求下列定积分:-|||-(int )_(dfrac {sqrt {2)}(2)}^1dfrac (sqrt {1-{x)^2}}({x)^2}dx

步骤 1：换元法
令 $x = \sin t$，则 $dx = \cos t dt$，且当 $x = \dfrac{\sqrt{2}}{2}$ 时，$t = \dfrac{\pi}{4}$；当 $x = 1$ 时，$t = \dfrac{\pi}{2}$。因此，原积分变为
$$
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\sqrt{1-\sin^2 t}}{\sin^2 t} \cos t dt = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos^2 t}{\sin^2 t} dt = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^2 t dt
$$
步骤 2：利用三角恒等式
利用三角恒等式 $\cot^2 t = \csc^2 t - 1$，原积分变为
$$
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\csc^2 t - 1) dt = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \csc^2 t dt - \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} dt
$$
步骤 3：计算积分
由于 $\int \csc^2 t dt = -\cot t + C$，则
$$
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \csc^2 t dt = -\cot t \Big|_{\frac{\pi}{4}}^{\frac{\pi}{2}} = -\cot \frac{\pi}{2} + \cot \frac{\pi}{4} = 0 + 1 = 1
$$
而
$$
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} dt = t \Big|_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}
$$
因此，原积分的值为
$$
1 - \frac{\pi}{4}
$$