习题4.2 换元积分法
1. 填空题
【题目】
$\int e^{3t}dt =$ ____；
$\int xe^{x^2}dx =$ ____；
$\int (3 - 2x)^{3}dx =$ ____；
$\int \frac{1}{1 - 2x} dx =$ ____；
$\int \frac{1}{\sqrt{x}(1 + x)} dx =$ ____；
$\int \frac{x}{\sqrt{2 - 3x^2}} dx =$ ____；
$\int \frac{\sin x}{\cos^3 x} dx =$ ____；
$\int \frac{1}{x^2} \sin \frac{1}{x} dx =$ ____；
9）设 $\int f(x)dx = x^2 + C$ ，则 $\int xf(1 - x^2)dx =$ ____.
【解析】
$\int e^{3t}dt = \frac{1}{3}\int e^{3t}d(3t) = \frac{1}{3}e^{3t} + C$.
$\int xe^{x^2}dx = \frac{1}{2}\int e^{x^2}d(x^2) = \frac{1}{2}e^{x^2} + C$.
$\int (3 - 2x)^3 dx = -\frac{1}{2}\int (3 - 2x)^3 d(3 - 2x) = -\frac{1}{8}(3 - 2x)^4 + C$.
$\int \frac{1}{1 - 2x} dx = -\frac{1}{2}\int \frac{d(1 - 2x)}{1 - 2x} = -\frac{1}{2}\ln|1 - 2x| + C$.
凑微分：$\frac{dx}{\sqrt{x}} = 2d(\sqrt{x})$。
$$\int \frac{1}{\sqrt{x}(1 + x)}dx = 2\int \frac{d(\sqrt{x})}{1 + (\sqrt{x})^2} = 2\arctan\sqrt{x} + C$$$\int \frac{x}{\sqrt{2 - 3x^2}}dx = -\frac{1}{6}\int (2 - 3x^2)^{-\frac{1}{2}}d(2 - 3x^2) = -\frac{1}{6} \cdot 2(2 - 3x^2)^{\frac{1}{2}} + C = -\frac{1}{3}\sqrt{2 - 3x^2} + C$.
$\int \frac{\sin x}{\cos^3 x}dx = -\int \cos^{-3} x d(\cos x) = -\frac{\cos^{-2} x}{-2} + C = \frac{1}{2}\sec^2 x + C$.
$\int \frac{1}{x^2}\sin\frac{1}{x}dx = -\int \sin\frac{1}{x}d\left(\frac{1}{x}\right) = \cos\frac{1}{x} + C$.
令 $u = 1 - x^2$，则 $du = -2xdx \implies xdx = -\frac{1}{2}du$。
$$\int x f(1 - x^2)dx = -\frac{1}{2}\int f(u)du$$已知 $\int f(u)du = u^2 + C$，故：
$$-\frac{1}{2}(u^2) + C = -\frac{1}{2}(1 - x^2)^2 + C$$
2. 计算下列各题
【题目】
$\int \sin^5 x\cos xdx;$
$\int \frac{x + 1}{x^2 + 2x + 5} dx;$
$\int \frac{f'(x)}{1 + f^2(x)} dx;$
$\int \frac{1}{e^x + e^{-x}} dx;$
$\int \left(\frac{(\arctan x)^{2}}{1 + x^{2}} + x \sqrt{1 - x^{2}}\right) dx;$
$\int \left(\frac{e^{\arccos x}}{\sqrt{1 - x^2}} +\frac{1}{\sqrt{\tan x}\cos^2x}\right)dx.$
【解析】
凑微分：$\cos x dx = d(\sin x)$。
$$\int \sin^5 x\cos xdx = \int \sin^5 x d(\sin x) = \frac{1}{6}\sin^6 x + C$$凑微分：$d(x^2 + 2x + 5) = (2x + 2)dx = 2(x + 1)dx$。
$$\int \frac{x + 1}{x^2 + 2x + 5}dx = \frac{1}{2}\int \frac{d(x^2 + 2x + 5)}{x^2 + 2x + 5} = \frac{1}{2}\ln(x^2 + 2x + 5) + C = \ln\sqrt{x^2 + 2x + 5} + C$$凑微分：$f'(x)dx = df(x)$。
$$\int \frac{df(x)}{1 + f^2(x)} = \arctan f(x) + C$$分子分母同乘 $e^x$：
$$\int \frac{e^x}{e^{2x} + 1}dx = \int \frac{d(e^x)}{(e^x)^2 + 1} = \arctan e^x + C$$分开积分：
- $\int \frac{(\arctan x)^2}{1 + x^2}dx = \int (\arctan x)^2 d(\arctan x) = \frac{1}{3}(\arctan x)^3$；
- $\int x\sqrt{1 - x^2}dx = -\frac{1}{2}\int (1 - x^2)^{\frac{1}{2}}d(1 - x^2) = -\frac{1}{3}(1 - x^2)^{\frac{3}{2}}$。 因此： $$\int \left(\frac{(\arctan x)^2}{1 + x^2} + x\sqrt{1 - x^2}\right)dx = \frac{1}{3}(\arctan x)^3 - \frac{1}{3}(1 - x^2)^{\frac{3}{2}} + C$$
分开积分：
- $\int \frac{e^{\arccos x}}{\sqrt{1 - x^2}}dx = -\int e^{\arccos x}d(\arccos x) = -e^{\arccos x}$；
- $\int \frac{1}{\sqrt{\tan x}\cos^2 x}dx = \int (\tan x)^{-\frac{1}{2}}d(\tan x) = 2\sqrt{\tan x}$。 因此： $$-e^{\arccos x} + 2\sqrt{\tan x} + C$$
3. 第二类换元积分法
【题目】
$\int \frac{x^2}{\sqrt{a^2 - x^2}} dx\ (a > 0);$
$\int \frac{1}{\sqrt{(x^2 + 1)^3}} dx;$
$\int \frac{1}{1 + \sqrt{1 - x^2}} dx;$
$\int \frac{1}{\sqrt{e^x + 2}} dx.$
【解析】
三角代换：令 $x = a\sin t\ (t \in (-\frac{\pi}{2}, \frac{\pi}{2}))$，则 $dx = a\cos t dt$，$\sqrt{a^2 - x^2} = a\cos t$。
$$\int \frac{a^2\sin^2 t}{a\cos t} \cdot a\cos t dt = a^2\int \sin^2 t dt = \frac{a^2}{2}\int (1 - \cos 2t)dt = \frac{a^2}{2}\left(t - \frac{1}{2}\sin 2t\right) + C$$$$= \frac{a^2}{2}(t - \sin t\cos t) + C = \frac{a^2}{2}\left(\arcsin\frac{x}{a} - \frac{x}{a}\sqrt{1 - \frac{x^2}{a^2}}\right) + C = \frac{a^2}{2}\left(\arcsin\frac{x}{a} - \frac{x}{a^2}\sqrt{a^2 - x^2}\right) + C$$三角代换：令 $x = \tan t\ (t \in (-\frac{\pi}{2}, \frac{\pi}{2}))$，则 $dx = \sec^2 t dt$，$\sqrt{x^2 + 1} = \sec t$。
$$\int \frac{\sec^2 t}{\sec^3 t}dt = \int \cos t dt = \sin t + C$$根据直角三角形辅助图，$\sin t = \frac{x}{\sqrt{1 + x^2}}$。故：
$$\frac{x}{\sqrt{1 + x^2}} + C$$三角代换：令 $x = \sin t\ (t \in [0, \frac{\pi}{2}))$，则 $dx = \cos t dt$，$\sqrt{1 - x^2} = \cos t$。
$$\int \frac{\cos t}{1 + \cos t}dt = \int \frac{(1 + \cos t) - 1}{1 + \cos t}dt = \int \left(1 - \frac{1}{2\cos^2\frac{t}{2}}\right)dt = t - \tan\frac{t}{2} + C$$利用半角正切公式 $\tan\frac{t}{2} = \frac{\sin t}{1 + \cos t} = \frac{x}{1 + \sqrt{1 - x^2}}$，且 $t = \arcsin x$：
$$\arcsin x - \frac{x}{1 + \sqrt{1 - x^2}} + C$$根式代换：令 $u = \sqrt{e^x + 2} > \sqrt{2}$，则 $u^2 = e^x + 2 \implies e^x = u^2 - 2$。 两边微分：$e^x dx = 2u du \implies dx = \frac{2u}{u^2 - 2}du$。
$$\int \frac{1}{u} \cdot \frac{2u}{u^2 - 2}du = 2\int \frac{du}{u^2 - 2} = \frac{2}{2\sqrt{2}}\ln\left|\frac{u - \sqrt{2}}{u + \sqrt{2}}\right| + C = \frac{1}{\sqrt{2}}\ln\left(\frac{\sqrt{e^x + 2} - \sqrt{2}}{\sqrt{e^x + 2} + \sqrt{2}}\right) + C$$