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FP3 2020 Oct Q4

A Level / Edexcel / FP3

IAL 2020 Oct Paper · Question 4

题目

Problem

4.

Let

In=∫xncos⁡x dx.I_n = \int x^n \cos x\,\mathrm{d}x.

(a) Show that, for n≥2n \ge 2,

In=xnsin⁡x+nxn−1cos⁡x−n(n−1)In−2.I_n = x^n\sin x + nx^{n-1}\cos x - n(n-1)I_{n-2}.
(4)

(b) Hence find the functions f(x)f(x) and g(x)g(x) such that

∫x4cos⁡x dx=f(x)sin⁡x+g(x)cos⁡x+c,\int x^4\cos x\,\mathrm{d}x = f(x)\sin x + g(x)\cos x + c,

where cc is an arbitrary constant.

(5)
题目中文翻译

设

In=∫xncos⁡x dxI_n = \int x^n \cos x\,\mathrm{d}x

(a) 证明当 n≥2n \ge 2 时

In=xnsin⁡x+nxn−1cos⁡x−n(n−1)In−2I_n = x^n\sin x + nx^{n-1}\cos x - n(n-1)I_{n-2}
(4)

(b) 因此求函数 f(x)f(x) 和 g(x)g(x)，使得

∫x4cos⁡x dx=f(x)sin⁡x+g(x)cos⁡x+c\int x^4\cos x\,\mathrm{d}x = f(x)\sin x + g(x)\cos x + c

其中 cc 为任意常数。

(5)

解答

(a)

解法一

思路

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对 InI_n 连续做两次分部积分。第一次把幂函数留作求导，使积分从 cos⁡x\cos x 变为 sin⁡x\sin x；第二次再把 sin⁡x\sin x 积回 cos⁡x\cos x，余下积分便正好是 In−2I_{n-2}。

答题过程

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Integrating by parts,

In= ∫xncos⁡x dx= xnsin⁡x−n∫xn−1sin⁡x dx.\begin{align*} I_n =&\,\int x^n\cos x\,\mathrm{d}x\\[4mm] =&\,x^n\sin x -n\int x^{n-1}\sin x\,\mathrm{d}x. \end{align*}

For the remaining integral, a second integration by parts gives

∫xn−1sin⁡x dx= −xn−1cos⁡x +(n−1)∫xn−2cos⁡x dx= −xn−1cos⁡x+(n−1)In−2.\begin{align*} \int x^{n-1}\sin x\,\mathrm{d}x =&\,-x^{n-1}\cos x\\[4mm] &\,\hspace{2pt}+(n-1) \int x^{n-2}\cos x\,\mathrm{d}x\\[4mm] =&\,-x^{n-1}\cos x+(n-1)I_{n-2}. \end{align*}

Substituting,

In= xnsin⁡x−n(−xn−1cos⁡x+(n−1)In−2)= xnsin⁡x+nxn−1cos⁡x−n(n−1)In−2.\begin{align*} I_n =&\,x^n\sin x -n\bigl(-x^{n-1}\cos x+(n-1)I_{n-2}\bigr)\\[4mm] =&\,\boxed{x^n\sin x+nx^{n-1}\cos x -n(n-1)I_{n-2}}. \end{align*}

解法二

思路

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把被积式写成 xn−1(xcos⁡x)x^{n-1}(x\cos x)。先求出 ∫xcos⁡x dx=xsin⁡x+cos⁡x\int x\cos x\,\mathrm{d}x=x\sin x+\cos x，再分部积分；随后仅对其中一个剩余积分再做一次分部积分，也可得到同一递推式。

答题过程

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First,

∫xcos⁡x dx=xsin⁡x+cos⁡x.\int x\cos x\,\mathrm{d}x=x\sin x+\cos x.

Now write

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

Integrating by parts,

In= xn−1(xsin⁡x+cos⁡x) −(n−1)∫xn−2(xsin⁡x+cos⁡x) dx= xnsin⁡x+xn−1cos⁡x −(n−1)∫xn−1sin⁡x dx −(n−1)In−2.\begin{align*} I_n =&\,x^{n-1}(x\sin x+\cos x)\\[4mm] &\,\hspace{2pt}-(n-1) \int x^{n-2}(x\sin x+\cos x)\,\mathrm{d}x\\[4mm] =&\,x^n\sin x+x^{n-1}\cos x\\[4mm] &\,\hspace{2pt}-(n-1) \int x^{n-1}\sin x\,\mathrm{d}x\\[4mm] &\,\hspace{4pt}-(n-1)I_{n-2}. \end{align*}

Also,

∫xn−1sin⁡x dx=−xn−1cos⁡x+(n−1)In−2.\int x^{n-1}\sin x\,\mathrm{d}x =-x^{n-1}\cos x+(n-1)I_{n-2}.

Therefore

In= xnsin⁡x+xn−1cos⁡x +(n−1)xn−1cos⁡x −(n−1)2In−2−(n−1)In−2= xnsin⁡x+nxn−1cos⁡x−n(n−1)In−2.\begin{align*} I_n =&\,x^n\sin x+x^{n-1}\cos x\\[4mm] &\,\hspace{2pt}+(n-1)x^{n-1}\cos x\\[4mm] &\,\hspace{4pt}-(n-1)^2I_{n-2}-(n-1)I_{n-2}\\[4mm] =&\,\boxed{x^n\sin x+nx^{n-1}\cos x -n(n-1)I_{n-2}}. \end{align*}

(b)

解法一

思路

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承接 (a) 的递推式，先把 I4I_4 降到 I2I_2，再降到 I0=sin⁡xI_0=\sin x。最后分别收集 sin⁡x\sin x 与 cos⁡x\cos x 的系数，即得到 f(x)f(x)、g(x)g(x)。

答题过程

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Using the reduction formula with n=4n=4,

I4=x4sin⁡x+4x3cos⁡x−12I2.I_4=x^4\sin x+4x^3\cos x-12I_2.

With n=2n=2,

I2=x2sin⁡x+2xcos⁡x−2I0.I_2=x^2\sin x+2x\cos x-2I_0.

Since

I0=∫cos⁡x dx=sin⁡x,I_0=\int\cos x\,\mathrm{d}x=\sin x,

we obtain

I4= x4sin⁡x+4x3cos⁡x −12(x2sin⁡x+2xcos⁡x−2sin⁡x)= (x4−12x2+24)sin⁡x +(4x3−24x)cos⁡x+c.\begin{align*} I_4 =&\,x^4\sin x+4x^3\cos x\\[4mm] &\,\hspace{2pt}-12 \bigl(x^2\sin x+2x\cos x-2\sin x\bigr)\\[4mm] =&\,(x^4-12x^2+24)\sin x\\[4mm] &\,\hspace{2pt}+(4x^3-24x)\cos x+c. \end{align*}

Therefore

f(x)=x4−12x2+24\boxed{f(x)=x^4-12x^2+24}

and

g(x)=4x3−24x.\boxed{g(x)=4x^3-24x}.