遇到两个根号式子相加减的定积分一般拆开来分别计算

一、题目

$$
I = \int_{0}^{1}\left[\sqrt{2 x-x^{2}}-\sqrt{\left(1-x^{2}\right)^{3}}\right] \mathrm{~ d} x=?
$$

难度评级：

二、解析

一、对于 $\int_{0}^{1} \sqrt{2 x-x^{2}} \mathrm{~ d} x$

$$
(x+1)^{2}=x^{2}+1+2 x \Rightarrow
$$

$$
(x-1)^{2}=x^{2}+1-2 x \Rightarrow
$$

$$
-(x-1)^{2}=2 x-x^{2}-1 \Rightarrow
$$

$$
+1-(x-1)^{2}=2 x-x^{2}
$$

$$
y=\sqrt{2 x-x^{2}} \Rightarrow y=\sqrt{1-(x-1)^{2}} \Rightarrow
$$

$$
(x-1)^{2}+y^{2}=1 \Rightarrow(1,0) \Rightarrow r=1 \Rightarrow
$$

$$
\int_{0}^{1} \sqrt{2 x-x^{2}}=\frac{1}{4} \pi r^{2}=\frac{\pi}{4}
$$

二、对于 $\int_{0}^{1} \sqrt{\left(1-x^{2}\right)^{3}} \mathrm{~ d} x$

$$
\int_{0}^{1} \sqrt{\left(1-x^{2}\right)^{3}} \mathrm{~ d} x \Rightarrow x=\sin t \Rightarrow
$$

$$
\sin t \in(0,1) \Rightarrow t \in\left(0, \frac{\pi}{2}\right)
$$

$$
1-x^{2}=\cos ^{2} t \Rightarrow
$$

$$
\int_{0}^{1} \sqrt{\left(1-x^{2}\right)^{3}} \mathrm{~ d} x=\int_{0}^{\frac{\pi}{2}} \cos ^{3} t \mathrm{~ d} (\sin t) \Rightarrow
$$

$$
\int_{0}^{\frac{\pi}{2}} \cos ^{4} t \mathrm{~ d} t=\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}=\frac{3 \pi}{16}
$$

综上可得：

$$
I=\frac{\pi}{4}-\frac{3 \pi}{16}=\frac{\pi}{16}
$$

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