问题
若通过三角代换计算积分 [$\textcolor{Orange}{\int \sqrt{x^{2} – a^{2}} \mathrm{d} x}$], 则应令 $\textcolor{Red}{x}$ $=$ $?$选项
[A]. $x$ $=$ $- a \sec t$[B]. $x$ $=$ $\sec t$
[C]. $x$ $=$ $a \sec t$
[D]. $x$ $=$ $a \csc t$
$$\int \textcolor{Red}{\sqrt{x^{2} – a^{2}}} \mathrm{d} \textcolor{Yellow}{x}$$ $$\textcolor{Green}{\xrightarrow[]{x = a \times \sec t}}$$ $$\int \textcolor{Red}{\sqrt{(a \sec t)^{2} – a^{2}}} \mathrm{d} \textcolor{Yellow}{(a \sec t)} =$$ $$\int \textcolor{Red}{\sqrt{(a^{2} \sec ^{2} t – a^{2})} \cdot a \sec t \tan t} \mathrm{d} \textcolor{Yellow}{t} =$$ $$\int \textcolor{Red}{\sqrt{a^{2} (\sec ^{2} t – 1)} \cdot a \sec t \tan t} \mathrm{d} \textcolor{Yellow}{t} =$$ $$\int \textcolor{Red}{a \sqrt{\sec ^{2} t – 1} \cdot a \sec t \tan t} \mathrm{d} \textcolor{Yellow}{t} =$$ $$\int \textcolor{Red}{a \sqrt{\tan ^{2} t} \cdot a \sec t \tan t} \mathrm{d} \textcolor{Yellow}{t} =$$ $$\int \textcolor{Red}{a \tan t \cdot a \sec t \tan t} \mathrm{d} \textcolor{Yellow}{t} =$$ $$\int \textcolor{Red}{a ^{2} \tan ^{2} t \cdot \sec t} \mathrm{d} \textcolor{Yellow}{t} =$$ $$\textcolor{Orange}{a ^{2}} \int \textcolor{Red}{\tan ^{2} t \cdot \sec t} \mathrm{d} \textcolor{Yellow}{t}.$$