📝 题目
例 11 求 $\displaystyle \int \sqrt{{a}^{2} - {x}^{2}}\mathrm{\;d}x\left( {a > 0}\right)$ .
💡 答案与解析
解 令 $x = a\sin t\left( {-\pi /2 \leq t \leq \pi /2}\right)$ ,我们得到
$$ \int \sqrt{{a}^{2} - {x}^{2}}\mathrm{\;d}x = {a}^{2}\int {\cos }^{2}t\mathrm{\;d}t $$
$$ = {a}^{2}\left( {\frac{t}{2} + \frac{\sin {2t}}{4}}\right) + C $$
$$ = \frac{1}{2}\left( {{a}^{2}t + {a}^{2}\sin t\cos t}\right) + C $$
$$ = \frac{1}{2}\left( {{a}^{2}\arcsin \frac{x}{a} + x\sqrt{{a}^{2} - {x}^{2}}}\right) + C. $$