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FP3 2023 Jan Q8

A Level / Edexcel / FP3

IAL 2023 Jan Paper · Question 8

题目

Problem

8.

In=∫cos⁡nx dxn≥0I_n=\int \cos^n x\,\mathrm{d}x \qquad n\ge 0

(a) Prove that, for n≥2n\ge 2,

In=1ncos⁡n−1xsin⁡x+n−1nIn−2I_n=\frac1n\cos^{n-1}x\sin x+\frac{n-1}{n}I_{n-2}
(4)

(b) Show that for positive even integers nn

∫0π/2cos⁡nx dx=n−1n⋅n−3n−2⋯56⋅34⋅12⋅π2\int_0^{\pi/2} \cos^n x\,\mathrm{d}x =\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdots\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}
(4)

(c) Hence determine the exact value of

∫0π/2cos⁡6xsin⁡2x dx\int_0^{\pi/2} \cos^6x\sin^2x\,\mathrm{d}x
(3)
题目中文翻译

8.

In=∫cos⁡nx dxn≥0I_n=\int \cos^n x\,\mathrm{d}x \qquad n\ge 0

(a) 证明当 n≥2n\ge 2 时

In=1ncos⁡n−1xsin⁡x+n−1nIn−2I_n=\frac1n\cos^{n-1}x\sin x+\frac{n-1}{n}I_{n-2}
(4)

(b) 证明当 nn 为正偶数时

∫0π/2cos⁡nx dx=n−1n⋅n−3n−2⋯56⋅34⋅12⋅π2\int_0^{\pi/2} \cos^n x\,\mathrm{d}x =\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdots\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}
(4)

(c) 因此求

∫0π/2cos⁡6xsin⁡2x dx\int_0^{\pi/2} \cos^6x\sin^2x\,\mathrm{d}x

的精确值。

(3)

解答

(a)

解法一：拆出一个 cos⁡x\cos x

思路

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把 cos⁡nx\cos^n x 写成 cos⁡n−1xcos⁡x\cos^{n-1}x\cos x，再进行分部积分。所得积分含有 sin⁡2x\sin^2x，用 sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x 将其化成 In−2−InI_{n-2}-I_n，最后整理出指定递推式。

答题过程

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Write

In=∫cos⁡n−1xcos⁡x dx.I_n=\int\cos^{n-1}x\cos x\,\mathrm{d}x.

Integrating by parts with

u=cos⁡n−1x,dv=cos⁡x dx,u=\cos^{n-1}x, \qquad \mathrm{d}v=\cos x\,\mathrm{d}x,

gives

du=−(n−1)cos⁡n−2xsin⁡x dx,v=sin⁡x.\mathrm{d}u =-(n-1)\cos^{n-2}x\sin x\,\mathrm{d}x, \qquad v=\sin x.

Therefore,

In= cos⁡n−1xsin⁡x +(n−1)∫cos⁡n−2xsin⁡2x dx= cos⁡n−1xsin⁡x +(n−1)∫cos⁡n−2x(1−cos⁡2x) dx= cos⁡n−1xsin⁡x+(n−1)(In−2−In).\begin{align*} I_n =&\,\cos^{n-1}x\sin x \\ &\,+(n-1)\int \cos^{n-2}x\sin^2x\,\mathrm{d}x \\ =&\,\cos^{n-1}x\sin x \\ &\,+(n-1)\int \cos^{n-2}x(1-\cos^2x)\,\mathrm{d}x \\ =&\,\cos^{n-1}x\sin x +(n-1)(I_{n-2}-I_n). \end{align*}

Hence

nIn=cos⁡n−1xsin⁡x+(n−1)In−2,nI_n =\cos^{n-1}x\sin x+(n-1)I_{n-2},

and so

In=1ncos⁡n−1xsin⁡x+n−1nIn−2.\boxed{ I_n=\frac1n\cos^{n-1}x\sin x +\frac{n-1}{n}I_{n-2} }.

解法二：先使用 cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x

思路

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先拆出 cos⁡2x\cos^2x 并使用平方恒等式，把 InI_n 写成 In−2I_{n-2} 减去另一个积分。对后一个积分分部积分，会产生含 InI_n 的项，移项后同样得到目标递推式。

答题过程

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

In= ∫cos⁡n−2xcos⁡2x dx= In−2−∫sin⁡2xcos⁡n−2x dx.\begin{align*} I_n =&\,\int\cos^{n-2}x\cos^2x\,\mathrm{d}x \\ =&\,I_{n-2} -\int\sin^2x\cos^{n-2}x\,\mathrm{d}x. \end{align*}

Let

J=∫sin⁡2xcos⁡n−2x dx.J=\int\sin^2x\cos^{n-2}x\,\mathrm{d}x.

For JJ, integrate by parts with

u=sin⁡x,dv=sin⁡xcos⁡n−2x dx.u=\sin x, \qquad \mathrm{d}v=\sin x\cos^{n-2}x\,\mathrm{d}x.

Then

du=cos⁡x dx,v=−cos⁡n−1xn−1.\mathrm{d}u=\cos x\,\mathrm{d}x, \qquad v=-\frac{\cos^{n-1}x}{n-1}.

Thus

J= −sin⁡xcos⁡n−1xn−1 +1n−1∫cos⁡nx dx= −sin⁡xcos⁡n−1xn−1+1n−1In.\begin{align*} J =&\,-\frac{\sin x\cos^{n-1}x}{n-1} \\ &\,+\frac1{n-1}\int\cos^n x\,\mathrm{d}x \\ =&\,-\frac{\sin x\cos^{n-1}x}{n-1} +\frac1{n-1}I_n. \end{align*}

Substituting this into In=In−2−JI_n=I_{n-2}-J,

In=In−2+sin⁡xcos⁡n−1xn−1−1n−1In.I_n =I_{n-2} +\frac{\sin x\cos^{n-1}x}{n-1} -\frac1{n-1}I_n.

Multiplying by n−1n-1 and collecting the InI_n terms,

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

Hence

In=1ncos⁡n−1xsin⁡x+n−1nIn−2.\boxed{ I_n=\frac1n\cos^{n-1}x\sin x +\frac{n-1}{n}I_{n-2} }.

(b)

解法一

思路

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把定积分另记为 JnJ_n，将 (a) 的递推式代入上下限。边界项在 00 与 π2\frac\pi2 都为零，因此每次只留下系数 n−1n\frac{n-1}{n}；反复递推到 J0J_0，再直接积分求出 J0=π2J_0=\frac\pi2。

答题过程

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Define

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

For even n≥2n\ge2, applying the reduction formula from part (a) gives

Jn= 1n[cos⁡n−1xsin⁡x]0π/2 +n−1nJn−2.\begin{align*} J_n =&\,\frac1n \bigl[\cos^{n-1}x\sin x\bigr]_0^{\pi/2} \\ &\,+\frac{n-1}{n}J_{n-2}. \end{align*}

At x=π2x=\frac\pi2, cos⁡n−1x=0\cos^{n-1}x=0, while at x=0x=0, sin⁡x=0\sin x=0. Hence

Jn=n−1nJn−2.J_n=\frac{n-1}{n}J_{n-2}.

Applying this repeatedly,

Jn= n−1nJn−2= n−1n⋅n−3n−2Jn−4= n−1n⋅n−3n−2⋯56⋅34⋅12J0.\begin{align*} J_n =&\,\frac{n-1}{n}J_{n-2} \\ =&\,\frac{n-1}{n}\cdot \frac{n-3}{n-2}J_{n-4} \\ =&\,\frac{n-1}{n}\cdot \frac{n-3}{n-2}\cdots \frac56\cdot\frac34\cdot\frac12J_0. \end{align*}

Finally,

J0=∫0π/21 dx=π2.J_0=\int_0^{\pi/2}1\,\mathrm{d}x =\frac\pi2.

Therefore,

∫0π/2cos⁡nx dx=n−1n⋅n−3n−2⋯56⋅34⋅12⋅π2.\boxed{ \int_0^{\pi/2}\cos^n x\,\mathrm{d}x =\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdots \frac56\cdot\frac34\cdot\frac12\cdot\frac\pi2 }.

(c)

解法一

思路

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必须承接 (b)。用 sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x 把原积分写成 J6−J8J_6-J_8，再分别使用 (b) 的乘积公式求出两个定积分并相减。

答题过程

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Using sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x and the result from part (b),

∫0π/2cos⁡6xsin⁡2x dx= J6−J8= 56⋅34⋅12⋅π2 −78⋅56⋅34⋅12⋅π2= 5π32−35π256= 5π256.\begin{align*} \int_0^{\pi/2}\cos^6x\sin^2x\,\mathrm{d}x =&\,J_6-J_8 \\ =&\,\frac56\cdot\frac34\cdot\frac12 \cdot\frac\pi2 \\ &\,-\frac78\cdot\frac56\cdot\frac34 \cdot\frac12\cdot\frac\pi2 \\ =&\,\frac{5\pi}{32}-\frac{35\pi}{256} \\ =&\,\boxed{\frac{5\pi}{256}}. \end{align*}