| 問題 |
三角代換法 |
範圍 |
利用的恆等式 |
| \(\sqrt {a - {x^2}} \) |
令\(x = \sqrt a \sin \theta \) |
\( - \frac{\pi }{2} \le \theta \le \frac{\pi }{2}\) |
\(1 - {\sin ^2}\theta = {\cos ^2}\theta \) |
| \(\sqrt {a + {x^2}} \) |
令\(x = \sqrt a \tan \theta \) |
\( - \frac{\pi }{2} \le \theta \le \frac{\pi }{2}\) |
\(1 + {\tan ^2}\theta = {\sec ^2}\theta \) |
| \(\sqrt {{x^2} - a} \) |
令\(x = \sqrt a \sec \theta \) |
\(0 \le \theta \le \frac{\pi }{2}\;or\;\frac{\pi }{2} \le \theta \le \frac{{3\pi }}{2}\) |
\({\sec ^2}\theta - 1 = {\tan ^2}\theta \) |
