被积函数 $\sqrt{a^{2} – x^{2}}$ 的三角代换方法(B006)

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].   $x$ $=$ $\cos t$

[B].   $x$ $=$ $a \cos t$

[C].   $x$ $=$ $\sin t$

[D].   $x$ $=$ $a \sin t$


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$$\int \textcolor{Red}{\sqrt{a^{2} – x^{2}}} \mathrm{d} \textcolor{Yellow}{x}$$ $$\textcolor{Green}{\xrightarrow[]{x = a \times \sin t}}$$ $$\int \textcolor{Red}{\sqrt{a^{2} – (a \sin t)^{2}}} \mathrm{d} \textcolor{Yellow}{(a \sin t)} =$$ $$\int \textcolor{Red}{\sqrt{a^{2}(1 – \sin ^{2} t)} \cdot a \cos t} \mathrm{d} \textcolor{Yellow}{t} =$$ $$\int \textcolor{Red}{a \sqrt{\cos ^{2} t} \cdot a \cos t} \mathrm{d} \textcolor{Yellow}{t} =$$ $$\int \textcolor{Red}{a^{2} \cdot \cos ^{2} t} \mathrm{d} \textcolor{Yellow}{t} =$$ $$\textcolor{Orange}{a^{2}} \int \textcolor{Red}{\cos ^{2} t} \mathrm{d} \textcolor{Yellow}{t}.$$其中,$a$ 为常数,且 $a^{2}$ $-$ $x^{2}$ $\neq$ $0$.