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P4 2024 Jan Q5

A Level / Edexcel / P4

IAL 2024 Jan Paper · Question 5

题目

Problem

(a) Find

∫x2cos⁡2x dx\int x^2\cos 2x\,\mathrm{d}x
(4)

(b) Hence solve the differential equation

dydt=(tcos⁡ty)2\frac{\mathrm{d}y}{\mathrm{d}t}=\bigg(\frac{t\cos t}{y}\bigg)^2

giving your answer in the form yn=f(t)y^n=f(t) where nn is an integer.

(5)
题目中文翻译

(a) 求

∫x2cos⁡2x dx\int x^2\cos 2x\,\mathrm{d}x
(4)

(b) 进而解微分方程

dydt=(tcos⁡ty)2\frac{\mathrm{d}y}{\mathrm{d}t}=\bigg(\frac{t\cos t}{y}\bigg)^2

并将答案写成 yn=f(t)y^n=f(t) 的形式,其中 nn 为整数。

(5)

解答

(a)

解法一

思路

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使用两次分部积分:先对 x2x^2 求导,再处理剩下的 xsin⁡2xx\sin2x。把两个积分分别记为 II 和 KK,有助于看清符号并保持每行较短。

答题过程

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Let

I=∫x2cos⁡2x dx.I=\int x^2\cos2x\,\mathrm{d}x.

For integration by parts, take u=x2u=x^2 and dv=cos⁡2x dx\mathrm{d}v=\cos2x\,\mathrm{d}x. Then du=2x dx\mathrm{d}u=2x\,\mathrm{d}x and v=12sin⁡2xv=\frac12\sin2x, so

I=12x2sin⁡2x−K,I=\frac12x^2\sin2x-K,

where

K=∫xsin⁡2x dx.K=\int x\sin2x\,\mathrm{d}x.

For KK, take u=xu=x and dv=sin⁡2x dx\mathrm{d}v=\sin2x\,\mathrm{d}x. Now du=dx\mathrm{d}u=\mathrm{d}x and v=−12cos⁡2xv=-\frac12\cos2x. Hence

K= −12xcos⁡2x +12∫cos⁡2x dx= −12xcos⁡2x+14sin⁡2x.\begin{align*} K=&\,-\frac12x\cos2x\\ &\,\hspace{2pt}+\frac12\int\cos2x\,\mathrm{d}x\\ =&\,-\frac12x\cos2x+\frac14\sin2x. \end{align*}

Substituting back,

I= 12x2sin⁡2x+12xcos⁡2x −14sin⁡2x+C.\begin{align*} I=&\,\frac12x^2\sin2x+\frac12x\cos2x\\ &\,\hspace{2pt}-\frac14\sin2x+C. \end{align*}

解法二

思路

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也可以用表格法组织分部积分。不断对 x2x^2 求导、对 cos⁡2x\cos2x 积分,再取带交替符号的对角乘积。

答题过程

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The repeated integration by parts table is

DIx2cos⁡2x2x12sin⁡2x2−14cos⁡2x0−18sin⁡2x\begin{array}{c|c} D&I\\ \hline x^2&\cos2x\\ 2x&\frac12\sin2x\\ 2&-\frac14\cos2x\\ 0&-\frac18\sin2x \end{array}

Take the three diagonal products with alternating signs:

x2(12sin⁡2x)=12x2sin⁡2x,x^2\bigg(\frac12\sin2x\bigg) =\frac12x^2\sin2x, −2x(−14cos⁡2x)=12xcos⁡2x,-2x\bigg(-\frac14\cos2x\bigg) =\frac12x\cos2x,

and

2(−18sin⁡2x)=−14sin⁡2x.2\bigg(-\frac18\sin2x\bigg) =-\frac14\sin2x.

Therefore

∫x2cos⁡2x dx= 12x2sin⁡2x +12xcos⁡2x −14sin⁡2x+C.\begin{align*} \int x^2\cos2x\,\mathrm{d}x =&\,\frac12x^2\sin2x\\ &\,\hspace{2pt}+\frac12x\cos2x\\ &\,\hspace{4pt}-\frac14\sin2x+C. \end{align*}

(b)

解法一

思路

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先分离变量,再用双角公式把右侧拆成两个积分。题目要求 Hence,因此明确调用 (a) 的原函数,并保留积分常数,最后写成 y3=f(t)y^3=f(t)。

答题过程

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Rearranging the differential equation,

y2dydt=t2cos⁡2t.y^2\frac{\mathrm{d}y}{\mathrm{d}t} =t^2\cos^2t.

Separate variables and integrate:

∫y2 dy=∫t2cos⁡2t dt.\int y^2\,\mathrm{d}y =\int t^2\cos^2t\,\mathrm{d}t.

The left side is 13y3\frac13y^3. On the right, use

cos⁡2t=12(1+cos⁡2t).\cos^2t=\frac12(1+\cos2t).

Let F(t)F(t) denote the antiderivative from part (a), with xx replaced by tt:

F(t)= 12t2sin⁡2t +12tcos⁡2t −14sin⁡2t.\begin{align*} F(t)=&\,\frac12t^2\sin2t\\ &\,\hspace{2pt}+\frac12t\cos2t\\ &\,\hspace{4pt}-\frac14\sin2t. \end{align*}

Then

∫t2cos⁡2t dt= 12∫t2 dt +12∫t2cos⁡2t dt= 16t3+12F(t)+C.\begin{align*} \int t^2\cos^2t\,\mathrm{d}t =&\,\frac12\int t^2\,\mathrm{d}t\\ &\,\hspace{2pt}+\frac12\int t^2\cos2t\,\mathrm{d}t\\ =&\,\frac16t^3+\frac12F(t)+C. \end{align*}

Thus

13y3=16t3+12F(t)+C.\frac13y^3=\frac16t^3+\frac12F(t)+C.

Multiply by 33 and absorb the constant:

y3=12t3+32F(t)+C.y^3=\frac12t^3+\frac32F(t)+C.

Substituting F(t)F(t) gives the required form:

y3= 12t3+34t2sin⁡2t +34tcos⁡2t −38sin⁡2t+C.\begin{align*} y^3=&\,\frac12t^3+\frac34t^2\sin2t\\ &\,\hspace{2pt}+\frac34t\cos2t\\ &\,\hspace{4pt}-\frac38\sin2t+C. \end{align*}

Here n=3n=3 and CC is an arbitrary constant.