2020年考研数二第03题解析：定积分、换元积分、三角函数积分

一、题目

二、解析

解法一

$$
\begin{aligned}
& \int_{0}^{1} \frac{\arcsin\sqrt{x}}{\sqrt{x \left( 1-x \right)}}\mathrm{~d} x \\ \\
\leadsto \ & \textcolor{gray}{\left( \sqrt{x} \right) ^{\prime} = \frac{1}{2} \frac{1}{\sqrt{x}} } \\ \\
= \ & \int_{0}^{1} \frac{\arcsin\sqrt{x}}{\sqrt{x} \cdot \sqrt{1-x}}\mathrm{~d} x \\ \\
= \ & \int_{0}^{1} \frac{2\arcsin\sqrt{x}}{\sqrt{1- \left( \sqrt{x} \right)^{2}}}\mathrm{~d} \left( \sqrt{x} \right) \\ \\
\leadsto \ & \textcolor{gray}{ \left( \arcsin \sqrt{x} \right) ^{\prime} = \frac{1}{\sqrt{1 – \left( \sqrt{x} \right)^{2}}} } \\ \\
= \ & 2\int_{0}^{1}\arcsin\sqrt{x} \mathrm{~d} \left( \arcsin \sqrt{x} \right) \\ \\
= \ & \left( \arcsin\sqrt{x} \right)^{2} \Bigg |_{0}^{1}=\left(\frac{\pi}{2}\right)^{2} \\ \\
= \ & \textcolor{springgreen}{ \frac{\pi^{2}}{4} }
\end{aligned}
$$

解法二

由于题目所给的式子中，被积函数中的 $\arcsin \sqrt{x}$ 比较复杂，所以，我们尝试直接对其做代换，即：

令 $t$ $=$ $\arcsin \sqrt{x}$, 则当 $x \in \left( 0, 1 \right)$ 时，$t \in \left( 0, \frac{\pi}{2} \right)$, 并且：

$$
\begin{aligned}
& \sin t = \sqrt{x} \leadsto x = \sin^{2} t \\ \\
& 1-x = 1 – \sin^{2} t = \cos^{2} t \\ \\
& \mathrm{d} x = 2 \left( \sin t \right) \left( \cos t \right) \mathrm{~d} t
\end{aligned}
$$

于是：

$$
\begin{aligned}
\int_{0}^{1} \frac{\arcsin \sqrt{x}}{\sqrt{x (1 – x)}} \mathrm{~d} x = \ & \int_{0}^{\frac{\pi}{2}} \frac{t}{\left( \sin t \right) \left( \cos t \right)} \cdot 2 \left( \sin t \right) \left( \cos t \right) \mathrm{~d} t \\ \\
= \ & \int_{0}^{\frac{\pi}{2}} 2 t \mathrm{~d} t \\ \\
= \ & t^{2} \Bigg| _{0}^{\frac{\pi}{2}} \\ \\
= \ & \textcolor{lightgreen}{ \frac{\pi^{2}}{4} }
\end{aligned}
$$

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