13.求不定积分 int dfrac (1)(sqrt {x(1-x))}dx

步骤 1：代换
令 $x = \dfrac{1}{2} + \dfrac{1}{2}\sin\theta$，则 $dx = \dfrac{1}{2}\cos\theta d\theta$。代入原积分，得到
$$\int \dfrac {1}{\sqrt {x(1-x)}}dx = \int \dfrac {1}{\sqrt {\left(\dfrac{1}{2} + \dfrac{1}{2}\sin\theta\right)\left(\dfrac{1}{2} - \dfrac{1}{2}\sin\theta\right)}}\cdot \dfrac{1}{2}\cos\theta d\theta$$
步骤 2：化简
化简上式，得到
$$\int \dfrac {1}{\sqrt {\left(\dfrac{1}{2} + \dfrac{1}{2}\sin\theta\right)\left(\dfrac{1}{2} - \dfrac{1}{2}\sin\theta\right)}}\cdot \dfrac{1}{2}\cos\theta d\theta = \int \dfrac {1}{\sqrt {\dfrac{1}{4} - \dfrac{1}{4}\sin^2\theta}}\cdot \dfrac{1}{2}\cos\theta d\theta$$
步骤 3：化简
化简上式，得到
$$\int \dfrac {1}{\sqrt {\dfrac{1}{4} - \dfrac{1}{4}\sin^2\theta}}\cdot \dfrac{1}{2}\cos\theta d\theta = \int \dfrac {1}{\sqrt {\dfrac{1}{4}\cos^2\theta}}\cdot \dfrac{1}{2}\cos\theta d\theta = \int d\theta$$
步骤 4：积分
对上式积分，得到
$$\int d\theta = \theta + c$$
步骤 5：回代
回代 $\theta = \arcsin(2x-1)$，得到
$$\theta + c = \arcsin(2x-1) + c$$