问题
若通过三角代换计算积分 [$\textcolor{Orange}{\int \sqrt{a^{2} + x^{2}} \mathrm{d} x}$], 则应令 $\textcolor{Red}{x}$ $=$ $?$选项
[A]. $x$ $=$ $- \tan t$[B]. $x$ $=$ $- a \tan t$
[C]. $x$ $=$ $\tan t$
[D]. $x$ $=$ $a \tan t$
$$\int \textcolor{Red}{\sqrt{a^{2} + x^{2}}} \mathrm{d} \textcolor{Yellow}{x}$$ $$\textcolor{Green}{\xrightarrow[]{x = a \times \tan t}}$$ $$\int \textcolor{Red}{\sqrt{a^{2} + (a \tan t)^{2}}} \mathrm{d} \textcolor{Yellow}{(a \tan t)}$$ $$\int \textcolor{Red}{\sqrt{a^{2}(1 + \tan ^{2} t)}} \cdot a \sec ^{2} t \mathrm{d} \textcolor{Yellow}{t}$$ $$\int \textcolor{Red}{a \sqrt{(1 + \tan ^{2} t)}} \cdot a \sec ^{2} t \mathrm{d} \textcolor{Yellow}{t}$$ $$\int \textcolor{Red}{a \sqrt{\sec ^{2} t}} \cdot a \sec ^{2} t \mathrm{d} \textcolor{Yellow}{t}$$ $$\int \textcolor{Red}{a \sec t} \cdot a \sec ^{2} t \mathrm{d} \textcolor{Yellow}{t}$$ $$\int \textcolor{Red}{a^{2} \sec ^{3} t} \mathrm{d} \textcolor{Yellow}{t}$$ $$\textcolor{Orange}{a^{2}} \int \textcolor{Red}{\sec ^{3} t} \mathrm{d} \textcolor{Yellow}{t}.$$