题目

Problem

Using the substitution $x=4\cosh\theta$ , show that

\int\frac{1}{(x^2-16)^{3/2}}\,\mathrm{d}x =\frac{ax}{\sqrt{x^2-16}}+c, \qquad |x|>4,

where $a$ is a constant to be determined and $c$ is an arbitrary constant.

(6)

题目中文翻译

使用代换 $x=4\cosh\theta$ ，证明

\int\frac{1}{(x^2-16)^{3/2}}\,\mathrm{d}x =\frac{ax}{\sqrt{x^2-16}}+c, \qquad |x|>4,

其中 $a$ 为待定常数， $c$ 为任意常数。

(6)

解答

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严格使用题目指定的双曲代换。由 $x=4\cosh\theta$ 可把 $x^2-16$ 化成 $16\sinh^2\theta$ ，积分随即化为 $\operatorname{cosech}^2\theta$ 的标准积分。最后将 $\coth\theta$ 换回 $x$ 。

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Let

x=4\cosh\theta.

Then

\mathrm{d}x=4\sinh\theta\,\mathrm{d}\theta

and

\begin{align*} x^2-16 =&\,16\cosh^2\theta-16\\[4mm] =&\,16\sinh^2\theta. \end{align*}

For $x>4$ , we may take $\theta>0$ , so $\sinh\theta>0$ . Hence

\begin{align*} \int\frac{1}{(x^2-16)^{3/2}}\,\mathrm{d}x =&\,\int \frac{4\sinh\theta} {(16\sinh^2\theta)^{3/2}}\,\mathrm{d}\theta\\[4mm] =&\,\frac1{16}\int \operatorname{cosech}^2\theta\,\mathrm{d}\theta\\[4mm] =&\,-\frac1{16}\coth\theta+c. \end{align*}

Also,

\cosh\theta=\frac{x}{4}

and

\sinh\theta =\sqrt{\cosh^2\theta-1} =\frac{\sqrt{x^2-16}}{4}.

Therefore

\begin{align*} \int\frac{1}{(x^2-16)^{3/2}}\,\mathrm{d}x =&\,-\frac1{16} \frac{\cosh\theta}{\sinh\theta}+c\\[4mm] =&\,-\frac{x}{16\sqrt{x^2-16}}+c. \end{align*}

To check the result on both intervals, differentiate:

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\bigg(-\frac{x}{16\sqrt{x^2-16}}\bigg) =&\,-\frac1{16}\bigg((x^2-16)^{-1/2} -x^2(x^2-16)^{-3/2}\bigg)\\ =&\,-\frac1{16}\frac{x^2-16-x^2}{(x^2-16)^{3/2}}\\ =&\,\frac1{(x^2-16)^{3/2}}. \end{align*}

Thus the same antiderivative form holds for $|x|>4$ , and

\boxed{a=-\frac1{16}}.