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IAL 2022 Jan FP3 Q6

A Level / Edexcel / FP3

IAL 2022 Jan Paper · Question 6

题目

Problem

In=∫exsin⁡nx dx,n∈Z, n≥0I_n=\int e^x\sin^n x\,\mathrm{d}x, \qquad n\in\mathbb Z,\ n\ge0

(a) Show that

In=exsin⁡n−1xn2+1(sin⁡x−ncos⁡x)+n(n−1)n2+1In−2n≥2I_n=\frac{e^x\sin^{n-1}x}{n^2+1}( \sin x-n\cos x )+\frac{n(n-1)}{n^2+1}I_{n-2} \qquad n\ge 2

(b) Hence find the exact value of

∫0π/2exsin⁡4x dx\int_0^{\pi/2}e^x\sin^4x\,\mathrm{d}x

giving your answer in the form Aeπ/2+BAe^{\pi/2}+B where AA and BB are rational numbers to be determined.

(10)
题目中文翻译

设

In=∫exsin⁡nx dx,n∈Z, n≥0I_n=\int e^x\sin^n x\,\mathrm{d}x, \qquad n\in\mathbb Z,\ n\ge0

(a) 证明

In=exsin⁡n−1xn2+1(sin⁡x−ncos⁡x)+n(n−1)n2+1In−2n≥2I_n=\frac{e^x\sin^{n-1}x}{n^2+1}( \sin x-n\cos x )+\frac{n(n-1)}{n^2+1}I_{n-2} \qquad n\ge 2

(b) 因此求

∫0π/2exsin⁡4x dx\int_0^{\pi/2}e^x\sin^4x\,\mathrm{d}x

的精确值，答案写成 Aeπ/2+BAe^{\pi/2}+B 的形式，其中 AA 和 BB 为待定有理数。

解答

(a)

解法一

思路

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先对 InI_n 分部积分一次，所得新积分再分部积分一次。第二次求导会产生 cos⁡2x\cos^2x，用 cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x 将它拆成 In−2I_{n-2} 与 InI_n，最后把所有 InI_n 项移到同一边。

答题过程

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Integrating by parts,

In= ∫exsin⁡nx dx= exsin⁡nx−n∫exsin⁡n−1xcos⁡x dx.\begin{align*} I_n =&\,\int e^x\sin^n x\,\mathrm{d}x \\ =&\,e^x\sin^n x -n\int e^x\sin^{n-1}x\cos x\,\mathrm{d}x. \end{align*}

Let

J=∫exsin⁡n−1xcos⁡x dx.J=\int e^x\sin^{n-1}x\cos x\,\mathrm{d}x.

Integrating JJ by parts gives

J= exsin⁡n−1xcos⁡x −∫exddx(sin⁡n−1xcos⁡x) dx= exsin⁡n−1xcos⁡x −∫ex((n−1)sin⁡n−2xcos⁡2x−sin⁡nx) dx.\begin{align*} J=&\,e^x\sin^{n-1}x\cos x \\ &\,-\int e^x \frac{\mathrm{d}}{\mathrm{d}x} \bigl(\sin^{n-1}x\cos x\bigr)\,\mathrm{d}x \\ =&\,e^x\sin^{n-1}x\cos x \\ &\,-\int e^x\bigl((n-1)\sin^{n-2}x\cos^2x -\sin^n x\bigr)\,\mathrm{d}x. \end{align*}

Since cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x,

J= exsin⁡n−1xcos⁡x −(n−1)(In−2−In)+In= exsin⁡n−1xcos⁡x−(n−1)In−2+nIn.\begin{align*} J =&\,e^x\sin^{n-1}x\cos x \\ &\,-(n-1)(I_{n-2}-I_n)+I_n \\ =&\,e^x\sin^{n-1}x\cos x -(n-1)I_{n-2}+nI_n. \end{align*}

Substituting this into the first integration-by-parts result,

In= exsin⁡nx−nexsin⁡n−1xcos⁡x +n(n−1)In−2−n2In.\begin{align*} I_n =&\,e^x\sin^n x-ne^x\sin^{n-1}x\cos x \\ &\,+n(n-1)I_{n-2}-n^2I_n. \end{align*}

Therefore,

(n2+1)In= exsin⁡n−1x(sin⁡x−ncos⁡x) +n(n−1)In−2,\begin{align*} (n^2+1)I_n =&\,e^x\sin^{n-1}x \bigl(\sin x-n\cos x\bigr) \\ &\,+n(n-1)I_{n-2}, \end{align*}

and hence

In=exsin⁡n−1xn2+1(sin⁡x−ncos⁡x)+n(n−1)n2+1In−2.\boxed{ I_n=\frac{e^x\sin^{n-1}x}{n^2+1} \bigl(\sin x-n\cos x\bigr) +\frac{n(n-1)}{n^2+1}I_{n-2} }.

(b)

解法一：先求原函数

思路

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必须承接 (a)。先令 n=4n=4，再对其中的 I2I_2 使用同一递推式，并以 I0=exI_0=e^x 收尾；得到完整原函数后才代入上下限。

答题过程

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From part (a),

I4=exsin⁡3x17(sin⁡x−4cos⁡x)+1217I2.I_4=\frac{e^x\sin^3x}{17} \bigl(\sin x-4\cos x\bigr)+\frac{12}{17}I_2.

Also,

I2=exsin⁡x5(sin⁡x−2cos⁡x)+25I0,I_2=\frac{e^x\sin x}{5} \bigl(\sin x-2\cos x\bigr)+\frac25I_0,

where

I0=∫ex dx=ex.I_0=\int e^x\,\mathrm{d}x=e^x.

Thus

I4= exsin⁡3x17(sin⁡x−4cos⁡x) +12exsin⁡x85(sin⁡x−2cos⁡x)+24ex85.\begin{align*} I_4 =&\,\frac{e^x\sin^3x}{17} \bigl(\sin x-4\cos x\bigr) \\ &\,+\frac{12e^x\sin x}{85} \bigl(\sin x-2\cos x\bigr) +\frac{24e^x}{85}. \end{align*}

Therefore,

∫0π/2exsin⁡4x dx= [exsin⁡3x17(sin⁡x−4cos⁡x) +12exsin⁡x85(sin⁡x−2cos⁡x) +24ex85]0π/2= eπ/217+12eπ/285+24eπ/285−2485= 4185eπ/2−2485.\begin{align*} \int_0^{\pi/2}e^x\sin^4x\,\mathrm{d}x =&\,\biggl[ \frac{e^x\sin^3x}{17} \bigl(\sin x-4\cos x\bigr) \\ &\,\hspace{8pt}+\frac{12e^x\sin x}{85} \bigl(\sin x-2\cos x\bigr) \\ &\,\hspace{16pt}+\frac{24e^x}{85} \biggr]_0^{\pi/2} \\ =&\,\frac{e^{\pi/2}}{17} +\frac{12e^{\pi/2}}{85} +\frac{24e^{\pi/2}}{85}-\frac{24}{85} \\ =&\,\boxed{ \frac{41}{85}e^{\pi/2}-\frac{24}{85} }. \end{align*}

Hence A=4185A=\frac{41}{85} and B=−2485B=-\frac{24}{85}.

解法二：直接递推定积分

思路

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把目标定积分记作 J4J_4，直接在每次使用 (a) 的递推式后代入上下限。这样不用先写出完整原函数，也能依次把 J4J_4 化为 J2J_2、J0J_0。

答题过程

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Define

Jn=∫0π/2exsin⁡nx dx.J_n=\int_0^{\pi/2}e^x\sin^n x\,\mathrm{d}x.

Applying the result from part (a) with n=4n=4,

J4=eπ/217+1217J2.J_4=\frac{e^{\pi/2}}{17}+\frac{12}{17}J_2.

Applying it again with n=2n=2,

J2=eπ/25+25J0.J_2=\frac{e^{\pi/2}}5+\frac25J_0.

Finally,

J0=∫0π/2ex dx=eπ/2−1.J_0=\int_0^{\pi/2}e^x\,\mathrm{d}x =e^{\pi/2}-1.

Hence

J2= eπ/25+25(eπ/2−1)= 35eπ/2−25,\begin{align*} J_2 =&\,\frac{e^{\pi/2}}5 +\frac25\bigl(e^{\pi/2}-1\bigr) \\ =&\,\frac35e^{\pi/2}-\frac25, \end{align*}

and therefore

J4= eπ/217+1217(35eπ/2−25)= 4185eπ/2−2485.\begin{align*} J_4 =&\,\frac{e^{\pi/2}}{17} +\frac{12}{17} \bigg(\frac35e^{\pi/2}-\frac25\bigg) \\ =&\,\boxed{ \frac{41}{85}e^{\pi/2}-\frac{24}{85} }. \end{align*}