CalcGospel 國際數學圖譜
返回

CIE 9231 2024 June Paper 21 Q4

A Level / CIE / FP2

CIE 9231 2024 June Paper 21 Paper · Question 4

题目

Problem

It is given that, for n≥0n\ge0,

In=∫0ln⁡3sech⁡nx dx.I_n=\int_0^{\ln3}\operatorname{sech}^n x\,\mathrm{d}x.

(a) Show that, for n≥2n\ge2,

(n−1)In=(35)n−2(45)+(n−2)In−2.(n-1)I_n=\left(\frac{3}{5}\right)^{n-2}\left(\frac{4}{5}\right)+(n-2)I_{n-2}.

You may use the result that

ddx(sech⁡x)=−tanh⁡xsech⁡x.\frac{\mathrm{d}}{\mathrm{d}x}(\operatorname{sech}x)=-\tanh x\operatorname{sech}x.
[5]

(b) Find the value of I4I_4.

[3]
题目中文翻译

已知当 n≥0n\ge0 时，

In=∫0ln⁡3sech⁡nx dxI_n=\int_0^{\ln3}\operatorname{sech}^n x\,\mathrm{d}x

(a) 证明当 n≥2n\ge2 时，

(n−1)In=(35)n−2(45)+(n−2)In−2(n-1)I_n=\left(\frac{3}{5}\right)^{n-2}\left(\frac{4}{5}\right)+(n-2)I_{n-2}

可使用以下结果：

ddx(sech⁡x)=−tanh⁡xsech⁡x\frac{\mathrm{d}}{\mathrm{d}x}(\operatorname{sech}x)=-\tanh x\operatorname{sech}x

(b) 求 I4I_4 的值。

解答