前言
下面这个式子连接了三角函数 $\sin$, $\cos$ 和 $\tan$, 即：
$$
\tan x = \frac{\sin x}{\cos x}
$$
因此，在遇到有关三角函数 $\tan$ 的积分时，我们可以尝试将其化作由三角函数 $\sin$ 和 $\cos$ 组成的等价表达式；类似地，在遇到有关三角函数 $\sin$ 和 $\cos$ 的积分时，我们可以尝试将其化作由三角函数 $\tan$ 组成的等价表达式.
化二为一
题目 1
已知 $\cos x \neq 0$, 则：
$$
\textcolor{lightgreen}{
\int \frac{\cos x}{\sin x + \cos x} \mathrm{~d} x = ?
}
$$
难度评级：
解析 1
首先，将被积函数的分子、分母同时除以 $\cos x$, 得：
$$
\begin{aligned}
\textcolor{lightgreen}{ \int \frac{\cos x}{\sin x + \cos x} \mathrm{~d} x } & = \int \frac{\frac{\cos x}{\cos x}}{\frac{\sin x}{\cos x} + \frac{\cos x}{\cos x}} \mathrm{~d} x \\ \\
& = \textcolor{orange}{ \int \frac{1}{\tan x+1} \mathrm{~d} x }
\end{aligned}
$$
接着，令 $t = \tan x$, 则：
$$
\textcolor{pink}{
x = \arctan t, \ \mathrm{d} x = \mathrm{d} \left( \arctan t \right) = \frac{1}{1+ t^{2}}
}
$$
于是：
$$
\begin{aligned}
\textcolor{orange}{ \int \frac{1}{1 + \tan x} \mathrm{~d} x } & = \int \frac{1}{1+t} \mathrm{~d} \left( \arctan t \right) \\ \\
& = \int \textcolor{magenta}{ \frac{1}{1 + t} \cdot \frac{1}{1 + t^{2}} } \mathrm{~d} t \\ \\
& = \textcolor{magenta}{ \frac{1}{2} } \int \left( \textcolor{magenta}{ \frac{1}{1 + t} – \frac{t – 1}{1 + t^{2}} } \right) \mathrm{~d} t \\ \\
& = \frac{1}{2} \int \left( \frac{1}{1 + t} – \frac{t}{1 + t^{2}} + \frac{1}{1 + t^{2}} \right) \mathrm{~d} t \\ \\
& = \frac{1}{2} \left[ \ln \left| 1 + t \right| – \frac{1}{2} \ln \left(1 + t^{2} \right) + \arctan t \right] + C \\ \\
& \leadsto \textcolor{gray}{t = \tan x} \\ \\
& = \frac{1}{2} \left[ \ln \left| 1 + \tan x \right| – \frac{1}{2} \ln \left( 1 + \tan^{2} x \right) + x \right] + C \\ \\
& = \frac{1}{2} \left[ \ln \left| 1 + \tan x \right| – \frac{1}{2} \ln \left(\frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x} \right) + x \right] + C \\ \\
& = \frac{1}{2} \left[ \ln \left| 1 + \tan x \right| – \ln \left(\frac{1}{\cos^{2} x} \right)^{\frac{1}{2}} + x \right] + C \\ \\
& = \frac{1}{2} \left( \ln \left| 1 + \tan x \right| – \ln \left| \sec x \right| + x \right) + C \\ \\
& = \textcolor{lightgreen}{ \frac{1}{2} (\ln \left| \cos x + \sin x \right| + x) + C }
\end{aligned}
$$
其中，$C$ 为任意常数.
上面从式子 $\textcolor{magenta}{ \frac{1}{1 + t} \cdot \frac{1}{1 + t^{2}} }$ 到式子 $\textcolor{magenta}{ \frac{1}{2} \left( \frac{1}{1 + t} – \frac{t – 1}{1 + t^{2}} \right) }$ 的转换过程，可以使用待定系数法实现求解：
令 $\frac{1}{1 + t} \cdot \frac{1}{1 + t^{2}}$ $=$ $\frac{a}{1+t} + \frac{bt + c}{1+t^{2}}$, 即 $\frac{1}{(1 + t) (1 + t^{2})}$ $=$ $\frac{a(1+t^{2}) + (1+t) (bt+c)}{(1+t)(1+t^{2})}$ $\leadsto$ $a+a t^{2} + bt + c + bt^{2} + ct$ $=$ $1$ $\leadsto$ $\begin{cases}
b+c=0 \\ a+b=0 \\ a+c=1
\end{cases}$ $\leadsto$ $\begin{cases}
a= \frac{1}{2} \\ b = \frac{-1}{2} \\ c = \frac{1}{2}
\end{cases}$, 因此：
$\frac{1}{1 + t} \cdot \frac{1}{1 + t^{2}}$ $=$ $\frac{\frac{1}{2}}{1+t} + \frac{\frac{-1}{2} t + \frac{1}{2}}{1+t^{2}}$ $=$ $\frac{1}{2} \left( \frac{1}{1+t^{2}} – \frac{t-1}{1+t^{2}} \right)$.
当然，由于 $\cot x = \frac{1}{\tan x}$, 所以，对于本题中的积分求解，如果已知 $\sin x \neq 0$, 我们也可以采用分子分母同时除以 $\sin x$ 的方式进行求解：
首先，将原式的分子分母同除以 $\sin x$, 得：
$$
\begin{aligned}
\textcolor{lightgreen}{ \int \frac{\cos x}{\sin x + \cos x} \mathrm{~d} x } & = \int \frac{\frac{\cos x}{\sin x}}{1 + \frac{\cos x}{\sin x}} \mathrm{~d} x \\ \\
& = \textcolor{orange}{ \int \frac{\cot x}{1 + \cot x} \mathrm{~d} x }
\end{aligned}
$$
接着，令 $u = \cot x$, 则：
$$
\textcolor{pink}{
x = \mathrm{arccot} \ u, \ \mathrm{d} x = \mathrm{d} \left( \mathrm{arccot} \ u \right) = \frac{-1}{1 + u^{2}} \mathrm{~d} u
}
$$
于是：
$$
\begin{aligned}
\textcolor{orange}{ \int \frac{\cot x}{1 + \cot x} \mathrm{~d} x } & = \int \frac{u}{1 + u} \mathrm{~d} \left( \mathrm{arccot} \ u \right) \\ \\
& = \int \frac{u}{1 + u} \cdot \frac{-1}{1 + u^{2}} \mathrm{~d} u \\ \\
& = \textcolor{magenta}{ – } \int \textcolor{magenta}{ \frac{u}{(1 + u)(1 + u^{2})} } \mathrm{~d} u \\ \\
& = – \int \left( \frac{\frac{-1}{2}}{1 + u} + \frac{\frac{1}{2} u + \frac{1}{2}}{1 + u^{2}} \right) \mathrm{~d} u \\ \\
& = \int \left( \frac{\frac{1}{2}}{1 + u} – \frac{\frac{1}{2} u + \frac{1}{2}}{1 + u^{2}} \right) \mathrm{~d} u \\ \\
& = \textcolor{magenta}{ \frac{1}{2} } \int \left( \textcolor{magenta}{ \frac{1}{1 + u} – \frac{u + 1}{1 + u^{2}} } \right) \mathrm{~d} u \\ \\
& = \frac{1}{2} \int \frac{1}{1 + u} \mathrm{~d} u – \frac{1}{2} \int \frac{u}{1 + u^{2}} \mathrm{~d} u – \frac{1}{2} \int \frac{1}{1 + u^{2}} \mathrm{~d} u \\ \\
& = \frac{1}{2} \ln \left| 1 + u \right| – \frac{1}{2} \cdot \frac{1}{2} \ln(1 + u^{2}) – \frac{1}{2} \arctan u + C \\ \\
& = \frac{1}{2} \ln \left| 1 + u \right| – \frac{1}{4} \ln(1 + u^{2}) – \frac{1}{2} \arctan u + C \\ \\
& \leadsto \textcolor{gray}{u = \cot x} \\ \\
& = \textcolor{tan}{ \frac{1}{2} \ln \left| 1 + \cot x \right| – \frac{1}{4} \ln(1 + \cot^{2} x) – \frac{1}{2} \arctan \left( \cot x \right) + C }
\end{aligned}
$$
又因为：
$$
\begin{aligned}
\textcolor{tan}{ \frac{1}{2} \ln \left| 1 + \cot x \right| } & = \frac{1}{2} \ln \left| \frac{\sin x + \cos x}{\sin x} \right| \\ \\ & = \textcolor{orangered}{ \frac{1}{2} \left( \ln\left| \sin x + \cos x \right| – \ln \sin x \right) } \\ \\
\textcolor{tan}{ \frac{-1}{4} \ln \left(1 + \cot^{2} x \right) } & = \frac{-1}{4} \ln \left( \frac{\sin^{2} x + \cos^{2} x}{\sin^{2} x} \right) \\ \\ & = \frac{-1}{4} \ln \left( \frac{1}{\sin^{2} x} \right) \\ \\ & = \frac{-1}{4} \ln \left( \sin x \right)^{-2} \\ \\ & = \frac{-1}{4} \cdot \left(-2 \ln \sin x \right) \\ \\ & = \textcolor{orangered}{ \frac{1}{2} \ln \sin x }
\end{aligned}
$$
同时，当 $x \in (0, \pi)$ 时，$\cot x = \tan \left( \frac{\pi}{2} – x \right)$, 所以：
$$
\begin{aligned}
\textcolor{tan}{ \frac{-1}{2} \arctan \left(\cot x \right) } & = \frac{-1}{2} \arctan \left[ \tan \left( \frac{\pi}{2} – x \right) \right] \\ \\
& = \frac{-1}{2} \left( \frac{\pi}{2} – x \right) \\ \\
& = \textcolor{orangered}{ \frac{x}{2} – \frac{\pi}{4} }
\end{aligned}
$$
综上可得：
$$
\begin{aligned}
& \textcolor{lightgreen}{ \int \frac{\cos x}{\sin x + \cos x} \mathrm{~d} x } \\ \\
& = \textcolor{tan}{ \frac{1}{2} \ln \left| 1 + \cot x \right| – \frac{1}{4} \ln(1 + \cot^{2} x) – \frac{1}{2} \arctan \left( \cot x \right) + C } \\ \\
& = \textcolor{orangered}{ \frac{1}{2} \left( \ln \left| \sin x + \cos x \right| – \ln \sin x \right) + \frac{1}{2} \ln \sin x + \frac{x}{2} – \frac{\pi}{4} } \\ \\
& = \textcolor{lightgreen}{ \frac{1}{2} \left( \ln \left| \sin x + \cos x \right| + x \right) + C }
\end{aligned}
$$
上面从式子 $\textcolor{magenta}{ \frac{-u}{(1 + u)(1 + u^{2})} }$ 到式子 $\textcolor{magenta}{ \frac{1}{2} \left( \frac{1}{1 + u} – \frac{u + 1}{1 + u^{2}} \right)}$ 的转换过程也可以使用前面介绍过的待定系数法实现求解：令 $\frac{-u}{(1 + u)(1 + u^{2})}$ $=$ $\frac{a}{1+u} + \frac{bu + c}{1+u^{2}}$ $\leadsto$ $\frac{-u}{(1 + u)(1 + u^{2})}$ $=$ $\frac{a \left( 1+u^{2} \right) + \left( 1+u \right) \left( bu+c \right)}{(1 + u)(1 + u^{2})}$ $\leadsto$ $a + au^{2} + bu + c + bu^{2} + cu = – u$ $\leadsto$ $\begin{cases}
a+b = 0 \\ b+c = -1 \\ a+c = 0
\end{cases}$ $\leadsto$ $\begin{cases}
a = \frac{1}{2} \\ b = \frac{-1}{2} \\ c = \frac{-1}{2}
\end{cases}$, 因此：
$\frac{-u}{(1 + u)(1 + u^{2})}$ $=$ $\frac{\frac{1}{2}}{1+u} + \frac{\frac{-1}{2} u – \frac{1}{2}}{1+u^{2}}$ $=$ $\frac{1}{2} \left( \frac{1}{1 + u} – \frac{u + 1}{1 + u^{2}} \right)$.
化一为二
题目 2
$$
\textcolor{lightgreen}{
\int \tan x \mathrm{~d} x = ?
}
$$
难度评级：
解析 2
$$
\begin{aligned}
\textcolor{lightgreen}{\int \tan x \mathrm{~d} x} & = \int \frac{\sin x}{\cos x} \mathrm{~d} x \\ \\
& = -\int \frac{\mathrm{d} (\cos x)}{\cos x} \\ \\
& = \textcolor{lightgreen}{ -\ln|\cos x| + C }
\end{aligned}
$$
其中，$C$ 为任意常数.