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IAL 2026 Jan FP3 Q7

A Level / Edexcel / FP3

IAL 2026 Jan 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.

Let

In=∫0π/4e4xtan⁡nx dxI_n = \int_0^{\pi/4} e^{4x}\tan^n x \, dx

(a) Use integration to prove that, for n≥1n \ge 1

In+1=kn−4nIn−In−1I_{n+1} = \frac{k}{n} - \frac{4}{n} I_n - I_{n-1}

where kk is a constant to be determined.

(b) Determine the exact value of I0I_0

Given that I1=3.7002I_1 = 3.7002 to 5 significant figures,

(c) use the answer to part (a) to determine the value of I4I_4 to 3 significant figures.

(4)
(2)
(3)
题目中文翻译

在本题中，你必须写出所有解题步骤。

完全依赖计算器技术求解是不可以的。

设

In=∫0π/4e4xtan⁡nx dxI_n = \int_0^{\pi/4} e^{4x}\tan^n x \, dx

(a) 利用积分证明，当 n≥1n \ge 1 时

In+1=kn−4nIn−In−1I_{n+1} = \frac{k}{n} - \frac{4}{n}I_n - I_{n-1}

其中 kk 为待定常数。

(b) 求 I0I_0 的精确值。

已知 I1=3.7002I_1 = 3.7002，精确到 5 位有效数字，

(c) 利用 (a) 的答案求 I4I_4，精确到 3 位有效数字。

解答

(a)

解法一

思路

展开

直接对 InI_n 分部积分，取 u=tan⁡nxu=\tan^n x、dv=e4x dx\mathrm{d}v=e^{4x}\,\mathrm{d}x。随后用 sec⁡2x=1+tan⁡2x\sec^2x=1+\tan^2x，剩下的积分正好拆成 In−1I_{n-1} 与 In+1I_{n+1}，整理后得到目标递推式。

答题过程

展开

Using integration by parts with

u=tan⁡nxu=\tan^n x

and

dv=e4x dx,\mathrm{d}v=e^{4x}\,\mathrm{d}x,

we have

du=ntan⁡n−1xsec⁡2x dx\mathrm{d}u=n\tan^{n-1}x\sec^2x\,\mathrm{d}x

and

v=14e4x.v=\frac14e^{4x}.

Therefore,

In= [14e4xtan⁡nx]0π/4 −n4∫0π/4e4xtan⁡n−1xsec⁡2x dx= eπ4−n4∫0π/4e4xtan⁡n−1x ×(1+tan⁡2x) dx= eπ4−n4(In−1+In+1).\begin{align*} I_n =&\,\left[\frac14e^{4x}\tan^n x\right]_0^{\pi/4}\\ &\,-\frac n4\int_0^{\pi/4} e^{4x}\tan^{n-1}x\sec^2x\,\mathrm{d}x\\ =&\,\frac{e^\pi}{4} -\frac n4\int_0^{\pi/4} e^{4x}\tan^{n-1}x\\ &\,\hspace{2pt}\times(1+\tan^2x)\,\mathrm{d}x\\ =&\,\frac{e^\pi}{4} -\frac n4\big(I_{n-1}+I_{n+1}\big). \end{align*}

Rearranging,

In+1=eπn−4nIn−In−1\boxed{ I_{n+1}=\frac{e^\pi}{n} -\frac4nI_n-I_{n-1}}

for n⩾1n\geqslant1. Hence

k=eπ.\boxed{k=e^\pi}.

解法二

思路

展开

官方评分资料也允许从 In+1I_{n+1} 出发。把一个 tan⁡2x\tan^2x 改写成 sec⁡2x−1\sec^2x-1，便分离出 In−1I_{n-1}；余下积分在分部积分时取 dv=tan⁡n−1xsec⁡2x dx\mathrm{d}v=\tan^{n-1}x\sec^2x\,\mathrm{d}x，可直接产生 InI_n，路线更快地对准目标下标。

答题过程

展开

Using tan⁡2x=sec⁡2x−1\tan^2x=\sec^2x-1,

In+1= ∫0π/4e4xtan⁡n−1xtan⁡2x dx= ∫0π/4e4xtan⁡n−1xsec⁡2x dx −In−1.\begin{align*} I_{n+1} =&\,\int_0^{\pi/4} e^{4x}\tan^{n-1}x\tan^2x\,\mathrm{d}x\\ =&\,\int_0^{\pi/4} e^{4x}\tan^{n-1}x\sec^2x\,\mathrm{d}x\\ &\,-I_{n-1}. \end{align*}

For the first integral, integrate by parts with

u=e4xu=e^{4x}

and

dv=tan⁡n−1xsec⁡2x dx.\mathrm{d}v=\tan^{n-1}x\sec^2x\,\mathrm{d}x.

Then

du=4e4x dx\mathrm{d}u=4e^{4x}\,\mathrm{d}x

and

v=1ntan⁡nx.v=\frac1n\tan^n x.

Hence

In+1= [1ne4xtan⁡nx]0π/4 −4n∫0π/4e4xtan⁡nx dx−In−1= eπn−4nIn−In−1.\begin{align*} I_{n+1} =&\,\left[\frac1n e^{4x}\tan^n x\right]_0^{\pi/4}\\ &\,-\frac4n\int_0^{\pi/4} e^{4x}\tan^n x\,\mathrm{d}x-I_{n-1}\\ =&\,\boxed{ \frac{e^\pi}{n}-\frac4nI_n-I_{n-1}}. \end{align*}

Thus k=eπk=e^\pi.

(b)

解法一

思路

展开

在 I0I_0 中，tan⁡0x=1\tan^0x=1，所以只需直接积分 e4xe^{4x} 并代入上下限。

答题过程

展开 I0= ∫0π/4e4x dx= [14e4x]0π/4= eπ−14.\begin{align*} I_0 =&\,\int_0^{\pi/4}e^{4x}\,\mathrm{d}x\\ =&\,\left[\frac14e^{4x}\right]_0^{\pi/4}\\ =&\,\boxed{\frac{e^\pi-1}{4}}. \end{align*}

(c)

解法一

思路

展开

由 (a) 的递推式，依次令 n=1,2,3n=1,2,3，便可从已知的 I0,I1I_0,I_1 逐步算到 I4I_4。中间保留足够小数位，最后才按题意取三位有效数字。

答题过程

展开

Using the reduction formula from part (a), with I1=3.7002I_1=3.7002,

I2= eπ−4I1−I0= eπ−4(3.7002)−eπ−14= 2.804719….\begin{align*} I_2 =&\,e^\pi-4I_1-I_0\\ =&\,e^\pi-4(3.7002)-\frac{e^\pi-1}{4}\\ =&\,2.804719\ldots. \end{align*}

Next,

I3= eπ2−2I2−I1= 2.260707….\begin{align*} I_3 =&\,\frac{e^\pi}{2}-2I_2-I_1\\ =&\,2.260707\ldots. \end{align*}

Finally,

I4= eπ3−43I3−I2= 1.894568…= 1.89(3 s.f.).\begin{align*} I_4 =&\,\frac{e^\pi}{3}-\frac43I_3-I_2\\ =&\,1.894568\ldots\\ =&\,\boxed{1.89}\qquad\text{(3 s.f.)}. \end{align*}