PART 19：例題-三角代換(求不定積分)

\(\int {\sqrt {16 - {x^2}} dx} \)

SOL:

(1) 變數假設就緒 令 \(x = 4\sin \theta \) ， \(dx = 4\cos \theta d\theta \)

(2)將 \(x\) 改為 \(\theta \) \(\int {\sqrt {16 - {x^2}} dx} \)  \( = \int {\sqrt {16 - {4^2}{{\sin }^2}\theta } 4\cos \theta d\theta } \)  \( = 4\int {\cos \theta 4\cos \theta d\theta } \)  \( = 16\int {{{\cos }^2}\theta d\theta } \)

(3)平方化倍角 \(16\int {{{\cos }^2}\theta d\theta } \)  \( = \frac{{16}}{2}\int {(1 + \cos 2\theta )d\theta } \) \( = 8(\theta  + \frac{1}{2}\sin 2\theta ) + C\)  \( = 8\theta  + 4\sin 2\theta  + C\)

(4) 將 \(\theta \) 還原為 \(x\) ，其中還使用了倍角公式 \(\sin 2\theta  = 2\sin \theta \cos \theta \)

\(8\theta  + 4\sin 2\theta  + C\)  \( = 8{\sin ^{ - 1}}\left( {\frac{x}{4}} \right) + 8\sin \theta \cos \theta  + C\)  \(= 8{\sin ^{ - 1}}\left( {\frac{x}{4}} \right) + 8\left( {\frac{x}{4}} \right)\left( {\frac{{\sqrt {16 - {x^2}} }}{4}} \right) + C\)

\( = 8{\sin ^{ - 1}}\left( {\frac{x}{4}} \right) + \frac{{x\sqrt {16 - {x^2}} }}{2} + C\)

註:  \(x = 4\sin \theta \) \( \Rightarrow \sin \theta  = \frac{x}{4}\) \(  \Rightarrow \theta  = {\sin ^{ - 1}}\left( {\frac{x}{4}} \right)\) ， \( \cos \theta  = \frac{{\sqrt {16 - {x^2}} }}{4}\)