詳解：利用三角函數的恆等式

\(1 - {\sin ^2}x = {\cos ^2}x\)

為了要脫離根號，令 \(x = 3\sin \theta \quad  \Rightarrow \quad \sqrt {9 - {x^2}}  = \sqrt {9 - 9{{\sin }^2}\theta }  = 3\sqrt {1 - {{\sin }^2}\theta }  = 3\cos \theta \)

但還要考慮 \(x = 3\sin \theta \quad  \Rightarrow \quad dx = 3\cos \theta d\theta \)

\(\int {\sqrt {9 - {x^2}} \;dx}  = \)\(\int {3\cos \theta \;dx}  = \)\(\int {3\cos \theta  \cdot 3\cos \theta \;d\theta }  = 9\int {{{\cos }^2}\theta \;d\theta } \)

使用平方化倍角公式

\({\cos ^2}\theta  = \frac{{1 + \cos 2\theta }}{2}\)

\(9\int {{{\cos }^2}\theta \;d\theta }  = \)\(9\int {\frac{{1 + \cos 2\theta }}{2}\;d\theta }  = \)\(\frac{9}{2}\int {(1 + \cos 2\theta )\;d\theta }  = \)\(\frac{9}{2}(\theta  + \frac{1}{2}\sin 2\theta ) + C\)

上式雖然已經解出積分，但表示法不可以 \(\theta \) 表示，現在將 \(\sin 2\theta \) 以倍角公式展開

\(\frac{9}{2}(\theta  + \frac{1}{2}\sin 2\theta ) + C = \frac{9}{2}\theta  + \frac{9}{2}\sin \theta \cos \theta  + C\)

\(x = 3\sin \theta \quad  \Rightarrow \quad \sin \theta  = \frac{x}{3}\quad \)，\(\cos \theta  =

\(\frac{{\sqrt {9 - {x^2}} }}{3}\)，\(\theta  = Si{n^{ - 1}}\frac{x}{3}\)

取代 \(\theta \) 與 \(\theta \) 的函數

\(\frac{9}{2}\theta  + \frac{9}{2}\sin \theta \cos \theta  + C\)\( = \frac{9}{2}Si{n^{ - 1}}\frac{x}{3} + \frac{9}{2}\frac{x}{3}\frac{{\sqrt {9 - {x^2}} }}{3} + C\)\( = \frac{9}{2}Si{n^{ - 1}}\frac{x}{3} + \frac{{x\sqrt {9 - {x^2}} }}{2} + C\)