人邮高数 第3章 第3-3-2题

[AI解答]

[AI解答]

以下是习题3-3中第2题各小题的详细解答。

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### （1）$\displaystyle{\int} \frac{1}{3+5 \cos x} \mathrm{~d} x$

**解**：利用万能代换 $\displaystyle t = \tan\frac{x}{2}$，则 $\displaystyle \cos x = \frac{1-t^2}{1+t^2}$，$\displaystyle \mathrm{d}x = \frac{2}{1+t^2}\mathrm{d}t$。

代入得： $$ \int \frac{1}{3+5\cdot\frac{1-t^2}{1+t^2}}\cdot\frac{2}{1+t^2}\mathrm{d}t = \int \frac{2}{3(1+t^2)+5(1-t^2)}\mathrm{d}t = \int \frac{2}{3+3t^2+5-5t^2}\mathrm{d}t = \int \frac{2}{8-2t^2}\mathrm{d}t = \int \frac{1}{4-t^2}\mathrm{d}t $$

分解为部分分式： $$ \frac{1}{4-t^2} = \frac{1}{(2-t)(2+t)} = \frac{1}{4}\left(\frac{1}{2-t} + \frac{1}{2+t}\right) $$

积分得： $$ \frac{1}{4}\left( -\ln|2-t| + \ln|2+t| \right) + C = \frac{1}{4}\ln\left|\frac{2+t}{2-t}\right| + C $$

代回 $\displaystyle t = \tan\frac{x}{2}$： $$ \boxed{\frac{1}{4}\ln\left|\frac{2+\tan\frac{x}{2}}{2-\tan\frac{x}{2}}\right| + C} $$

难度：★★★☆☆

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### （2）$\displaystyle{\int} \cos^{5} x \mathrm{~d} x$

**解**：将 $\cos^5 x = \cos^4 x \cos x = (1-\sin^2 x)^2 \cos x$，令 $u = \sin x$，$\mathrm{d}u = \cos x \mathrm{d}x$。

$$ \int (1-u^2)^2 \mathrm{d}u = \int (1 - 2u^2 + u^4)\mathrm{d}u = u - \frac{2}{3}u^3 + \frac{1}{5}u^5 + C $$

代回 $u = \sin x$： $$ \boxed{\sin x - \frac{2}{3}\sin^3 x + \frac{1}{5}\sin^5 x + C} $$

难度：★★☆☆☆

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### （3）$\displaystyle{\int} \sin^{2} x \cos^{4} x \mathrm{~d} x$

**解**：利用倍角公式降幂： $$ \sin^2 x = \frac{1-\cos 2x}{2},\quad \cos^4 x = \left(\frac{1+\cos 2x}{2}\right)^2 $$

乘积为： $$ \frac{1-\cos 2x}{2} \cdot \frac{(1+\cos 2x)^2}{4} = \frac{1}{8}(1-\cos 2x)(1+2\cos 2x+\cos^2 2x) $$

先展开： $$ = \frac{1}{8}\left[1+2\cos 2x+\cos^2 2x - \cos 2x - 2\cos^2 2x - \cos^3 2x\right] $$ $$ = \frac{1}{8}\left[1 + \cos 2x - \cos^2 2x - \cos^3 2x\right] $$

再用 $\displaystyle \cos^2 2x = \frac{1+\cos 4x}{2}$，$\displaystyle \cos^3 2x = \frac{\cos 6x + 3\cos 2x}{4}$，代入整理得： $$ = \frac{1}{8}\left[1 + \cos 2x - \frac{1+\cos 4x}{2} - \frac{\cos 6x + 3\cos 2x}{4}\right] $$ $$ = \frac{1}{8}\left[1 + \cos 2x - \frac12 - \frac12\cos 4x - \frac14\cos 6x - \frac34\cos 2x\right] $$ $$ = \frac{1}{8}\left[\frac12 + \frac14\cos 2x - \frac12\cos 4x - \frac14\cos 6x\right] $$

积分： $$ \frac{1}{8}\left(\frac12 x + \frac14\cdot\frac{\sin 2x}{2} - \frac12\cdot\frac{\sin 4x}{4} - \frac14\cdot\frac{\sin 6x}{6}\right) + C $$ $$ = \frac{x}{16} + \frac{\sin 2x}{64} - \frac{\sin 4x}{64} - \frac{\sin 6x}{192} + C $$

$$ \boxed{\frac{x}{16} + \frac{\sin 2x}{64} - \frac{\sin 4x}{64} - \frac{\sin 6x}{192} + C} $$

难度：★★★★☆

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### （4）$\displaystyle{\int} \frac{\sin^{3} x}{\cos^{4} x} \mathrm{~d} x$

**解**：改写为： $$ \int \frac{\sin^2 x}{\cos^4 x} \sin x \mathrm{d}x = \int \frac{1-\cos^2 x}{\cos^4 x} \sin x \mathrm{d}x $$

令 $u = \cos x$，$\mathrm{d}u = -\sin x \mathrm{d}x$： $$ = -\int \frac{1-u^2}{u^4} \mathrm{d}u = -\int (u^{-4} - u^{-2})\mathrm{d}u = -\left( -\frac{1}{3}u^{-3} + u^{-1} \right) + C $$ $$ = \frac{1}{3u^3} - \frac{1}{u} + C $$

代回 $u = \cos x$： $$ \boxed{\frac{1}{3\cos^3 x} - \frac{1}{\cos x} + C} $$

难度：★★☆☆☆

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### （5）$\displaystyle{\int} \frac{\sin^{5} x}{\cos^{4} x} \mathrm{~d} x$

**解**：$\sin^5 x = \sin x (1-\cos^2 x)^2$，令 $u = \cos x$，$\mathrm{d}u = -\sin x \mathrm{d}x$： $$ \int \frac{(1-u^2)^2}{u^4}(-\mathrm{d}u) = -\int \frac{1 - 2u^2 + u^4}{u^4}\mathrm{d}u = -\int (u^{-4} - 2u^{-2} + 1)\mathrm{d}u $$ $$ = -\left( -\frac{1}{3u^3} + \frac{2}{u} + u \right) + C = \frac{1}{3u^3} - \frac{2}{u} - u + C $$

代回： $$ \boxed{\frac{1}{3\cos^3 x} - \frac{2}{\cos x} - \cos x + C} $$

难度：★★☆☆☆

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### （6）$\displaystyle{\int} \cos 4x \cos 6x \mathrm{~d} x$

**解**：利用积化和差： $$ \cos A \cos B = \frac12[\cos(A+B) + \cos(A-B)] $$ 此处 $A=4x, B=6x$： $$ \cos 4x \cos 6x = \frac12[\cos 10x + \cos(-2x)] = \frac12(\cos 10x + \cos 2x) $$

积分： $$ \frac12\left( \frac{\sin 10x}{10} + \frac{\sin 2x}{2} \right) + C = \frac{\sin 10x}{20} + \frac{\sin 2x}{4} + C $$

$$ \boxed{\frac{\sin 10x}{20} + \frac{\sin 2x}{4} + C} $$

难度：★☆☆☆☆

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### （7）$\displaystyle{\int} \frac{1}{1+\sin x} \mathrm{~d} x$

**解**：利用恒等式： $$ \frac{1}{1+\sin x} = \frac{1-\sin x}{1-\sin^2 x} = \frac{1-\sin x}{\cos^2 x} = \sec^2