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IAL 2021 June FP3 Q5

A Level / Edexcel / FP3

IAL 2021 June Paper · Question 5

题目

Problem

Let

In=secnxdxn0.I_n = \int \sec^n x\,dx \qquad n \ge 0.

(a) Prove that for n2n \ge 2

(n1)In=tanxsecn2x+(n2)In2.(n-1)I_n = \tan x\,\sec^{n-2}x + (n-2)I_{n-2}.

(b) Hence, showing each step of your working, find the exact value of

0π/4sec6xdx.\int_0^{\pi/4} \sec^6 x\,dx.
(10)
题目中文翻译

In=secnxdxn0I_n = \int \sec^n x\,\mathrm{d}x \qquad n \ge 0

(a) 证明当 n2n \ge 2

(n1)In=tanxsecn2x+(n2)In2.(n-1)I_n = \tan x\,\sec^{n-2}x + (n-2)I_{n-2}.

(b) 因此,写出每一步的计算,求

0π/4sec6xdx\int_0^{\pi/4} \sec^6 x\,dx

的精确值。

解答

(a)

解法一:直接分部积分

思路

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这是官方主要路线。把 secnx\sec^n x 拆成 secn2xsec2x\sec^{n-2}x\sec^2x,令 secn2x\sec^{n-2}x 为待求导部分、sec2x\sec^2x 为待积分部分。分部积分后用 tan2x=sec2x1\tan^2x=\sec^2x-1,便能同时产生 InI_nIn2I_{n-2}

答题过程

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Write

In=secn2xsec2xdx.I_n=\int\sec^{n-2}x\sec^2x\,\mathrm{d}x.

Using integration by parts with

u=secn2x,dv=sec2xdx,u=\sec^{n-2}x, \qquad \mathrm{d}v=\sec^2x\,\mathrm{d}x,

we have

du=(n2)secn2xtanxdx,v=tanx.\mathrm{d}u =(n-2)\sec^{n-2}x\tan x\,\mathrm{d}x, \qquad v=\tan x.

Therefore,

In=tanxsecn2x(n2)secn2xtan2xdx=tanxsecn2x(n2)secn2x(sec2x1)dx=tanxsecn2x(n2)In+(n2)In2.\begin{align*} I_n =&\,\tan x\sec^{n-2}x \\ &\,-(n-2)\int \sec^{n-2}x\tan^2x\,\mathrm{d}x \\ =&\,\tan x\sec^{n-2}x \\ &\,-(n-2)\int \sec^{n-2}x(\sec^2x-1)\,\mathrm{d}x \\ =&\,\tan x\sec^{n-2}x -(n-2)I_n+(n-2)I_{n-2}. \end{align*}

Hence

(n1)In=tanxsecn2x+(n2)In2.\boxed{ (n-1)I_n =\tan x\sec^{n-2}x+(n-2)I_{n-2} }.

解法二:先使用 sec2x=1+tan2x\sec^2x=1+\tan^2x

思路

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官方替代路线先把积分拆成 In2I_{n-2} 与一个含 tan2x\tan^2x 的积分,再只对后者分部积分。由于中间需要除以 n2n-2,这条推导适用于 n>2n>2n=2n=2 必须另行验证。

答题过程

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For n>2n>2,

In=secn2x(1+tan2x)dx=In2+tan2xsecn2xdx.\begin{align*} I_n =&\,\int\sec^{n-2}x (1+\tan^2x)\,\mathrm{d}x \\ =&\,I_{n-2} +\int\tan^2x\sec^{n-2}x\,\mathrm{d}x. \end{align*}

For the second integral, use integration by parts with

u=tanx,dv=tanxsecn2xdx.u=\tan x, \qquad \mathrm{d}v=\tan x\sec^{n-2}x\,\mathrm{d}x.

Then

du=sec2xdx,v=secn2xn2.\mathrm{d}u=\sec^2x\,\mathrm{d}x, \qquad v=\frac{\sec^{n-2}x}{n-2}.

Thus

tan2xsecn2xdx=tanxsecn2xn21n2secnxdx=tanxsecn2xn2Inn2.\begin{align*} &\,\int\tan^2x\sec^{n-2}x\,\mathrm{d}x \\ =&\,\frac{\tan x\sec^{n-2}x}{n-2} -\frac1{n-2}\int\sec^n x\,\mathrm{d}x \\ =&\,\frac{\tan x\sec^{n-2}x}{n-2} -\frac{I_n}{n-2}. \end{align*}

Substituting this into the earlier expression and multiplying by n2n-2 gives

(n1)In=tanxsecn2x+(n2)In2.(n-1)I_n =\tan x\sec^{n-2}x+(n-2)I_{n-2}.

When n=2n=2, the formula becomes I2=tanxI_2=\tan x, which follows directly from sec2xdx=tanx\int\sec^2x\,\mathrm{d}x=\tan x. Hence the result holds for all n2n\ge2.

(b)

解法一

思路

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为避免把不定积分常数带入递推,定义带固定上下限的 JnJ_n。将 (a) 的递推式在 00π/4\pi/4 之间取值,依次由 J2J_2 算出 J4J_4,再算出 J6J_6;这样完整体现题目要求的每一步。

答题过程

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Let

Jn=0π/4secnxdx.J_n=\int_0^{\pi/4}\sec^n x\,\mathrm{d}x.

The reduction formula gives

(n1)Jn=[tanxsecn2x]0π/4+(n2)Jn2.(n-1)J_n =\left[\tan x\sec^{n-2}x\right]_0^{\pi/4} +(n-2)J_{n-2}.

First,

J2=[tanx]0π/4=1.J_2=\left[\tan x\right]_0^{\pi/4}=1.

For n=4n=4,

3J4=[tanxsec2x]0π/4+2J2=(1)(2)+2(1)=4,\begin{align*} 3J_4 =&\,\left[\tan x\sec^2x\right]_0^{\pi/4} +2J_2 \\ =&\,(1)(2)+2(1) \\ =&\,4, \end{align*}

so J4=43J_4=\frac43.

For n=6n=6,

5J6=[tanxsec4x]0π/4+4J4=(1)(4)+4(43)=283.\begin{align*} 5J_6 =&\,\left[\tan x\sec^4x\right]_0^{\pi/4} +4J_4 \\ =&\,(1)(4)+4\bigg(\frac43\bigg) \\ =&\,\frac{28}{3}. \end{align*}

Therefore,

0π/4sec6xdx=J6=2815.\boxed{ \int_0^{\pi/4}\sec^6x\,\mathrm{d}x =J_6=\frac{28}{15} }.