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IAL 2022 Oct Q7

A Level / Edexcel / P4

IAL 2022 Oct Paper · Question 7

题目

Problem

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

Solutions relying entirely on calculator technology are not acceptable.

(i) Use the substitution u=ex−3u=e^x-3 to show that

∫ln⁡5ln⁡74e3xex−3 dx=a+bln⁡2\int_{\ln 5}^{\ln 7}\frac{4e^{3x}}{e^x-3}\,dx=a+b\ln 2

where aa and bb are constants to be found.

(7)

(ii) Show, by integration, that

∫3excos⁡2x dx=pexsin⁡2x+qexcos⁡2x+c\int 3e^x\cos 2x\,dx=pe^x\sin 2x+qe^x\cos 2x+c

where pp and qq are constants to be found and cc is an arbitrary constant.

(5)
题目中文翻译

本题中你必须写出解题过程的所有步骤。

完全依赖计算器技术的解法不被接受。

(i) 令 u=ex−3u=e^x-3，证明

∫ln⁡5ln⁡74e3xex−3 dx=a+bln⁡2\int_{\ln 5}^{\ln 7}\frac{4e^{3x}}{e^x-3}\,dx=a+b\ln 2

其中 a,ba,b 为待求常数。

(ii) 用积分证明

∫3excos⁡2x dx=pexsin⁡2x+qexcos⁡2x+c\int 3e^x\cos 2x\,dx=pe^x\sin 2x+qe^x\cos 2x+c

其中 p,qp,q 为待求常数，cc 为任意常数。

解答

(i)

解法一

思路

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按题目指定令 u=ex−3u=e^x-3。需要同时改写 exe^x、dx\mathrm{d}x 和上下限；换元后被积函数可展开为 4u+24+36u4u+24+\frac{36}{u}，于是积分结果自然包含 ln⁡u\ln u。

答题过程

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Let

u=ex−3.u=e^x-3.

Then

ex=u+3e^x=u+3

and

dudx=ex=u+3,\frac{\mathrm{d}u}{\mathrm{d}x} =e^x=u+3,

so

dx=duu+3.\mathrm{d}x=\frac{\mathrm{d}u}{u+3}.

The limits become

x=ln⁡5⟹u=2x=\ln5\Longrightarrow u=2

and

x=ln⁡7⟹u=4.x=\ln7\Longrightarrow u=4.

Therefore,

 ∫ln⁡5ln⁡74e3xex−3 dx= ∫244(u+3)3uduu+3= ∫244(u+3)2u du= ∫24(4u+24+36u) du.\begin{align*} &\,\int_{\ln5}^{\ln7} \frac{4e^{3x}}{e^x-3}\,\mathrm{d}x\\ =&\,\int_2^4 \frac{4(u+3)^3}{u} \frac{\mathrm{d}u}{u+3}\\ =&\,\int_2^4 \frac{4(u+3)^2}{u}\,\mathrm{d}u\\ =&\,\int_2^4 \bigg(4u+24+\frac{36}{u}\bigg)\,\mathrm{d}u. \end{align*}

Hence

 ∫ln⁡5ln⁡74e3xex−3 dx= [2u2+24u+36ln⁡u]24= (128+36ln⁡4)−(56+36ln⁡2)= 72+36ln⁡2.\begin{align*} &\,\int_{\ln5}^{\ln7} \frac{4e^{3x}}{e^x-3}\,\mathrm{d}x\\ =&\,\big[2u^2+24u+36\ln u\big]_2^4\\ =&\,\big(128+36\ln4\big) -\big(56+36\ln2\big)\\ =&\,72+36\ln2. \end{align*}

Thus

a=72,b=36.\boxed{a=72,\qquad b=36}.

(ii)

解法一

思路

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连续使用两次分部积分。第一次会产生 ∫exsin⁡2x dx\int e^x\sin2x\,\mathrm{d}x；第二次处理这个积分后，原积分会再次出现。把它移到等式同一边，即可解出原积分。

答题过程

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Let

I=∫3excos⁡2x dx.I=\int3e^x\cos2x\,\mathrm{d}x.

Using integration by parts,

I= 32exsin⁡2x−32∫exsin⁡2x dx.\begin{align*} I =&\,\frac32e^x\sin2x -\frac32\int e^x\sin2x\,\mathrm{d}x. \end{align*}

Applying integration by parts again,

∫exsin⁡2x dx= −12excos⁡2x+12∫excos⁡2x dx= −12excos⁡2x+16I.\begin{align*} \int e^x\sin2x\,\mathrm{d}x =&\,-\frac12e^x\cos2x +\frac12\int e^x\cos2x\,\mathrm{d}x\\ =&\,-\frac12e^x\cos2x+\frac16I. \end{align*}

Substituting this into the expression for II,

I= 32exsin⁡2x−32(−12excos⁡2x+16I)= 32exsin⁡2x+34excos⁡2x−14I.\begin{align*} I =&\,\frac32e^x\sin2x -\frac32\bigg( -\frac12e^x\cos2x+\frac16I \bigg)\\ =&\,\frac32e^x\sin2x +\frac34e^x\cos2x-\frac14I. \end{align*}

Therefore,

54I= 32exsin⁡2x+34excos⁡2x,I= 65exsin⁡2x+35excos⁡2x+c.\begin{align*} \frac54I =&\,\frac32e^x\sin2x +\frac34e^x\cos2x,\\ I =&\,\frac65e^x\sin2x +\frac35e^x\cos2x+c. \end{align*}

Hence

p=65,q=35.\boxed{p=\frac65,\qquad q=\frac35}.