}+2\ln7-2\ln5 \\ =&\,\boxed{-\frac{12}{35}+\ln\bigg(\frac{49}{25}\bigg)}. \end{align*}

Thus a=−1235a=-\frac{12}{35} and b=4925b=\frac{49}{25}.