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CIE 9709 2026 June Paper 33 Q6

A Level / CIE / P3

CIE 9709 2026 June Paper 33 Paper · Question 6

题目

Problem

Use the substitution u=xu=\sqrt{x} to find the exact value of

∫116π214π2cos⁡(3x)x dx.\int_{\frac1{16}\pi^2}^{\frac14\pi^2}\frac{\cos(3\sqrt{x})}{\sqrt{x}}\,\mathrm{d}x.
(6)
题目中文翻译

使用代换 u=xu=\sqrt{x},求

∫116π214π2cos⁡(3x)x dx\int_{\frac1{16}\pi^2}^{\frac14\pi^2}\frac{\cos(3\sqrt{x})}{\sqrt{x}}\,\mathrm{d}x

的精确值。

(6)
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

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按题目要求令 u=xu=\sqrt{x}。上下限也随之改变,微分中的 uu 与分母的平方根相消。

答题过程

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Let u=xu=\sqrt{x}. Then x=u2x=u^2 and dx=2u du\mathrm{d}x=2u\,\mathrm{d}u. The limits become u=π/4u=\pi/4 and u=π/2u=\pi/2. The transformed integrand is (cos⁡3u/u)(2u)=2cos⁡3u(\cos3u/u)(2u)=2\cos3u. Hence the integral is

∫π/4π/22cos⁡3u du= [23sin⁡3u]π/4π/2= 23(−1−22)= −23−23.\begin{align*} \int_{\pi/4}^{\pi/2}2\cos3u\,\mathrm{d}u =&\,\bigg[\frac23\sin3u\bigg]_{\pi/4}^{\pi/2}\\ =&\,\frac23\bigg(-1-\frac{\sqrt2}{2}\bigg)\\ =&\,-\frac23-\frac{\sqrt2}{3}. \end{align*}