CalcGospel 國際數學圖譜
返回

CIE 9709 2024 June Paper 31 Q8

A Level / CIE / P3

CIE 9709 2024 June Paper 31 Paper · Question 8

题目

Problem

Use the substitution u=1−sin⁡xu=1-\sin x to find the exact value of

∫π32πsin⁡2x1−sin⁡x dx.\int_\pi^{\frac32\pi}\frac{\sin2x}{\sqrt{1-\sin x}}\,\mathrm{d}x.

Give your answer in the form a+b2a+b\sqrt2 where aa and bb are rational numbers to be determined.

(7)
题目中文翻译

使用代换 u=1−sin⁡xu=1-\sin x，求

∫π32πsin⁡2x1−sin⁡x dx\int_\pi^{\frac32\pi}\frac{\sin2x}{\sqrt{1-\sin x}}\,\mathrm{d}x

的精确值。答案以 a+b2a+b\sqrt2 的形式表示，其中 aa 和 bb 为待求的有理数。

(7)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

解法一

思路

展开

指定换元后，把二倍角拆成 2sin⁡xcos⁡x2\sin x\cos x，使余弦与微分一起替换。注意原下限对应 u=1u=1。

答题过程

展开

Let u=1−sin⁡xu=1-\sin x. Then du=−cos⁡x dx\mathrm{d}u=-\cos x\,\mathrm{d}x and sin⁡x=1−u\sin x=1-u.

The limits are

x=π ⇒u=1,x=3π2 ⇒u=2.\begin{align*} x=\pi&\,\Rightarrow u=1,\\[4mm] x=\frac{3\pi}{2}&\,\Rightarrow u=2. \end{align*}

Using sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x,

I= −2∫121−uu du= ∫12(2u1/2−2u−1/2) du= [43u3/2−4u1/2]12= (823−42) −(43−4)= 83−432.\begin{align*} I=&\,-2\int_1^2\frac{1-u}{\sqrt u}\,\mathrm{d}u\\ =&\,\int_1^2(2u^{1/2}-2u^{-1/2})\,\mathrm{d}u\\ =&\,\bigg[\frac43u^{3/2}-4u^{1/2}\bigg]_1^2\\ =&\,\bigg(\frac{8\sqrt2}{3}-4\sqrt2\bigg)\\[4mm] &\,\hspace{2pt}-\bigg(\frac43-4\bigg)\\[4mm] =&\,\boxed{\frac83-\frac43\sqrt2}. \end{align*}