第六次作业参考答案

9.1 第六次作业参考答案

9.1.1 习题6.2

例 9.1.1 （习题6.2.1）计算以下不定积分：

(1): $\int x \sqrt{x \sqrt{x}} d x$
(2): $\int \frac{1}{x^{2}} \cos \frac{1}{x} d x$
(3): $\int \frac{1}{e^{x} + e^{- x}} d x$
(4): $\int \sqrt{1 + \cos 2 x} d x$
(5): $\int \frac{d x}{\sqrt{1 + x^{2}}}$

解 (1) $\begin{matrix} (1) & \int x \sqrt{x \sqrt{x}} d x = \int x^{7 / 4} d x = \frac{4}{11} x^{11 / 4} + C \end{matrix}$

(2) $\begin{matrix} (2) & \int \frac{1}{x^{2}} \cos \frac{1}{x} d x = - \int \cos \frac{1}{x} d (\frac{1}{x}) = - \sin \frac{1}{x} + C \end{matrix}$

(3) $\begin{matrix} (3) & \int \frac{1}{e^{x} + e^{- x}} d x = \int \frac{d (e^{x})}{e^{2 x} + 1} = \arctan (e^{x}) + C \end{matrix}$

(4) 设 $x \in [- \frac{π}{2}, \frac{π}{2}]$ ，则 $\begin{matrix} (4) & \int \sqrt{1 + \cos 2 x} d x = \int \sqrt{2} \cos x d x = \sqrt{2} \sin x + C \end{matrix}$

(5) $\begin{matrix} (5) & \int \frac{d x}{\sqrt{1 + x^{2}}} \overset{x = \tan t}{=} \int \sec t d t = \ln | \sec t + \tan t | + C = \ln (x + \sqrt{1 + x^{2}}) + C \end{matrix}$ $◻$

例 9.1.2 （习题6.2.2）计算以下不定积分，其中 $n$ 是正整数， $a, b, α, β, λ$ 都是非零实数：

(1): $\int e^{λ x} \sin x d x$
(2): $\int \sin α x \cos β x d x$
(3): $\int \tan x d x$
(4): $\int \frac{\sin x d x}{a \sin x + b \cos x}$ 、 $\int \frac{\cos x d x}{a \sin x + b \cos x}$
(5): $\int x^{4} \ln^{2} x d x$
(6): $\int \sin (\ln x) d x$

解 (1) 设 $[e^{λ x} (A \sin x + B \cos x)]^{'} = e^{λ x} \sin x$ ，则 $\begin{matrix} (6) & e^{λ x} \sin x = e^{λ x} [(λ A - B) \sin x + (A + λ B) \cos x] ⟹ A = \frac{λ}{λ^{2} + 1}, B = - \frac{1}{λ^{2} + 1} \end{matrix}$ 因此 $\begin{matrix} (7) & \int e^{λ x} \sin x d x = \frac{e^{λ x} (λ \sin x - \cos x)}{λ^{2} + 1} + C \end{matrix}$

(2) 设 $α^{2} \neq β^{2}$ ，则 $\begin{matrix} (8) & \begin{aligned} \int \sin α x \cos β x d x & = \frac{1}{2} \int [\sin (α + β) x + \sin (α - β) x] d x \\ = - \frac{\cos (α + β) x}{2 (α + β)} - \frac{\cos (α - β) x}{2 (α - β)} + C \end{aligned} \end{matrix}$

(3) $\begin{matrix} (9) & \int \tan x d x = - \int \frac{d (\cos x)}{\cos x} = - \ln | \cos x | + C \end{matrix}$

(4) 设第一、二个被积函数的原函数分别为 $F_{1}, F_{2}$ ，则 $\begin{matrix} (10) & \begin{aligned} a F_{1} + b F_{2} & = \int d x = x + C_{1}, \\ - b F_{1} + a F_{2} & = \int \frac{a \cos x - b \sin x}{a \sin x + b \cos x} d x = \ln | a \sin x + b \cos x | + C_{2} \end{aligned} \end{matrix}$ 因此 $\begin{matrix} (11) & \begin{aligned} \int \frac{\sin x d x}{a \sin x + b \cos x} & = \frac{a x - b \ln | a \sin x + b \cos x |}{a^{2} + b^{2}} + C \\ \int \frac{\cos x d x}{a \sin x + b \cos x} & = \frac{b x + a \ln | a \sin x + b \cos x |}{a^{2} + b^{2}} + C \end{aligned} \end{matrix}$

(5) $\begin{matrix} (12) & \int x^{4} \ln^{2} x d x = \frac{1}{5} x^{5} \ln^{2} x - \frac{2}{5} \int x^{4} \ln x d x = \frac{1}{5} x^{5} \ln^{2} x - \frac{2}{25} x^{5} \ln x + \frac{2}{125} x^{5} + C \end{matrix}$

(6) $\begin{matrix} (13) & \int \sin (\ln x) d x \overset{x = e^{t}}{=} \int e^{t} \sin t d t = \frac{1}{2} x (\sin (\ln x) - \cos (\ln x)) + C \end{matrix}$ $◻$

9.1.2 习题6.3

例 9.1.3 （习题6.3.4）计算以下不定积分：

(1): $\int \frac{d x}{x^{2} (1 + x^{2})^{2}} d x$
(2): $\int \frac{x^{3}}{(1 - x^{2})^{3}} d x$
(3): $\int \sqrt[3]{\frac{x + 3}{1 - x}} d x$
(4): $\int \sqrt{1 + x^{2}} d x$
(5): $\int \frac{1}{\sqrt{x} + \sqrt[3]{x}} d x$
(6): $\int \frac{x^{2} - x + 1}{\sqrt{x^{2} + 2 x + 2}} d x$
(7): $\int x^{3} \sqrt{1 - x^{2}} d x$
(8): $\int \frac{d x}{\cos^{3} x} d x$
(9): $\int \frac{\sin^{5} x}{\cos^{4} x} d x$
(10): $\int \frac{d x}{x \sqrt[3]{1 + x^{5}}}$
(11): $\int \frac{d x}{3 + 2 \sin x}$
(12): $\int \frac{d x}{\cos^{4} x}$
(13): $\int \frac{d x}{\sqrt[4]{1 + x^{4}}}$
(14): $\int \frac{\sqrt[3]{1 + \sqrt[4]{x}}}{\sqrt{x}} d x$

解 (1) 设 $\begin{matrix} (14) & \frac{1}{x^{2} (1 + x^{2})^{2}} = \frac{A}{x^{2}} + \frac{E}{1 + x^{2}} + \frac{F}{(1 + x^{2})^{2}} ⟹ 1 = A (1 + x^{2})^{2} + E x^{2} (1 + x^{2}) + F x^{2} \end{matrix}$ 令 $x = 0$ 可得 $A = 1$ ，令 $x = i$ 可得 $F = - 1$ 。等式两边同时求导并代入 $x = i$ 可得 $\begin{matrix} (15) & 0 = 0 + 0 + 2 E x^{3} + 2 F x ⟹ E = F = - 1 \end{matrix}$ 因此 $\begin{matrix} (16) & \int \frac{d x}{x^{2} (1 + x^{2})^{2}} = - \frac{1}{x} - \frac{x}{2 (1 + x^{2})} - \frac{3}{2} \arctan x + C \end{matrix}$ 其中 $\begin{matrix} (17) & \begin{aligned} I_{n} := & \int \frac{d x}{(1 + x^{2})^{n}} = \frac{x}{(1 + x^{2})^{n}} + 2 n \int \frac{x^{2} d x}{(1 + x^{2})^{n + 1}} = \frac{x}{(1 + x^{2})^{n}} + 2 n (I_{n} - I_{n + 1}) \\ ⟹ I_{n + 1} = \frac{x}{2 n (1 + x^{2})^{n}} + \frac{2 n - 1}{2 n} I_{n} \end{aligned} \end{matrix}$

(2) 设 $\begin{matrix} (18) & \frac{x^{3}}{(1 - x^{2})^{3}} = \frac{A x}{1 - x^{2}} + \frac{B x}{(1 - x^{2})^{2}} + \frac{D x}{(1 - x^{2})^{3}} ⟹ x^{2} = A (1 - x^{2})^{2} + B (1 - x^{2}) + D \end{matrix}$ 令 $x = 1$ 可得 $D = 1$ 。等式两边同时求导可得 $\begin{matrix} (19) & 2 x = - 4 A x (1 - x^{2}) - 2 B x ⟹ A = 0, B = - 1 \end{matrix}$ 因此 $\begin{matrix} (20) & \int \frac{x^{3}}{(1 - x^{2})^{3}} d x = - \frac{1}{2 (1 - x^{2})} + \frac{1}{4 (1 - x^{2})^{2}} + x + C \end{matrix}$

(3) 令 $t = \sqrt[3]{\frac{x + 3}{1 - x}}$ ，则 $x = \frac{t^{3} - 3}{t^{3} + 1}$ ， $d x = \frac{12 t^{2}}{(1 + t^{3})^{2}} d t$ ，此时 $\begin{matrix} (21) & \begin{aligned} \int \sqrt[3]{\frac{x + 3}{1 - x}} d x = 12 \int \frac{t^{3} d t}{(1 + t^{3})^{2}} = - \frac{4 t}{1 + t^{3}} + \frac{4 \sqrt{3}}{3} \arctan \frac{2 t - 1}{\sqrt{3}} + \frac{2}{3} \ln \frac{(t + 1)^{2}}{t^{2} - t + 1} + C \\ = (x - 1) \sqrt[3]{\frac{x + 3}{1 - x}} + \frac{4 \sqrt{3}}{3} \arctan \frac{2 \sqrt[3]{\frac{x + 3}{1 - x}} - 1}{\sqrt{3}} + 2 \ln (1 + \sqrt[3]{\frac{x + 3}{1 - x}}) + \frac{2}{3} \ln (1 - x) + C \end{aligned} \end{matrix}$

(4) $\begin{matrix} (22) & \int \sqrt{1 + x^{2}} d x \overset{x = \tan t}{=} \int \sec^{3} t d t = \frac{1}{2} [x \sqrt{1 + x^{2}} + \ln (x + \sqrt{1 + x^{2}})] + C \end{matrix}$

(5) $\begin{matrix} (23) & \int \frac{1}{\sqrt{x} + \sqrt[3]{x}} d x \overset{x = t^{6}}{=} \int \frac{6 t^{5}}{t^{3} + t^{2}} d t = 6 \sqrt[6]{x} - 3 \sqrt[3]{x} + 2 \sqrt{x} - 6 \ln (1 + \sqrt[6]{x}) + C \end{matrix}$

(6) 令 $t = x + 1$ ，则 $\begin{matrix} (24) & \begin{aligned} \int \frac{x^{2} - x + 1}{\sqrt{x^{2} + 2 x + 2}} d x = \int \frac{t^{2} + 1 - 3 t + 2}{\sqrt{t^{2} + 1}} d t = \int \sqrt{1 + t^{2}} d t + \int \frac{2 d t}{\sqrt{1 + t^{2}}} - 3 \sqrt{1 + t^{2}} \\ = \frac{1}{2} [t \sqrt{1 + t^{2}} + \ln (t + \sqrt{1 + t^{2}})] + 2 \ln (t + \sqrt{1 + t^{2}}) - 3 \sqrt{1 + t^{2}} + C \\ = \frac{1}{2} (x + 1) \sqrt{x^{2} + 2 x + 2} + \frac{5}{2} \ln (x + 1 + \sqrt{x^{2} + 2 x + 2}) - 3 \sqrt{x^{2} + 2 x + 2} + C \end{aligned} \end{matrix}$

(7) 令 $x = \sin t$ 、 $u = \cos t = \sqrt{1 - x^{2}}$ ，则 $\begin{matrix} (25) & \begin{aligned} \int x^{3} \sqrt{1 - x^{2}} d x & = \int \sin^{3} t \cos^{2} t d t = \int (u^{2} - 1) u^{2} d u = \frac{1}{5} {(1 - x^{2})}^{5 / 2} - \frac{1}{3} {(1 - x^{2})}^{3 / 2} + C \end{aligned} \end{matrix}$

(8) $\begin{matrix} (26) & \begin{aligned} \int \sec^{3} x d x & = \int \sec x d (\tan x) = \sec x \tan x - \int \sec x \tan^{2} x d x \\ = \sec x \tan x + \int \sec x d x - \int \sec^{3} x d x \\ = \frac{1}{2} (\sec x \tan x + \ln | \sec x + \tan x |) + C \end{aligned} \end{matrix}$

(9) 令 $t = \cos x$ ，则 $\begin{matrix} (27) & \int \frac{\sin^{5} x}{\cos^{4} x} d x = - \int \frac{(1 - t^{2})^{2} d t}{t^{4}} = \frac{1}{3 \cos^{3} x} - \frac{2}{\cos x} - \cos x + C \end{matrix}$

(10) 令 $t = \sqrt[3]{1 + x^{5}}$ ，则 $\frac{d x}{x} = \frac{5 x^{4} d x}{5 x^{5}} = \frac{3 t^{2} d t}{5 (t^{3} - 1)}$ $\begin{matrix} (28) & \int \frac{d x}{x \sqrt[3]{1 + x^{5}}} = \frac{3}{5} \int \frac{t d t}{t^{3} - 1} = \frac{\sqrt{3}}{5} \arctan \frac{1 + 2 \sqrt[3]{1 + x^{5}}}{\sqrt{3}} + \frac{3}{10} \ln | \sqrt[3]{1 + x^{5}} - 1 | - \frac{1}{2} \ln | x | + C \end{matrix}$

(11) 利用万能公式可得 $\begin{matrix} (29) & \int \frac{d x}{3 + 2 \sin x} = \frac{2}{\sqrt{5}} \arctan \frac{2 + 3 \tan \frac{x}{2}}{\sqrt{5}} + C \end{matrix}$

(12) $\begin{matrix} (30) & \int \sec^{4} x d x = \int (1 + \tan^{2} x) d (\tan x) = \tan x + \frac{1}{3} \tan^{3} x + C \end{matrix}$

(13) 令 $t = x^{- 4}$ 、 $u = \sqrt[4]{1 + t}$ ，则 $\begin{matrix} (31) & \int \frac{d x}{\sqrt[4]{1 + x^{4}}} = - \frac{1}{4} \int \frac{d t}{t \sqrt[4]{1 + t}} = - \int \frac{u^{2} d u}{u^{4} - 1} = \frac{1}{2} \arctan \frac{x}{\sqrt[4]{1 + x^{4}}} + \frac{1}{4} \ln \frac{x + \sqrt[4]{1 + x^{4}}}{x - \sqrt[4]{1 + x^{4}}} + C \end{matrix}$

(14) 令 $t = \sqrt[3]{1 + \sqrt[4]{x}}$ ，则 $\begin{matrix} (32) & \int \frac{\sqrt[3]{1 + \sqrt[4]{x}}}{\sqrt{x}} d x = 12 \int (t^{3} - 1) t^{3} d t = \frac{12}{7} {(1 + \sqrt[4]{x})}^{7 / 3} - 3 {(1 + \sqrt[4]{x})}^{4 / 3} + C \end{matrix}$ $◻$

9.1.3 习题7.1

例 9.1.4 （习题7.1.5）设 $g \in R [a, b]$ 满足 $\int_{a}^{b} g (x) d x > 0$ ，证明： $\exists ξ \in (a, b)$ 使得 $g$ 在 $ξ$ 处连续且 $g (ξ) > 0$ 。

证明采用反证法¹。假设 $\forall ξ \in (a, b)$ ，若 $g$ 在 $ξ$ 处连续则 $g (ξ) \leq 0$ 。由于 $g \in R [a, b]$ ，故

$g$ 有界，即 $\exists M > 0$ 使得 $| g (x) | \leq M$ 。
$g$ 在 $[a, b]$ 上的间断点集为零测集，即 $\forall δ > 0$ ，存在总长度为 $δ$ 的有限开区间集 $A$ 覆盖间断点集。

故 $\forall ε > 0$ ，取 $δ = \frac{ε}{M}$ ，则 $\begin{matrix} (33) & \int_{a}^{b} g (x) d x = \int_{A} g (x) d x + \int_{[a, b] ∖ A} g (x) d x \leq \int_{A} M d x + \int_{[a, b] ∖ A} 0 d x \leq M δ = ε \end{matrix}$ 令 $ε \to 0^{+}$ 可得 $\begin{matrix} (34) & \int_{a}^{b} g (x) d x \leq 0 \end{matrix}$ 与题设矛盾！故假设不成立，即 $\exists ξ \in (a, b)$ 使得 $g$ 在 $ξ$ 处连续且 $g (ξ) > 0$ 。 $◻$

另证仍然采用反证法，但是绕过Lebesgue法则。根据例9.3.15， $f$ 的连续点集在 $[a, b]$ 上稠密，故对 $[a, b]$ 的任意划分，在每一个子区间中总能取到连续点作为标志点。由于 $f$ 在连续点处非正，因此Riemann和非正，作为Riemann和极限的定积分也必定非正，与题设矛盾。 $◻$

例 9.1.5 （习题7.1.6）设 $f \in R [0, \frac{π}{2}]$ ，证明： $\begin{matrix} (35) & lim_{n \to + \infty} \int_{0}^{π / 2} f (x) \sin^{n} x d x = 0 \end{matrix}$

证明由 $f \in R [a, b]$ 知 $f$ 有界，故 $\exists M > 0$ 使得 $| f (x) | \leq M$ 。 $\forall ε > 0$ ，注意到 $\begin{matrix} (36) & \int_{0}^{π / 2} \sin^{n} x d x \leq \int_{0}^{(π - ε) / 2} \sin^{n} \frac{π - ε}{2} d x + \int_{(π - ε) / 2}^{π / 2} d x < \frac{π}{2} \cos^{n} \frac{ε}{2} + \frac{ε}{2} \overset{?}{\leq} ε \end{matrix}$ 取 $n$ 满足下式即可 $\begin{matrix} (37) & n \geq \frac{\ln (ε / π)}{\ln [\cos (ε / 2)]} \end{matrix}$ 此时 $\begin{matrix} (38) & | \int_{0}^{π / 2} f (x) \sin^{n} x d x | \leq M \int_{0}^{π / 2} \sin^{n} x d x < M ε ⟹ lim_{n \to + \infty} \int_{0}^{π / 2} f (x) \sin^{n} x d x = 0 \end{matrix}$ $◻$

另证实际上，由Wallis公式可得 $\begin{matrix} (39) & \int_{0}^{π / 2} \sin^{n} x d x = {\begin{cases} \frac{π}{2} \frac{(n - 1)!!}{n!!}, & n \in even \\ \frac{(n - 1)!!}{n!!}, & n \in odd \end{cases} \end{matrix}$ 令 $a_{n} = \frac{(n - 1)!!}{n!!}$ ，则 $\begin{matrix} (40) & a_{n} a_{n + 1} = \frac{1}{n + 1} \to 0, a_{2 n} = \exp \sum_{k = 1}^{n} \ln (1 - \frac{1}{2 k}) \leq \exp \sum_{k = 1}^{n} - \frac{1}{2 k} \to 0 \end{matrix}$ 亦即 $lim_{n \to + \infty} a_{n} = 0$ 。因此 $\begin{matrix} (41) & | \int_{0}^{π / 2} f (x) \sin^{n} x d x | \leq M \int_{0}^{π / 2} \sin^{n} x d x \leq \frac{M π}{2} \frac{(n - 1)!!}{n!!} \to 0 \end{matrix}$ $◻$

例 9.1.6 （习题7.1.8）设 $f : [a, b] \to [α, β]$ 严格单调， $g = f^{- 1}$ 。讨论 $\int_{a}^{b} f (x) d x$ 和 $\int_{α}^{β} g (y) d y$ 之间的关系，并证明你的结论。

证明本题与9.3.3类似，画图可知结论为

$f$ 严格增： $\int_{a}^{b} f (x) d x + \int_{α}^{β} g (y) d y = b f (b) - a f (a) = b β - a α$ 。
$f$ 严格减： $\int_{a}^{b} f (x) d x - \int_{α}^{β} g (y) d y = b f (b) - a f (a) = b α - a β$ 。

换元法需要 $f$ 可微；若 $f$ 仅仅可积，我们需要通过定义来证明。

对 $[a, b]$ 的任意划分 $P : a = x_{0} < x_{1} < \dots < x_{n} = b$ 。若 $f$ 严格增，则 $f (P) : α = f (x_{0}) < f (x_{1}) < \dots < f (x_{n}) = β$ 亦是 $[α, β]$ 的划分，因此两者的Riemann和满足 $\begin{matrix} (42) & \begin{aligned} S (f, P, {x_{k}}_{k = 1}^{n}) + S (f^{- 1}, f (P), {f (x_{k})}_{k = 0}^{n - 1}) \\ = \sum_{k = 1}^{n} [f (x_{k}) (x_{k} - x_{k - 1}) + f^{- 1} (f (x_{k - 1})) (f (x_{k}) - f (x_{k - 1}))] \\ = \sum_{k = 1}^{n} [x_{k} f (x_{k}) - x_{k - 1} f (x_{k - 1})] = b f (b) - a f (a) \end{aligned} \end{matrix}$ 令 $∥ P ∥ \to 0$ ，即得 $\begin{matrix} (43) & \int_{a}^{b} f (x) d x + \int_{α}^{β} g (y) d y = b f (b) - a f (a) = b β - a α \end{matrix}$

若 $f$ 严格减，则 $f (P) : α = f (x_{0}) > f (x_{1}) > \dots > f (x_{n}) = β$ 亦是 $[α, β]$ 的划分，因此两者的Riemann和满足 $\begin{matrix} (44) & \begin{aligned} S (f, P, {x_{k}}_{k = 1}^{n}) - S (f^{- 1}, f (P), {f (x_{k})}_{k = 0}^{n - 1}) \\ = \sum_{k = 1}^{n} [f (x_{k}) (x_{k} - x_{k - 1}) - f^{- 1} (f (x_{k - 1})) (f (x_{k - 1}) - f (x_{k}))] \\ = \sum_{k = 1}^{n} [x_{k} f (x_{k}) - x_{k - 1} f (x_{k - 1})] = b f (b) - a f (a) \end{aligned} \end{matrix}$ 令 $∥ P ∥ \to 0$ ，即得 $\begin{matrix} (45) & \int_{a}^{b} f (x) d x - \int_{α}^{β} g (y) d y = b f (b) - a f (a) = b α - a β \end{matrix}$ $◻$

9.1.4 习题7.4

例 9.1.7 （习题7.4.1）计算以下定积分：

(1): $\int_{- π}^{π} \sin m x \sin n x d x$ ，其中 $m, n \in N^{*}$ 。
(2): $\int_{0}^{π / 2} \frac{\sin (2 m - 1) x}{\sin x} d x$ ，其中 $m \in N^{*}$ 。

解 (1) 当 $m = n$ 时，有 $\begin{matrix} (46) & \int_{- π}^{π} \sin^{2} n x d x = \int_{- π}^{π} \frac{1 - \cos 2 n x}{2} d x = π \end{matrix}$ 当 $m \neq n$ 时，有 $\begin{matrix} (47) & \begin{aligned} \int_{- π}^{π} \sin m x \sin n x d x & = \frac{1}{2} \int_{- π}^{π} [\cos (m - n) x - \cos (m + n) x] d x \\ = \frac{1}{2} {[\frac{\sin (m - n) x}{m - n} - \frac{\sin (m + n) x}{m + n}]}_{- π}^{π} = 0 \end{aligned} \end{matrix}$ 因此 $\begin{matrix} (48) & \int_{- π}^{π} \sin m x \sin n x d x = π δ_{m n} \end{matrix}$

(2) 注意到 $\begin{matrix} (49) & \sin (2 m - 1) x - \sin x = \sum_{k = 1}^{m - 1} [\sin (2 k + 1) x - \sin (2 k - 1) x] = 2 \sin x \sum_{k = 1}^{m - 1} \cos 2 k x \end{matrix}$ 因此 $\begin{matrix} (50) & \int_{0}^{π / 2} \frac{\sin (2 m - 1) x}{\sin x} d x = \int_{0}^{π / 2} (1 + 2 \sum_{k = 1}^{m - 1} \cos 2 k x) d x = \frac{π}{2} \end{matrix}$ $◻$

补充同时注意到 $\begin{matrix} (51) & \sin 2 m x = \sum_{k = 0}^{m - 1} [\sin (2 k + 2) x - \sin (2 k) x] = 2 \sin x \sum_{k = 0}^{m - 1} \cos (2 k + 1) x \end{matrix}$ 因此 $\begin{matrix} (52) & \int_{0}^{π / 2} \frac{\sin 2 m x}{\sin x} d x = 2 \sum_{k = 0}^{m - 1} \int_{0}^{π / 2} \cos (2 k + 1) x d x = 2 \sum_{k = 0}^{m - 1} \frac{(- 1)^{k}}{2 k + 1} \end{matrix}$

例 9.1.8 （习题7.4.2）计算以下定积分：

(1): $\int_{0}^{a} \sqrt{a^{2} - x^{2}} d x$
(2): $\int_{1}^{2} \frac{\sqrt{x^{2} - 1}}{x^{4}} d x$
(3): $\int_{0}^{π} \frac{x \sin x}{1 + \cos^{2} x} d x$
(4): $\int_{0}^{1} \frac{\ln (1 + x)}{1 + x^{2}} d x$

解 (1) 令 $x = a \cos θ$ ，则 $\begin{matrix} (53) & \int_{0}^{a} \sqrt{a^{2} - x^{2}} d x = \int_{0}^{π / 2} a^{2} \sin^{2} θ d θ = \frac{π}{4} a^{2} \end{matrix}$ 本题也可以直接通过几何意义得到答案。

(2) 令 $x = \sec θ$ ，则 $\begin{matrix} (54) & \int_{1}^{2} \frac{\sqrt{x^{2} - 1}}{x^{4}} d x = \int_{0}^{\arccos \frac{1}{2}} \sin^{2} θ \cos θ d θ = {\frac{1}{3} \sin^{3} θ |}_{0}^{\arccos \frac{1}{2}} = \frac{\sqrt{3}}{8} \end{matrix}$

(3) 令 $x = π - t$ ，则 $\begin{matrix} (55) & \int_{0}^{π} \frac{x \sin x}{1 + \cos^{2} x} d x = \int_{0}^{π} \frac{(π - t) \sin t}{1 + \cos^{2} t} d t = \frac{π}{2} \int_{0}^{π} \frac{\sin t}{1 + \cos^{2} t} d t = - {\frac{π}{2} \arctan \cos t |}_{0}^{π} = \frac{π^{2}}{4} \end{matrix}$

(4) 令 $x = \tan θ$ ，则 $\begin{matrix} (56) & \int_{0}^{1} \frac{\ln (1 + x)}{1 + x^{2}} d x = \int_{0}^{π / 4} \ln (1 + \tan θ) d θ \overset{φ = π / 4 - θ}{=} \int_{0}^{π / 4} \ln (1 + \frac{1 - \tan φ}{1 + \tan φ}) d φ \end{matrix}$ 因此 $\begin{matrix} (57) & \int_{0}^{π / 4} \ln (1 + \tan θ) d θ = \frac{1}{2} \int_{0}^{π / 4} \ln 2 d θ = \frac{π}{8} \ln 2 \end{matrix}$ $◻$

¹ $\neg (A \land B) = A \to \neg B$ 。