高等数学预习-定积分
参考资料
https://space.bilibili.com/42428180
https://space.bilibili.com/1035929235
基本定义与性质
设 \(f(x)\) 是定义在区间 \([a, b]\) 上的函数，对于区间 \([a, b]\) 插入分点 \(x_{i}(i = 0, 1, \cdots, n)\)，使它们满足
\[T : a = x_{0} < x_{1} < x_{2} < \cdots < x_{n - 1} < x_{n} = b
\]
我们称之为对区间 \([a, b]\) 的一种 分割，并记之为 \(T\)。又记 \(\Delta{x_{i}} = x_{i} - x_{i - 1}(i = 1, \dots, n)\)，\(\lambda(T) = \max\{ \Delta{x_{i}} | i = 1, \cdots, n \}\)。
若对任意的分割 \(T\) 及任意选取的中间值 \(\xi_{i} \in [x_{i - 1}, x_{i}]\)，下列极限
\[\lim\limits_{\lambda(T) \to 0}{\sum\limits_{i = 1}^{n}f(\xi_{i})\Delta{x_{i}}}
\]
都存在，则称这个极限值为 \(f\) 在 \([a, b]\) 上的 定积分，记为
\[\int_{a}^{b}{f(x)\text{d}x}
\]
并称 \(f\) 在区间 \([a, b]\) 上 可积。
这里的 \(f\) 称为 被积函数，而 \(a\) 与 \(b\) 分别称为积分的 下限 和 上限，\(x\) 称为 积分变量，\([a, b]\) 称为 积分区间。
定积分定义中的和 \(\sum\limits_{i = 1}^{n}{f(\xi_{i})\Delta{x_{i}}}\) 称为 黎曼和，上述积分称作 黎曼积分，而上述函数的可积也称作 黎曼可积。
在上述定积分的定义中要求积分上限 \(b\) 大于积分下限 \(a\)。为了今后运算方便，约定
\(\int_{a}^{b}{f(x)\text{d}x} = 0\)，当 \(a = b\)
\(\int_{a}^{b}{f(x)\text{d}x} = -\int_{b}^{a}{f(x)\text{d}x}\)，当 \(b < a\)。
定积分有如下性质。
定积分只依赖于被积函数及积分区间，而与定积分中所使用的变量符号无关。比如，
\[\int_{a}^{b}{f(x)\text{d}x} = \int_{a}^{b}{f(t)\text{d}t}
\]
若 \(y = f(x)\) 在 \([a, b]\) 上可积，则对于任意常数 \(C\)，函数 \(y = Cf(x)\) 也可积，且有
\[\int_{a}^{b}{Cf(x)\text{d}x} = C\int_{a}^{b}{f(x)\text{d}x}
\]
若 \(y = f(x)\) 在 \([a, b]\) 上可积，则对于任意 \(c \in (a, b)\) 而言，函数 \(y = f(x)\) 在 \([a, c]\) 及在 \([c, b]\) 上都可积，且
\[\int_{a}^{b}{f(x)\text{d}x} = \int_{a}^{c}{f(x)\text{d}x} + \int_{c}^{b}{f(x)\text{d}x}
\]
当 \(c \not\in (a, b)\) 时，只要 \(y = f(x)\) 在积分区间上都可积，上述公式也成立。
若 \(y = f(x)\) 及 \(y = g(x)\) 在 \([a, b]\) 上可积，则 \(y = f(x) \pm g(x)\) 在 \([a, b]\) 上可积，且
\[\int_{a}^{b}{[f(x) \pm g(x)]\text{d}x} = \int_{a}^{b}{f(x)\text{d}x} \pm \int_{a}^{b}{g(x)\text{d}x}
\]
若 \(y = f(x)\) 在 \([a, b]\) 上可积，且 \(f(x) \ge 0, \forall x \in [a, b]\)，则
\[\int_{a}^{b}{f(x)\text{d}x} \ge 0
\]
若 \(y = f(x)\) 及 \(y = g(x)\) 在 \([a, b]\) 上可积，且 \(f(x) \ge g(x), \forall x \in [a, b]\)，则
\[\int_{a}^{b}{f(x)\text{d}x} \ge \int_{a}^{b}{g(x)\text{d}x}
\]
若 \(y = f(x)\) 在 \([a, b]\) 上可积，则 \(y = |f(x)|\) 在 \([a, b]\) 上也可积，且
\[\left| \int_{a}^{b}{f(x)\text{d}x} \right| \le \int_{a}^{b}{|f(x)|\text{d}x}
\]
若 \(y = f(x)\) 在 \([a, b]\) 上可积，且 \(M, m\) 分别是 \(f(x)\) 在 \([a, b]\) 上的最大值和最小值，则
\[m(b - a) \le \int_{a}^{b}{f(x)\text{d}x} \le M(b - a)
\]
定积分中值定理：
若函数 \(f(x)\) 在区间 \([a, b]\) 上连续，则在 \([a, b]\) 内至少存在一点 \(\xi\)，使得
\[\colorbox{yellow}{$\int_{a}^{b}{f(x)\text{d}x} = f(\xi)(b - a)$}
\]
其中 \(f(\xi) = \frac{\int_{a}^{b}{f(x)\text{d}x}}{b - a}\) 称为函数 \(f(x)\) 在区间 \([a, b]\) 上的平均值。
证明：
因为 \(f(x)\) 在 \([a, b]\) 上连续，它在 \([a, b]\) 上就有最大值与最小值，分别记为 \(M\) 和 \(m\)，即
\[m \le f(x) \le M, \forall x \in [a, b]
\]
所以
\[\int_{a}^{b}{m\text{d}x} \le \int_{a}^{b}{f(x)\text{d}x} \le \int_{a}^{b}{M\text{d}x} \\
\]
由此推出
\[m \le \frac{\int_{a}^{b}{f(x)\text{d}x}}{b - a} \le M
\]
再由连续函数的介值定理（若 \(f(x)\) 在区间 \([a, b]\) 上连续，则它必能取到介于最小值和最大值间的任意值），存在 \(\xi \in [a, b]\)，使得
\[f(\xi) = \frac{\int_{a}^{b}{f(x)\text{d}x}}{b - a}
\]
证毕。
下面是一些例题。
用定积分的定义计算 \(\int_{0}^{1}{x^{2}\text{d}x}\)。
\(f(x) = x^{2}\) 在 \([0, 1]\) 上连续可积。不妨对 \([0, 1]\) 进行 \(n\) 等分，每段取右端点值，即
\(\int_{0}^{1}{x^{2}\text{d}x} = \lim\limits_{n \to \infty}{\sum\limits_{i = 1}^{n}\frac{1}{n}f(\frac{i}{n})} = \lim\limits_{n \to \infty}\frac{1}{n^{3}}\sum\limits_{i = 1}^{n}i^{2} = \lim\limits_{n \to \infty}{\frac{1}{n^{3}} \times \frac{n(n + 1)(2n + 1)}{6}} = \frac{1}{3}\)。
求 \(\lim\limits_{n \to \infty}{\frac{1}{n^{2}}\left( \sqrt{n^{2} - 1^{2}} + \sqrt{n^{2} - 2^{2}} + \cdots + \sqrt{n^{2} - n^{2}} \right)}\)。
\(\lim\limits_{n \to \infty}{\frac{1}{n^{2}}\sum\limits_{i = 1}^{n}\sqrt{n^{2} - i^{2}}} = \lim\limits_{n \to \infty}\frac{1}{n}\sum\limits_{i = 1}^{n}\sqrt{1 - \left( \frac{i}{n} \right)^{2}} = \int_{0}^{1}{\sqrt{1 - x^{2}}\text{d}x} = \frac{\pi}{4}\)。
其中 \(\int_{0}^{1}{\sqrt{1 - x^{2}}\text{d}x}\) 相当于是单位圆在第一象限的面积。
变限积分
设函数 \(f(x)\) 在 \([a, b]\) 上可积，\(x\) 是 \([a, b]\) 上任意一点，则称
\[\Phi(x) = \int_{a}^{x}{f(t)\text{d}t}(a \le x \le b)
\]
为积分变上限函数，而称
\[\Phi(x) = \int_{x}^{b}{f(t)\text{d}t}(a \le x \le b)
\]
为积分变下限函数。
由于 \(\int_{x}^{b}{f(t)\text{d}t} = -\int_{b}^{x}{f(t)\text{d}t}\)，所以只需考察变上限积分。
设函数 \(f(x)\) 在 \([a, b]\) 上连续，则其变上限积分
\[\Phi(x) = \int_{a}^{x}{f(t)\text{d}t}
\]
是 \([a, b]\) 上的连续函数，且在 \((a, b)\) 内可导，并有
\[\colorbox{yellow}{$\Phi^{'}(x) = f(x), \forall x \in (a, b)$}
\]
证明：
由积分中值定理，对于任意 \(c \in [a, b)\) 及 \(x \in (c, b)\)，存在 \(\xi \in (c, x)\) 使得 \(\Phi(x) - \Phi(c) = \int_{c}^{x}{f(t)\text{d}t} = f(\xi)(x - c)\)，当 \(x \to c + 0\) 时，\(\Phi(x) - \Phi(c) = f(\xi)(x - c) \to 0\)，所以 \(\lim\limits_{x \to c + 0}\Phi(x) = \Phi(c)\)，由此证明了 \(\Phi(x)\) 在 \(c\) 处右连续，又由于 \(c\) 是 \([a, b)\) 上任意一点，所以 \(\Phi(x)\) 在 \([a, b)\) 中每一点都右连续。
由积分中值定理，对于任意 \(c \in (a, b]\) 及 \(x \in (a, c)\)，存在 \(\xi \in (x, c)\) 使得 \(\Phi(c) - \Phi(x) = \int_{x}^{c}{f(t)\text{d}t} = f(\xi)(c - x)\)，当 \(x \to c - 0\) 时，\(\Phi(c) - \Phi(x) = f(\xi)(c - x) \to 0\)，所以 \(\lim\limits_{x \to c - 0}\Phi(x) = \Phi(c)\)，由此证明了 \(\Phi(x)\) 在 \(c\) 处左连续，又由于 \(c\) 是 \((a, b]\) 上任意一点，所以 \(\Phi(x)\) 在 \((a, b]\) 中每一点都左连续。
综上，\(\Phi(x)\) 是 \([a, b]\) 上的连续函数。
由积分中值定理，对于任意 \(c \in (a, b)\) 及 \(x \in (a, b)\)，存在 \(\xi\) 在 \(c\) 和 \(x\) 之间，使得 \(\Phi(x) - \Phi(c) = \int_{c}^{x}{f(t)\text{d}t} = f(\xi)(x - c)\)，也即 \(f(\xi) = \frac{\Phi(x) - \Phi(c)}{x - c}\)，当 \(x \to c\) 时，\(\xi \to c\)，所以 \(\Phi^{'}(c) = \lim\limits_{x \to c}\frac{\Phi(x) - \Phi(c)}{x - c} = \lim\limits_{x \to c}f(\xi) = f(c)\)。
证毕。
这个定理揭示了 连续函数的变上限积分是该函数的一个原函数。
变限积分将定积分与不定积分联系在了一起。
变限积分求导公式：
设函数 \(f(x)\) 在 \([a, b]\) 上连续，\(\phi(x)\) 与 \(\psi(x)\) 在 \([a, b]\) 上可导，则积分变限函数 \(F(x) = \int_{\phi(x)}^{\psi(x)}{f(t)\text{d}t}\) 在 \([a, b]\) 上可导，且
\[\colorbox{yellow}{$F^{'}(x) = f[\psi(x)]\psi^{'}(x) - f[\phi(x)]\phi^{'}(x)$}
\]
推导：
\(F(x) = \int_{\phi(x)}^{a}{f(t)\text{d}t} + \int_{a}^{\psi(x)}{f(t)\text{d}t} = \int_{a}^{\psi(x)}{f(t)\text{d}t} - \int_{a}^{\phi(x)}{f(t)\text{d}t}\)，
则 \(F^{'}(x) = [f(\psi(x))]^{'} - [f(\phi(x))]^{'} = f[\psi(x)]\psi^{'}(x) - f[\phi(x)]\phi^{'}(x)\)。
下面是一些例题。
\(F(x) = \int_{0}^{\sqrt{x}}{\sin{t^{2}}\text{d}t}\)，求 \(F^{'}(x)\)。
令 \(\Phi(x) = \int_{0}^{x}{\sin{t^{2}}\text{d}t}\)，则 \(F(x) = \Phi(\sqrt{x})\)，\(\Phi^{'}(x) = \sin{x^{2}}\)。
所以 \(F^{'}(x) = \left[ \Phi(\sqrt{x}) \right]^{'} = \frac{\Phi^{'}(\sqrt{x})}{2\sqrt{x}} = \frac{\sin{x}}{2\sqrt{x}}\)。
求 \(\lim\limits_{x \to 0}{\frac{x^{2} - \int_{0}^{x^{2}}{\cos{t^{2}}\text{d}t}}{\sin^{10}{x}}}\)。
\[\begin{align}
\lim\limits_{x \to 0}{\frac{x^{2} - \int_{0}^{x^{2}}{\cos{t^{2}}\text{d}t}}{\sin^{10}{x}}}
&= \lim\limits_{x \to 0}{\frac{x^{2} - \int_{0}^{x^{2}}{\cos{t^{2}}\text{d}t}}{x^{10}}} \\
&= \lim\limits_{x \to 0}{\frac{2x - \cos{(x^{2})^{2}} \times (x^{2})^{'}}{10x^{9}}} \\
&= \lim\limits_{x \to 0}{\frac{1 - \cos{x^{4}}}{5x^{8}}} \\
&= \lim\limits_{x \to 0}{\frac{\frac{1}{2}(x^{4})^{2}}{5x^{8}}} \\
&= \frac{1}{10} \\
\end{align}
\]
求 \(\frac{\text{d}}{\text{d}x}\int_{x^{2}}^{x^{3}}\frac{\text{d}t}{\sqrt{1 + t^{4}}}\)。
\(\frac{\text{d}}{\text{d}x}\int_{x^{2}}^{x^{3}}\frac{\text{d}t}{\sqrt{1 + t^{4}}} = \frac{3x^{2}}{\sqrt{1 + x^{12}}} - \frac{2x}{\sqrt{1 + x^{8}}}\)。
求 \(\frac{\text{d}}{\text{d}x}\int_{\sin{x}}^{\cos{x}}{\cos(\pi t^{2})\text{d}t}\)。
\[\begin{align}
\frac{\text{d}}{\text{d}x}\int_{\sin{x}}^{\cos{x}}{\cos(\pi t^{2})\text{d}t}
&= \cos(\pi \cos^{2}{x}) \times (-\sin{x}) - \cos(\pi \sin^{2}{x}) \times \cos{x} \\
&= \cos(\pi(1 - \sin^{2}{x})) \times (-\sin{x}) - \cos(\pi \sin^{2}{x}) \times \cos{x} \\
&= \cos(\pi \sin^{2}{x})(\sin{x} - \cos{x}) \\
\end{align}
\]
微积分基本定理
微积分基本定理（牛顿-莱布尼茨公式）：
设函数 \(f\) 在区间 \([a, b]\) 上连续，又设函数 \(F\) 是 \(f\) 在 \((a, b)\) 上的一个原函数，也即
\[F^{'}(x) = f(x), \forall x \in (a, b)
\]
且 \(F\) 在 \([a, b]\) 上连续，则
\[\int_{a}^{b}{f(x)\text{d}x} = F(b) - F(a)
\]
也可以写成
\[\int_{a}^{b}{\frac{\text{d}F}{\text{d}x}\text{d}x} = F(b) - F(a)
\]
或
\[\int_{a}^{b}{\text{d}F} = F(x)\Bigg{|}_{a}^{b}
\]
证明：
令 \(\Phi(x) = \int_{a}^{x}{f(t)\text{d}t}\)，根据积分变上限函数的性质知 \(\Phi^{'}(x) = f(x), \forall x \in (a, b)\)。
这样，\(F^{'}(x) - \Phi^{'}(x) = 0\)，于是 \(F(x) = \Phi(x) + C, \forall x \in (a, b)\)，其中 \(C\) 是一个常数。
又因为 \(F(x), \Phi(x)\) 在 \([a, b]\) 上连续，所以 \(F(a) = \lim\limits_{x \to a + 0}{F(x)} = \lim\limits_{x \to a + 0}(\Phi(x) + C) = \Phi(a) + C = C\)，同理 \(F(b) = \Phi(b) + C\)。
于是 \(\int_{a}^{b}{f(x)\text{d}x} = \Phi(b) = F(b) - C = F(b) - F(a)\)。证毕。
下面是一些例题。
求 \(\int_{0}^{1}{\frac{1}{\sqrt{1 + x^{2}}}\text{d}x}\)。
\(\int_{0}^{1}{\frac{1}{\sqrt{1 + x^{2}}}\text{d}x} = \ln(x + \sqrt{1 + x^{2}})\Big{|}_{0}^{1} = \ln(1 + \sqrt{2})\)。
求 \(\int_{0}^{\frac{\pi}{4}}{\tan^{2}{\theta}\text{d}\theta}\)。
\(\int_{0}^{\frac{\pi}{4}}{\tan^{2}{\theta}\text{d}\theta} = \int_{0}^{\frac{\pi}{4}}{(\sec^{2}{\theta} - 1){\theta}\text{d}\theta} = (\tan{\theta} - \theta)\Big{|}_{0}^{\frac{\pi}{4}} = 1 = \frac{\pi}{4}\)。
定积分的分部积分法
设 \(u(x)\) 和 \(v(x)\) 在 \([a, b]\) 上可导且其导函数在 \([a, b]\) 上连续，则
\[\int_{a}^{b}{u(x)v^{'}(x)\text{d}x} = u(x)v(x)\Bigg{|}_{a}^{b} - \int_{a}^{b}{u^{'}(x)v(x)\text{d}x}
\]
也即
\[\int_{a}^{b}{u(x)\text{d}v(x)} = u(x)v(x)\Bigg{|}_{a}^{b} - \int_{a}^{b}{v(x)\text{d}u(x)}
\]
如果方便，也可以直接先通过不定积分求出原函数，然后再代入定积分求解。
但一定要有这样的意识：计算一个数往往比计算一个函数更简便。
下面是一些例题。
求 \(\int_{1}^{2}{x\ln{x}\text{d}x}\)。
法一：\(\int_{1}^{2}{x\ln{x}\text{d}x} = \frac{1}{2}\int_{1}^{2}{\ln{x}\text{d}x^{2}} = \frac{1}{2}x^{2}\ln{x}\Big{|}_{1}^{2} - \frac{1}{2}\int_{1}^{2}x^{2}\text{d}\ln{x} = 2\ln{2} - \frac{1}{2}\int_{1}^{2}{x\text{d}x} = 2\ln{2} - \frac{1}{4}x^{2}\Big{|}_{1}^{2} = 2\ln{2} - \frac{3}{4}\)。
法二：先用不定积分的分部积分法求出 \(\int{x\ln{x}\text{d}x} = \frac{1}{2}x^{2}\ln{x} - \frac{1}{4}x^{2} + C\)，得 \(\int_{1}^{2}{x\ln{x}\text{d}x} = \left( \frac{1}{2}x^{2}\ln{x} - \frac{1}{4}x^{2} \right)\Big{|}_{1}^{2} = 2\ln{2} - \frac{3}{4}\)。
求定积分 \(I_{n} = \int_{0}^{\frac{\pi}{2}}{\sin^{n}{x}\text{d}x}\)，其中 \(n \in N\)。（Wallis公式）
\[\begin{align}
I_{n} &= -\int_{0}^{\frac{\pi}{2}}{\sin^{n - 1}{x}\text{d}(\cos{x})} \\
&= -(\sin^{n - 1}{x}\cos{x})\Big{|}_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}}{\cos{x}\text{d}(\sin^{n - 1}{x})} \\
&= (n - 1)\int_{0}^{\frac{\pi}{2}}{\sin^{n - 2}{x}\cos^{2}{x}\text{d}x} \\
&= (n - 1)\int_{0}^{\frac{\pi}{2}}{\sin^{n - 2}{x}(1 - \sin^{2}{x})\text{d}x} \\
&= (n - 1)I_{n - 2} - (n - 1)I_{n} \\
\end{align}
\]
整理得递推公式
\[I_{n} = \frac{n - 1}{n}I_{n - 2}
\]
而 \(I_{0} = \int_{0}^{\frac{\pi}{2}}{\text{d}x} = \frac{\pi}{2}\)，\(I_{1} = \int_{0}^{\frac{\pi}{2}}{\sin{x}\text{d}x} = (-\cos{x})\Big{|}_{0}^{\frac{\pi}{2}} = 1\)，所以
\[\begin{align}
\colorbox{yellow}{$
I_{n} =
\begin{cases}
\frac{(n - 1)!!}{n!!} \times \frac{\pi}{2}, & n = 2k, k \in N \\
\frac{(n - 1)!!}{n!!}, &n = 2k + 1, k \in N \\
\end{cases}
$}
\end{align}
\]
用同样的方法可推导处 Wallis公式 的推广形式：记 \(I_{n, m} = \int_{0}^{\frac{\pi}{2}}{\sin^{m}{x}\cos^{n}{x}\text{d}x}\)，则
\[\begin{align}
\colorbox{yellow}{$
I_{n, m} =
\begin{cases}
\frac{(m - 1)!!(n - 1)!!}{(m + n)!!} \times \frac{\pi}{2}, & m, n 均为偶数 \\
\frac{(m - 1)!!(n - 1)!!}{(m + n)!!}, & \text{otherwise} \\
\end{cases}
$}
\end{align}
\]
定积分的换元法
设 \(f(x)\) 在 \([A, B]\) 上连续，又设 \(\varphi(t)\) 在 \([\alpha, \beta]\) 上有连续的导函数，且当 \(t\) 在 \([\alpha, \beta]\) 中变动时 \(\varphi(t)\) 在 \([A, B]\) 内变动。
假定 \(\varphi(\alpha) = a, \varphi(\beta) = b\)，\(a, b \in [A, B]\)，则
\[\int_{a}^{b}{f(x)\text{d}x} = \int_{\alpha}^{\beta}{f(\varphi(t))\varphi^{'}(t)\text{d}t}
\]
定积分换元法中极其重要的一点是：不能搞错上下标换元的对应关系。
下面是一些例题。
求 \(\int_{0}^{\frac{\pi}{2}}{\cos^{5}{x}\sin{x}\text{d}x}\)。
\(I = \int_{0}^{\frac{\pi}{2}}{\cos^{5}{x}\text{d}(\cos{x})} = \int_{0}^{1}{t^{5}\text{d}t} = \frac{1}{6}t^{6}\Big{|}_{0}^{1} = \frac{1}{6}\)。
计算椭圆 \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \le 1 (a > 0, b > 0)\) 的面积 \(S\)。
\(S = 2\int_{-a}^{a}{b\sqrt{1 - \frac{x^{2}}{a^{2}}}\text{d}x}\)。令 \(x = a\sin{t}(-\frac{\pi}{2} \le t \le \frac{\pi}{2})\)，则
\[\begin{align}
S &= 2b\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\cos{t}\text{d}(a\sin{t})} \\
&= 2ab\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\cos^{2}{t}\text{d}t} \\
&= ab\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{(1 + \cos{2t})\text{d}t} \\
&= ab\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\text{d}t} + \frac{1}{2}ab\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\cos{2t}\text{d}(2t)} \\
&= ab\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\text{d}t} + \frac{1}{2}ab\int_{-\pi}^{\pi}{\cos{t}\text{d}(t)} \\
&= abt\Big{|}_{-\frac{\pi}{2}}^{\frac{\pi}{2}} + \frac{1}{2}ab\sin{t}\Big{|}_{-\pi}^{\pi} \\
&= ab\pi \\
\end{align}
\]
证明：\(\int_{0}^{\frac{\pi}{2}}{\cos^{n}{x}\text{d}x} = \int_{0}^{\frac{\pi}{2}}{\sin^{n}{x}\text{d}x}\)。
令 \(x = \frac{\pi}{2} - t(0 \le t \le \frac{\pi}{2})\)，则 \(\int_{0}^{\frac{\pi}{2}}{\cos^{n}{x}\text{d}x} = \int_{\frac{\pi}{2}}^{0}{\sin^{n}{t}\text{d}(\frac{\pi}{2} - t)} = -\int_{\frac{\pi}{2}}^{0}{\sin^{n}{t}\text{d}t} = \int_{0}^{\frac{\pi}{2}}{\sin^{n}{t}\text{d}t} = \int_{0}^{\frac{\pi}{2}}{\sin^{n}{x}\text{d}x}\)。
练习1
求 \(\int_{0}^{1}{\frac{x^{2}\arcsin{x}}{\sqrt{1 - x^{2}}}\text{d}x}\)。
令 \(x = \sin{t}(0 \le t \le \frac{\pi}{2})\)，则
\[\begin{align}
I &= \int_{0}^{\frac{\pi}{2}}{\frac{t\sin^{2}{t}}{\cos{t}}\text{d}(\sin{t})} \\
&= \int_{0}^{\frac{\pi}{2}}{t\sin^{2}{t}\text{d}t} \\
&= \int_{0}^{\frac{\pi}{2}}{t \times \frac{1 - \cos{2t}}{2}\text{d}t} \\
&= \frac{1}{2}\int_{0}^{\frac{\pi}{2}}{t\text{d}t} - \frac{1}{8}\int_{0}^{\frac{\pi}{2}}{2t\cos{2t}\text{d}{2t}} \\
&= \frac{1}{4}t^{2}\Big{|}_{0}^{\frac{\pi}{2}} - \frac{1}{8}\int_{0}^{\pi}{t\cos{t}\text{d}t} \\
&= \frac{\pi^{2}}{16} - \frac{1}{8}\int_{0}^{\pi}{t\text{d}(\sin{t})} \\
&= \frac{\pi^{2}}{16} - \frac{1}{8}\left( t\sin{t}\Big{|}_{0}^{\pi} - \int_{0}^{\pi}{\sin{t}\text{d}t} \right) \\
&= \frac{\pi^{2}}{16} - \frac{1}{8}(t\sin{t} + \cos{t})\Big{|}_{0}^{\pi} \\
&= \frac{\pi^{2}}{16} + \frac{1}{4} \\
\end{align}
\]
求 \(\int_{0}^{3}{\arcsin{\sqrt{\frac{x}{1 + x}}}\text{d}x}\)。
令 \(\sqrt{\frac{x}{1 + x}} = t\)，反解得 \(x = \frac{1}{1 - t^{2}} - 1\)。\(\text{d}x = \text{d}\left( \frac{1}{1 - t^{2}} - 1 \right) = \text{d}\left( \frac{1}{1 - t^{2}} \right)\)。则
\[\begin{align}
I &= \int_{0}^{\frac{\sqrt{3}}{2}}{\arcsin{t}\text{d}\left( \frac{1}{1 - t^{2}} \right)} \\
&= \frac{\arcsin{t}}{1 - t^{2}}\Bigg{|}_{0}^{\frac{\sqrt{3}}{2}} - \int_{0}^{\frac{\sqrt{3}}{2}}{\frac{1}{1 - t^{2}}\text{d}(\arcsin{t})} \\
&= \frac{4\pi}{3} - \int_{0}^{\frac{\sqrt{3}}{2}}{\frac{\text{d}t}{(1 - t^{2})^{\frac{3}{2}}}} \\
\end{align}
\]
令 \(t = \sin{u}(0 \le u \le \frac{\pi}{3})\)，则 \(\int_{0}^{\frac{\sqrt{3}}{2}}{\frac{\text{d}t}{(1 - t^{2})^{\frac{3}{2}}}} = \int_{0}^{\frac{\pi}{3}}{\frac{\text{d}(\sin{u})}{\cos^{3}{u}}} = \int_{0}^{\frac{\pi}{3}}{\frac{\text{d}u}{\cos^{2}{u}}} = \tan{u}\Big{|}_{0}^{\frac{\pi}{3}} = \sqrt{3}\)，故 \(I = \frac{4\pi}{3} - \sqrt{3}\)。
求 \(\int_{1}^{16}{\arctan{\sqrt{\sqrt{x} - 1}}\text{d}x}\)。
令 \(\sqrt{\sqrt{x} - 1} = t\)，反解得 \(x = (t^{2} + 1)^{2}\)，则
\[\begin{align}
I &= \int_{0}^{\sqrt{3}}{\arctan{t}\text{d}(t^{2} + 1)^{2}} \\
&= (t^{2} + 1)^{2}\arctan{t}\Big{|}_{0}^{\sqrt{3}} - \int_{0}^{\sqrt{3}}{(t^{2} + 1)^{2}\text{d}(\arctan{t})} \\
&= \frac{16\pi}{3} - \int_{0}^{\sqrt{3}}{(t^{2} + 1)\text{d}t} \\
&= \frac{16\pi}{3} - \left( \frac{1}{3}t^{3} + t \right)\Bigg{|}_{0}^{\sqrt{3}} \\
&= \frac{16\pi}{3} - 2\sqrt{3} \\
\end{align}
\]
求定积分的基本技巧与注意事项
可利用几何定义计算积分。
\(\int_{-a}^{a}{f(x)\text{d}x} = \int_{-a}^{0}(f(x) + f(-x))\text{d}x\)。
若 \(f(x)\) 为奇函数：\(\int_{-a}^{a}{f(x)\text{d}x} = 0\)。
若 \(f(x)\) 为偶函数：\(\int_{-a}^{a}{f(x)\text{d}x} = 2\int_{0}^{a}{f(x)\text{d}x}\)。
利用周期性：
若 \(T\) 是函数 \(f(x)\) 的一个周期，则
\[\colorbox{yellow}{$
\int_{a}^{a + nT}{f(x)\text{d}x} = n\int_{0}^{T}{f(x)\text{d}x}
$}
\]
Wallis公式（见上文）
\[\colorbox{yellow}{$
\int_{0}^{\frac{\pi}{2}}{f(\sin{x})\text{d}x} = \int_{0}^{\frac{\pi}{2}}{f(\cos{x})\text{d}x}
$}
\]
推导：
\(\int_{0}^{\frac{\pi}{2}}{f(\sin{x})\text{d}x} = \int_{0}^{\frac{\pi}{2}}{f(\cos(\frac{\pi}{2} - x))\text{d}x} \xlongequal{t = \frac{\pi}{2} - x} \int_{\frac{\pi}{2}}^{0}{f(\cos{t})\text{d}(\frac{\pi}{2} - t)} = \int_{0}^{\frac{\pi}{2}}f(\cos{t})\text{d}t = \int_{0}^{\frac{\pi}{2}}{f(\cos{x})\text{d}x}\)。
一个常见的积分公式：
\[\colorbox{yellow}{$
\int_{0}^{\pi}{xf(\sin{x})\text{d}x} = \pi\int_{0}^{\frac{\pi}{2}}{f(\sin{x})\text{d}x} $}
\]
推导：
\(\int_{0}^{\pi}{xf(\sin{x})\text{d}x} = \int_{0}^{\pi}{(\pi - x)f(\sin{x})\text{d}x} = \pi\int_{0}^{\pi}{f(\sin{x})\text{d}x} - \int_{0}^{\pi}{xf(\sin{x})\text{d}x}\)，
移项得 \(\int_{0}^{\pi}{xf(\sin{x})\text{d}x} = \frac{\pi}{2}\int_{0}^{\pi}{f(\sin{x})\text{d}x} = \pi\int_{0}^{\frac{\pi}{2}}{f(\sin{x})\text{d}x}\)。
使用牛顿-莱布尼茨公式时要注意原函数在积分区间上是有定义的。
练习2
求 \(\int_{-1}^{1}{(\arctan{\frac{1}{x}})^{'}\text{d}x}\)。
法一：\(I = \arctan{\frac{1}{x}}\Big{|}_{-1}^{0^{-}} + \arctan{\frac{1}{x}}\Big{|}_{0^{+}}^{1} = [-\frac{\pi}{2} - (-\frac{\pi}{4})] + (\frac{\pi}{4} - \frac{\pi}{2}) = -\frac{\pi}{2}\)。（\(x = 0\) 不是可去间断点）
法二：\(I = \int_{-1}^{1}{\frac{-\frac{1}{x^{2}}}{1 + \left( \frac{1}{x} \right)^{2}}\text{d}x} = -\int_{-1}^{1}\frac{\text{d}x}{x^{2} + 1}(x \neq 0) = -\arctan{x}\Big{|}_{-1}^{1} = -\frac{\pi}{2}\)。（\(x = 0\) 是可去间断点）
求 \(\int_{0}^{2\pi}{\frac{\text{d}x}{1 + \cos^{2}{x}}}\)。
\[\begin{align}
I &= \int_{0}^{2\pi}{\frac{\text{d}(\tan{x})}{\sec^{2}{x} + 1}} \\
&= \int_{0}^{2\pi}{\frac{\text{d}(\tan{x})}{\tan^{2}{x} + 2}} \\
&= \frac{1}{\sqrt{2}}\arctan{\frac{\tan{x}}{\sqrt{2}}}\Bigg{|}_{0}^{\frac{\pi}{2}^{-}} + \frac{1}{\sqrt{2}}\arctan{\frac{\tan{x}}{\sqrt{2}}}\Bigg{|}_{\frac{\pi}{2}^{+}}^{\frac{3\pi}{2}^{-}} + \frac{1}{\sqrt{2}}\arctan{\frac{\tan{x}}{\sqrt{2}}}\Bigg{|}_{\frac{3\pi}{2}^{+}}^{2\pi} \\
&= \frac{1}{\sqrt{2}}(\frac{\pi}{2} - 0) + \frac{1}{\sqrt{2}}[\frac{\pi}{2} - (-\frac{\pi}{2})] + \frac{1}{\sqrt{2}}[0 - (-\frac{\pi}{2})] \\
&= \sqrt{2}\pi \\
\end{align}
\]
注意：\(\int_{0}^{2\pi}{\frac{\text{d}(\tan{x})}{\tan^{2}{x} + 2}} = \frac{1}{\sqrt{2}}\arctan{\frac{\tan{x}}{\sqrt{2}}}\Bigg{|}_{0}^{2\pi} = 0\) 是错误的，因为 \(x = \frac{\pi}{2}\) 和 \(x = \frac{3\pi}{2}\) 时 \(\tan{x}\) 无定义，被积函数和原函数不连续。
求 \(\int_{-2}^{2}{x\ln(1 + e^{x})\text{d}x}\)。
\[\begin{align}
I &= \int_{0}^{2}{[x\ln(1 + e^{x}) - x\ln(1 + e^{-x})]\text{d}x} \\
&= \int_{0}^{2}{x\ln{\left( \frac{e^{x} + 1}{1 + e^{-x}} \right)}\text{d}x} \\
&= \int_{0}^{2}{x^{2}\text{d}x} \\
&= \frac{1}{3}x^{3}\Big{|}_{0}^{2} \\
&= \frac{8}{3} \\
\end{align}
\]
求 \(\int_{-1}^{1}{\frac{x + 1}{1 + \sqrt[3]{x^{2}}}\text{d}x}\)。
\[\begin{align}
I &= \int_{-1}^{1}{\frac{x}{1 + \sqrt[3]{x^{2}}}\text{d}x} + \int_{-1}^{1}{\frac{1}{1 + \sqrt[3]{x^{2}}}\text{d}x} \\
&= 2\int_{0}^{1}{\frac{1}{1 + \sqrt[3]{x^{2}}}\text{d}x} \\
&\xlongequal{t = \sqrt[3]{x}} 2\int_{0}^{1}{\frac{1}{1 + t^{2}}\text{d}t^{3}} \\
&= 2\int_{0}^{1}\frac{3t^{2}}{1 + t^{2}\text{d}t} \\
&= 6\int_{0}^{1}{\left( 1 - \frac{1}{1 + t^{2}} \right)\text{d}t} \\
&= 6(t - \arctan{t})\Big{|}_{0}^{1} \\
&= 6 - \frac{3\pi}{2} \\
\end{align}
\]
求 \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{(\arctan{e^{x}})\sin^{2}{x}\text{d}x}\)。
先证：
\[\begin{align}
\colorbox{yellow}{$
\arctan{x} + \arctan{\frac{1}{x}} =
\begin{cases}
\frac{\pi}{2}, & x > 0 \\
-\frac{\pi}{2}, & x < 0 \\
\end{cases}
$}
\end{align}
\]
设 \(f(x) = \arctan{x} + \arctan{\frac{1}{x}}\)，则 \(f^{'}(x) = \frac{1}{1 + x^{2}} + \frac{1}{1 + \frac{1}{x^{2}}} \times \left( -\frac{1}{x^{2}} \right) = 0\)，\(f(x)\) 分别在 \((-\infty, 0), (0, \infty)\) 上连续，
故 \(x > 0\) 时，\(f(x) = 2\arctan{1} = \frac{\pi}{2}\)；\(x < 0\) 时，\(f(x) = 2\arctan(-1) = -\frac{\pi}{2}\)。证毕。
\[\begin{align}
I &= \int_{0}^{\frac{\pi}{2}}{\sin^{2}{x}(\arctan{e^{x}} + \arctan{e^{-x}})\text{d}x} \\
&= \frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}{\sin^{2}{x}\text{d}x} \\
&= \frac{\pi}{2} \times \frac{1!!}{2!!} \times \frac{\pi}{2} \\
&= \frac{\pi^{2}}{8} \\
\end{align}
\]
求 \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\frac{\sin^{2}{x}}{1 + e^{x}}\text{d}x}\)。
\[\begin{align}
I &= \int_{0}^{\frac{\pi}{2}}{\left( \frac{\sin^{2}{x}}{1 + e^{x}} + \frac{\sin^{2}{x}}{1 + e^{-x}} \right)\text{d}x} \\
&= \int_{0}^{\frac{\pi}{2}}{\left( \frac{\sin^{2}{x}}{1 + e^{x}} + \frac{e^{x}\sin^{2}{x}}{1 + e^{x}} \right)\text{d}x} \\
&= \int_{0}^{\frac{\pi}{2}}{\sin^{2}{x}\text{d}x} \\
&= \frac{1!!}{2!!}\times \frac{\pi}{2} \\
&= \frac{\pi}{4} \\
\end{align}
\]
求 \(\int_{0}^{\pi}{\sqrt{1 - \sin{x}}\text{d}x}\)。
\[\begin{align}
I &= \int_{0}^{\pi}{\sqrt{\sin^{2}{\frac{x}{2}} + \cos^{2}{\frac{x}{2}} - 2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}\text{d}x} \\
&= \int_{0}^{\pi}{\left| \sin{\frac{x}{2}} - \cos{\frac{x}{2}} \right| \text{d}x} \\
&= \sqrt{2}\int_{0}^{\pi}{\left| \sin(\frac{x}{2} - \frac{\pi}{4}) \right| \text{d}x} \\
&\xlongequal{t = \frac{x}{2} - \frac{\pi}{4}}2\sqrt{2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}{|\sin{t}| \text{d}t} \\
&= 4\sqrt{2}\int_{0}^{\frac{\pi}{4}}{\sin{t}\text{d}t} \\
&= 4\sqrt{2}(-\cos{t})\Big{|}_{0}^{\frac{\pi}{4}} \\
&= 4\sqrt{2} - 4 \\
\end{align}
\]
求 \(\int_{e^{-2n\pi}}^{1}{\left| \frac{\text{d}}{\text{d}x} \cos(\ln{\frac{1}{x}}) \right| \text{d}x}\)。
\[\begin{align}
I &= \int_{e^{-2n\pi}}^{1}{\left| \frac{\text{d}}{\text{d}x}\cos(\ln{x}) \right|\text{d}x} \\
&= \int_{e^{-2n\pi}}^{1}{\left| -\sin(\ln{x}) \times \frac{1}{x} \right| \text{d}x} \\
&= \int_{e^{-2n\pi}}^{1}{\left| \frac{1}{x}\sin(\ln{x}) \right|\text{d}x} \\
&= \int_{e^{-2n\pi}}^{1}{\left| \sin(\ln{x}) \right| \text{d}(\ln{x})} \\
&\xlongequal{t = \ln{x}}\int_{-2n\pi}^{0}{|\sin{t}| \text{d}t} \\
&= 2n\int_{0}^{\pi}{|\sin{t}|\text{d}t} \\
&= 2n\int_{0}^{\pi}{\sin{t}\text{d}t} \\
&= 2n(-\cos{t})\Big{|}_{0}^{\pi} \\
&= 4n \\
\end{align}
\]
求 \(\int_{0}^{\pi}{x\sqrt{\sin^{2}{x} - \sin^{4}{x}}\text{d}x}\)。
\[\begin{align}
I &= \pi\int_{0}^{\frac{\pi}{2}}{\sqrt{\sin^{2}{x} - \sin^{4}{x}}\text{d}x} \\
&= \pi\int_{0}^{\frac{\pi}{2}}{\sin{x}\cos{x}\text{d}x} \\
&= \frac{\pi}{4}\int_{0}^{\frac{\pi}{2}}{\sin{2x}\text{d}(2x)} \\
&\xlongequal{t = 2x}\frac{\pi}{4}\int_{0}^{\pi}{\sin{t}\text{d}t} \\
&= \frac{\pi}{4}(-\cos{t})\Big{|}_{0}^{\pi} \\
&= \frac{\pi}{2} \\
\end{align}
\]
区间再现
设 \(f(x)\) 为连续函数，则
\[\colorbox{yellow}{$
\int_{a}^{b}{f(x)\text{d}x} = \int_{a}^{b}{f(a + b - x)\text{d}x} = \frac{1}{2}\int_{a}^{b}{[f(x) + f(a + b - x)]\text{d}x}
$}
\]
下面是一些例题。
求 \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos^{n}{x}}{\sin^{n}{x} + \cos^{n}{x}}\text{d}x}\)。
\(I = \int_{0}^{\frac{\pi}{2}}{\frac{\cos^{n}{x}}{\sin^{n}{x} + \cos^{n}{x}}\text{d}x} = \int_{0}^{\frac{\pi}{2}}{\frac{\cos^{n}(\frac{\pi}{2} - x)}{\sin^{n}(\frac{\pi}{2} - x) + \cos^{n}(\frac{\pi}{2} - x)}\text{d}x} = \int_{0}^{\frac{\pi}{2}}{\frac{\sin^{n}{x}}{\sin^{n}{x} + \cos^{n}{x}}\text{d}x}\)，
所以 \(2I = \int_{0}^{\frac{\pi}{2}}{\frac{\cos^{n}{x}}{\sin^{n}{x} + \cos^{n}{x}}\text{d}x} + \int_{0}^{\frac{\pi}{2}}{\frac{\sin^{n}{x}}{\sin^{n}{x} + \cos^{n}{x}}\text{d}x} = \int_{0}^{\frac{\pi}{2}}{\text{d}x} = \frac{\pi}{2}\)，所以 \(I = \frac{\pi}{4}\)。
求 \(\int_{0}^{+\infty}{\frac{\text{d}x}{(1 + x^{2})(1 + x^{n})}}\)。
令 \(x = \tan{t}(0 \le t < \frac{\pi}{2})\)，则 \(I = \int_{0}^{\frac{\pi}{2}}{\frac{\text{d}(\tan{t})}{(1 + \tan^{2}{t})(1 + \tan^{n}{t})}} = \int_{0}^{\frac{\pi}{2}}{\frac{\text{d}t}{1 + \tan^{n}{t}}} = \frac{\pi}{4}\)。
求 \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}{\frac{\cos^{2}{x}}{x(\pi - 2x)}\text{d}x}\)。
记 \(f(x) = \frac{\cos^{2}{x}}{x(\pi - 2x)}\)，则 \(f(\frac{\pi}{2} - x) = \frac{\cos^{2}(\frac{\pi}{2} - x)}{(\frac{\pi}{2} - x)[\pi - 2(\frac{\pi}{2} - x)]} = \frac{\sin^{2}{x}}{x(\pi - 2x)}\)，\(f(x) + f(\frac{\pi}{2} - x) = \frac{1}{x(\pi - 2x)}\)。
\[\begin{align}
I &= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}{f(x)\text{d}x} \\
&= \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}{[f(x) + f(\frac{\pi}{6} + \frac{\pi}{3} - x)]\text{d}x} \\
&= \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\text{d}x}{x(\pi - 2x)} \\
&= \frac{1}{4}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\text{d}x}{x(\frac{\pi}{2} - x)} \\
&= \frac{1}{4} \times \frac{2}{\pi}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}{\left( \frac{1}{x} - \frac{1}{x - \frac{\pi}{2}} \right)\text{d}x} \\
&= \frac{1}{2\pi}\left[ \ln|x| - \ln\left|x - \frac{\pi}{2}\right| \right]\Bigg{|}_{\frac{\pi}{6}}^{\frac{\pi}{3}} \\
&= \frac{\ln{2}}{\pi} \\
\end{align}
\]
求 \(\int_{0}^{\frac{\pi}{4}}{\frac{x}{\cos{x}\cos(\frac{\pi}{4} - x)}\text{d}x}\)。
\[\begin{align}
I &= \frac{1}{2}\int_{0}^{\frac{\pi}{4}}{\left(\frac{x}{\cos{x}\cos(\frac{\pi}{4} - x)} + \frac{\frac{\pi}{4} - x}{\cos(\frac{\pi}{4} - x)\cos{x}}\right)\text{d}x} \\
&= \frac{\pi}{8}\int_{0}^{\frac{\pi}{4}}{\frac{1}{\cos{x}\cos(\frac{\pi}{4} - x)}\text{d}x} \\
&= \frac{\pi}{8} \times \frac{1}{\sin{\frac{\pi}{4}}} \times \int_{0}^{\frac{\pi}{4}}{\frac{\sin[x + (\frac{\pi}{4} - x)]}{\cos{x}\cos(\frac{\pi}{4} - x)}\text{d}x} \\
&= \frac{\sqrt{2}\pi}{8}\int_{0}^{\frac{\pi}{4}}{\frac{\sin{x}\cos(\frac{\pi}{4} - x) + \sin(\frac{\pi}{4} - x)\cos{x}}{\cos{x}\cos(\frac{\pi}{4} - x)}\text{d}x} \\
&= \frac{\sqrt{2}\pi}{8}\int_{0}^{\frac{\pi}{4}}{\left( \frac{\sin{x}}{\cos{x}} + \frac{\sin(\frac{\pi}{4} - x)}{\cos(\frac{\pi}{4} - x)} \right)\text{d}x} \\
&= \frac{\sqrt{2}\pi}{8}\left( -\int_{0}^{\frac{\pi}{4}}{\frac{\text{d}\cos{x}}{\cos{x}}} + \int_{0}^{\frac{\pi}{4}}{\frac{\text{d}\cos(\frac{\pi}{4} - x)}{\cos(\frac{\pi}{4} - x)}} \right) \\
&= \frac{\sqrt{2}\pi}{8}\ln{\left| \frac{\cos(\frac{\pi}{4} - x)}{\cos{x}} \right|}\Bigg{|}_{0}^{\frac{\pi}{4}} \\
&= \frac{\sqrt{2}\pi\ln{2}}{8} \\
\end{align}
\]
求 \(\int_{0}^{\frac{\pi}{2}}{\ln{\sin{x}}\text{d}x}\)。
求 \(\int_{0}^{\frac{\pi}{2}}{\frac{x}{\tan{x}}\text{d}x}\)。
求 \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}\ln{\cos{x}}}{1 + \sin{x} + \cos{x}}\text{d}x}\)。
\[\begin{align}
I_{1} &= \frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\left( \ln{\sin{x}} + \ln{\sin(\frac{\pi}{2} - x)} \right) \text{d}x} \\
&= \frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\ln(\sin{x}\cos{x})\text{d}x} \\
&= \frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\ln{\frac{\sin{2x}}{2}}\text{d}x} \\
&= \frac{1}{4}\int_{0}^{\frac{\pi}{2}}{\ln\sin{2x}\text{d}(2x)} - \frac{\ln{2}}{2}x\Bigg{|}_{0}^{\frac{\pi}{2}} \\
&\xlongequal{t = 2x} \frac{1}{4}\int_{0}^{\pi}{\ln{\sin{t}}\text{d}t} - \frac{\pi\ln{2}}{4} \\
&= \frac{1}{2}I_{1} - \frac{\pi\ln{2}}{4} \\
\end{align}
\]
所以 \(\frac{1}{2}I_{1} = -\frac{\pi\ln{2}}{4}\)，得 \(I_{1} = -\frac{\pi\ln{2}}{2}\)。
\[\begin{align}
I_{2} &= \int_{0}^{\frac{\pi}{2}}{\frac{x\cos{x}}{\sin{x}}\text{d}x} \\
&= \int_{0}^{\frac{\pi}{2}}{\frac{x}{\sin{x}}\text{d}\sin{x}} \\
&= \int_{0}^{\frac{\pi}{2}}{x\text{d}(\ln{\sin{x}})} \\
&= x\ln{\sin{x}}\Big{|}_{0^{+}}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}}{\ln{\sin{x}}\text{d}x} \\
&= -I_{1} = \frac{\pi\ln{2}}{2} \\
\end{align}
\]
\[\begin{align}
I_{3} &= \int_{0}^{\frac{\pi}{2}}{\left[ \frac{\cos{x}\ln{\cos{x}}}{1 + \sin{x} + \cos{x}} + \frac{\cos(-x)\ln\cos(-x)}{1 + \sin(-x) + \cos(-x)} \right] \text{d}x} \\
&= \int_{0}^{\frac{\pi}{2}}{\cos{x}\ln{\cos{x}}\left( \frac{1}{1 + \sin{x} + \cos{x}} + \frac{1}{1 - \sin{x} + \cos{x}} \right) \text{d}x} \\
&= \int_{0}^{\frac{\pi}{2}}{\frac{(2 + 2\cos{x})\cos{x}\ln{\cos{x}}}{(1 + \cos{x})^{2} - \sin^{2}{x}} \text{d}x} \\
&= \int_{0}^{\frac{\pi}{2}}{\ln\cos{x}\text{d}x} \\
&= \int_{0}^{\frac{\pi}{2}}{\ln\cos(\frac{\pi}{2} - x)\text{d}x} \\
&= I_{1} = -\frac{\pi\ln{2}}{2} \\
\end{align}
\]
求 \(\int_{0}^{1}{\frac{\ln(1 + x)}{1 + x^{2}}\text{d}x}\)。
求 \(\int_{0}^{1}{\frac{\arctan{x}}{x + 1}\text{d}x}\)。
令 \(x = \tan{t}(0 \le t \le \frac{\pi}{4})\)，则
\[\begin{align}
I_{1} &= \int_{0}^{\frac{\pi}{4}}{\frac{\ln(1 + \tan{t})}{\sec^{2}{t}}\text{d}(\tan{t})} \\
&= \int_{0}^{\frac{\pi}{4}}{\ln(1 + \tan{t})\text{d}t} \\
&= \frac{1}{2}\int_{0}^{\frac{\pi}{4}}{\left[ \ln(1 + \tan{t}) + \ln(1 + \tan(\frac{\pi}{4} - t)) \right]\text{d}t} \\
&= \frac{1}{2}\int_{0}^{\frac{\pi}{4}}{\ln{2}\text{d}t} \\
&= \frac{\ln{2}}{2}x\Bigg{|}_{0}^{\frac{\pi}{4}} \\
&= \frac{\pi\ln2}{8} \\
\end{align}
\]
\[\begin{align}
I_{2} &= \int_{0}^{1}{\arctan{x}\text{d}\ln(1 + x)} \\
&= \ln(1 + x)\arctan{x}\Big{|}_{0}^{1} - \int_{0}^{1}{\ln(1 + x)\text{d}(\arctan{x})} \\
&= \frac{\pi\ln{2}}{4} - I_{1} = \frac{\pi\ln{2}}{8} \\
\end{align}
\]
求 \(\int_{0}^{n\pi}{x|\sin{x}|}\text{d}x\)，其中 \(n\) 为正整数。
\[\begin{align}
I &= \int_{0}^{n\pi}{(n\pi - x)|\sin{x}|\text{d}x} \\
&= n\pi\int_{0}^{n\pi}{|\sin{x}|\text{d}x} - I \\
&= \frac{n\pi}{2}\int_{0}^{n\pi}{|\sin{x}|\text{d}x} \\
&= \frac{n^{2}\pi}{2}\int_{0}^{\pi}{\sin{x}\text{d}x} \\
&= n^{2}\pi\int_{0}^{\frac{\pi}{2}}{\sin{x}\text{d}x} \\
&= n^{2}\pi(-\cos{x})\Big{|}_{0}^{\frac{\pi}{2}} \\
&= n^{2}\pi \\
\end{align}
\]
求 \(\int_{0}^{1}{\frac{\arcsin{\sqrt{x}}}{\sqrt{x^{2} - x + 1}}\text{d}x}\)。
先证：
\[\colorbox{yellow}{$\arcsin{\sqrt{x}} + \arcsin{\sqrt{1 - x}} = \frac{\pi}{2}(0 \le x \le 1)$}
\]
设 \(f(x) = \arcsin{\sqrt{x}} + \arcsin{\sqrt{1 - x}}(0 \le x \le 1)\)，则 \(f^{'}(x) = \frac{1}{\sqrt{1 - x}} \times \frac{1}{2\sqrt{x}} + \frac{1}{\sqrt{1 - (1 - x)}} \times (-\frac{1}{2\sqrt{1 - x}}) = 0\)。
又因为 \(f(x)\) 在 \([0, 1]\) 上连续，所以 \(f(x) = f(0) = \frac{\pi}{2}\)。
\[\begin{align}
I &= \frac{1}{2}\int_{0}^{1}{\left( \frac{\arcsin{\sqrt{x}}}{\sqrt{x^{2} - x + 1}} + \frac{\arcsin{\sqrt{1 - x}}}{\sqrt{(1 - x)^{2} - (1 - x) + 1}} \right)\text{d}x} \\
&= \frac{1}{2}\int_{0}^{1}{\frac{\arcsin{\sqrt{x}} + \arcsin{\sqrt{1 - x}}}{\sqrt{x^{2} - x + 1}}\text{d}x} \\
&= \frac{\pi}{4}\int_{0}^{1}{\frac{\text{d}x}{\sqrt{x^{2} - x + 1}}} \\
&= \frac{\pi}{4}\int_{0}^{1}\frac{\text{d}x}{\sqrt{\left( x - \frac{1}{2} \right)^{2} + \left( \frac{\sqrt{3}}{2} \right)^{2}}} \\
&= \frac{\pi}{4} \times \ln\left( x - \frac{1}{2} + \sqrt{\left( x - \frac{1}{2} \right)^{2} + \left( \frac{\sqrt{3}}{2} \right)^{2}} \right) \Bigg{|}_{0}^{1} \\
&= \frac{\pi\ln{3}}{4} \\
\end{align}
\]
求 \(\int_{0}^{1}{x\arcsin{(2\sqrt{x - x^{2}})}\text{d}x}\)。
\[\begin{align}
I &= \frac{1}{2}\int_{0}^{1}{\left( x\arcsin(2\sqrt{x - x^{2}}) + (1 - x)\arcsin(2\sqrt{(1 - x) - (1 - x)^{2}}) \right)\text{d}x} \\
&= \frac{1}{2}\int_{0}^{1}{\arcsin(2\sqrt{x - x^{2}})\text{d}x} \\
&= \frac{1}{2}\int_{0}^{1}{\arcsin\left( 2\sqrt{\frac{1}{4} - \left(x - \frac{1}{2} \right)^{2}} \right) \text{d}x} \\
&\xlongequal{x - \frac{1}{2} = \frac{1}{2}\cos{t}} \frac{1}{2}\int_{\pi}^{0}{\arcsin(\sin{t})\text{d}(\frac{1}{2}\cos{t} + \frac{1}{2})} \\
&= \frac{1}{4}\int_{0}^{\pi}{\sin{t}\arcsin(\sin{t})\text{d}t} \\
&= \frac{1}{4}\int_{0}^{\frac{\pi}{2}}{t\sin{t}\text{d}t} + \frac{1}{4}\int_{\frac{\pi}{2}}^{\pi}{(\pi - t)\sin{t}\text{d}t} \\
&= \frac{1}{2}\int_{0}^{\frac{\pi}{2}}{t\sin{t}\text{d}t} \\
&= -\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{t\text{d}\cos{t}} \\
&= -\frac{1}{2}t\cos{t}\Big{|}_{0}^{\frac{\pi}{2}} + \frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{t}\text{d}t} \\
&= 0 + \frac{1}{2}\sin{t}\Big{|}_{0}^{\frac{\pi}{2}} \\
&= \frac{1}{2} \\
\end{align}
\]