新接触到的积分的代换

许久没接触与继续学习积分的技巧了,把最近接触到的个别积分代换技巧来小结一下。

1. 万能公式代换

I = \int{\frac{\mathrm{d}x}{(1+\cos x)^2}}

实际上,关于$ R(\sin x, \cos x)\mathrm{d}x $的积分,都可令$ t= \tan \frac{x}{2}, x \in (-\pi,\pi) $来使其有理化。
在本例中,我们便可由如下关系求其积分$ \begin{cases}\cos x &=\dfrac{1-t^2}{1+t^2} \\x &=2\arctan t\\ \mathrm{d}x &=\dfrac{2\mathrm{d}t}{1+t^2} \Leftrightarrow \mathrm{d}t=\dfrac{1}{2}\mathrm{d}x\sec^2\dfrac{x}{2}\end{cases} $

\begin{aligned}
\therefore I & = \int{\frac{\mathrm{d}x}{(1+\cos x)^2}} \\
& = \int{\frac{2}{(1+t^2)(1+\frac{1-t^2}{1+t^2})^2}\mathrm{d}t} \\
& = \int{\dfrac{1+t^2}{2}\mathrm{d}t} \\
& = \dfrac{t}{2} + \dfrac{t^3}{6} + C \\
& = \frac{1}{2}\tan\dfrac{x}{2} +  \frac{1}{6}\tan^3\dfrac{x}{2} + C.
\end{aligned}

2. 双曲代换(也可三角代换)

I = \int{\frac{x^2}{a-bx^2}\mathrm{d}x}

二话不多说,先分离常数,转化为

\begin{aligned}
I &= \dfrac{a}{b}\int{\dfrac{\mathrm{d}x}{a-bx^2}}-\dfrac{1}{b}\int{\mathrm{d}x} \\
&= -\dfrac{a}{b}\int{\dfrac{\mathrm{d}x}{bx^2-a}}-\dfrac{x}{b} \\
&= -\dfrac{1}{b}\int{\dfrac{\mathrm{d}x}{\frac{b}{a} x^2-1}}-\dfrac{x}{b}.
\end{aligned}

$\frac{\sqrt b}{\sqrt a}x=\cosh t$,则$x = \frac{\sqrt a}{\sqrt b}\cosh t, \mathrm{d}x = \frac{\sqrt a}{\sqrt b}\sinh t\mathrm{d}t$
原式转化为
$\begin{aligned}I & = -\dfrac{1}{b}\int{\dfrac{\frac{\sqrt a}{\sqrt b}\sinh t}{\cosh^2 t-1}\mathrm{d}t}-\dfrac{x}{b}\\ & = -\dfrac{\sqrt a}{b\sqrt b}\int{\dfrac{\mathrm{d}t}{\sinh t}}-\dfrac{x}{b}\end{aligned}$
已知$\int{\dfrac{\mathrm{d}t}{\sinh t}} = \ln\tanh\dfrac{t}{2}.$且根据之前的代换式,我们可以知道

\begin{aligned}
\ln \tanh\dfrac{t}{2} & = \dfrac{1}{2}\ln\dfrac{\sinh^2 t}{(1+\cosh t)^2}\\
& = \dfrac{1}{2}\ln\dfrac{\cosh^2 t - 1}{(1+\cosh t)^2} \\
& = \frac{1}{2}\ln\dfrac{\frac{b}{a}x^2 - 1}{(1+\frac{\sqrt b}{\sqrt a}x)^2} \\
& = \dfrac{1}{2}\ln\dfrac{(\frac{\sqrt b}{\sqrt a}x+1)(\frac{\sqrt b}{\sqrt a}x-1)}{(\frac{\sqrt b}{\sqrt a}x+1)^2} \\
& = \dfrac{1}{2}\ln\dfrac{\frac{\sqrt b}{\sqrt a}x-1}{\frac{\sqrt b}{\sqrt a}x+1} \\
& = -\dfrac{1}{2}\ln\dfrac{\sqrt b x+\sqrt a}{\sqrt b x-\sqrt a}.\\
\therefore I & = \dfrac{\sqrt a}{2b\sqrt b}\ln\dfrac{\sqrt b x+\sqrt a}{\sqrt b x-\sqrt a} -\dfrac{x}{b} +C.
\end{aligned}

实际上,也可直接利用原先的三角代换,来达到相同的效果。

习题

  1. $\int{\dfrac{\sin^2 x \cos x}{\sin x+\cos x} \mathrm{d}x}$
  2. $\int{\dfrac{\mathrm{d}x}{a+b\cos x}}$
  3. $\int{\sqrt{x^2+x}\mathrm{d}x}$
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