题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Use the substitution $x=4\sin\theta$ to find the exact value of

\int_2^{2\sqrt{3}}\frac{1}{(16-x^2)^{\frac{3}{2}}}\,\mathrm{d}x

(6)

题目中文翻译

本题必须写出全部解题步骤。

不接受完全依赖计算器技术的解法。

使用代换 $x=4\sin\theta$ ，求

\int_2^{2\sqrt{3}}\frac{1}{(16-x^2)^{\frac{3}{2}}}\,\mathrm{d}x

的精确值。

(6)

解答

展开

按题目要求令 $x=4\sin\theta$ ，变换微分与上下限。在所选区间余弦为正，根式化为余弦的三次方，积分变为 $\sec^2\theta$ 。

展开

Let

I=\int_2^{2\sqrt3}(16-x^2)^{-3/2}\,\mathrm{d}x.

Use the required substitution $x=4\sin\theta$ :

\frac{\mathrm{d}x}{\mathrm{d}\theta}=4\cos\theta, \qquad \mathrm{d}x=4\cos\theta\,\mathrm{d}\theta.

Choose the branch $0<\theta<\pi/2$ . The limits become:

$x$	$\sin\theta=x/4$	$\theta$
$2$	$1/2$	$\pi/6$
$2\sqrt3$	$\sqrt3/2$	$\pi/3$

Thus $\cos\theta>0$ throughout the integration interval. Using $1-\sin^2\theta=\cos^2\theta$ ,

\begin{align*} 16-x^2=&\,16\cos^2\theta,\\ (16-x^2)^{3/2}=&\,64\cos^3\theta. \end{align*}

The transformed integral is

\begin{align*} I=&\,\int_{\pi/6}^{\pi/3} \frac{4\cos\theta}{64\cos^3\theta}\,\mathrm{d}\theta\\ =&\,\frac1{16}\int_{\pi/6}^{\pi/3}\sec^2\theta\,\mathrm{d}\theta\\ =&\,\frac1{16}[\tan\theta]_{\pi/6}^{\pi/3}. \end{align*}

At the upper and lower limits, $\tan(\pi/3)=\sqrt3$ and $\tan(\pi/6)=1/\sqrt3$ . Hence

\begin{align*} I=&\,\frac1{16}\bigg(\sqrt3-\frac1{\sqrt3}\bigg)\\ =&\,\frac1{16}\cdot\frac2{\sqrt3}\\ =&\,\frac1{8\sqrt3}\\ =&\,\boxed{\frac{\sqrt3}{24}}. \end{align*}