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}}}\,dx

(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}}}\,dx

的精确值。

解答

Use the substitution

x=4\sin\theta

Then

\frac{dx}{d\theta}=4\cos\theta

dx=4\cos\theta\,d\theta

Now

16-x^2=16-16\sin^2\theta

16-x^2=16\cos^2\theta

Hence

(16-x^2)^{3/2} =(16\cos^2\theta)^{3/2} =64\cos^3\theta

Since the limits will give $0<\theta<\dfrac{\pi}{2}$ , we can take $\cos\theta>0$ .

Change the limits.

When

x=2

we have

2=4\sin\theta

\sin\theta=\frac12

and hence

\theta=\frac{\pi}{6}

When

x=2\sqrt3

we have

2\sqrt3=4\sin\theta

\sin\theta=\frac{\sqrt3}{2}

and hence

\theta=\frac{\pi}{3}

Therefore

\int_2^{2\sqrt3}\frac{1}{(16-x^2)^{3/2}}\,dx = \int_{\pi/6}^{\pi/3} \frac{4\cos\theta}{64\cos^3\theta}\,d\theta

\int_{\pi/6}^{\pi/3} \frac{4\cos\theta}{64\cos^3\theta}\,d\theta = \frac{1}{16}\int_{\pi/6}^{\pi/3}\sec^2\theta\,d\theta

Thus

\frac{1}{16}\int_{\pi/6}^{\pi/3}\sec^2\theta\,d\theta = \frac{1}{16}\left[\tan\theta\right]_{\pi/6}^{\pi/3}

Hence

\begin{aligned} \int_2^{2\sqrt3}\frac{1}{(16-x^2)^{3/2}}\,dx &=\frac{1}{16}\left(\tan\frac{\pi}{3}-\tan\frac{\pi}{6}\right) \\ &=\frac{1}{16}\left(\sqrt3-\frac{1}{\sqrt3}\right) \end{aligned}

Now

\sqrt3-\frac{1}{\sqrt3} =\frac{3-1}{\sqrt3} =\frac{2}{\sqrt3}

Therefore

\frac{1}{16}\left(\sqrt3-\frac{1}{\sqrt3}\right) =\frac{1}{8\sqrt3} =\frac{\sqrt3}{24}

So the exact value is

\boxed{\frac{\sqrt3}{24}}