📝 题目
17． $\displaystyle{\int}\left(x^{2}-1\right) \sin 2 x \mathrm{~d} x$ ．
💡 答案与解析
[AI解答]
[AI解答] 我们要求不定积分： $$ \int (x^2 - 1) \sin 2x \, \mathrm{d}x $$
**步骤1：使用分部积分法** 设 $$ u = x^2 - 1, \quad \mathrm{d}v = \sin 2x \, \mathrm{d}x $$ 则 $$ \mathrm{d}u = 2x \, \mathrm{d}x, \quad v = \int \sin 2x \, \mathrm{d}x = -\frac{1}{2} \cos 2x $$
由分部积分公式 $\int u \, \mathrm{d}v = uv - \int v \, \mathrm{d}u$，得 $$ \int (x^2 - 1) \sin 2x \, \mathrm{d}x = (x^2 - 1) \cdot \left( -\frac{1}{2} \cos 2x \right) - \int \left( -\frac{1}{2} \cos 2x \right) \cdot 2x \, \mathrm{d}x $$ 化简： $$ = -\frac{1}{2}(x^2 - 1) \cos 2x + \int x \cos 2x \, \mathrm{d}x $$
**步骤2：再次分部积分** 对 $\int x \cos 2x \, \mathrm{d}x$，设 $$ u = x, \quad \mathrm{d}v = \cos 2x \, \mathrm{d}x $$ 则 $$ \mathrm{d}u = \mathrm{d}x, \quad v = \frac{1}{2} \sin 2x $$ 于是 $$ \int x \cos 2x \, \mathrm{d}x = x \cdot \frac{1}{2} \sin 2x - \int \frac{1}{2} \sin 2x \, \mathrm{d}x = \frac{x}{2} \sin 2x - \frac{1}{2} \int \sin 2x \, \mathrm{d}x $$ 而 $$ \int \sin 2x \, \mathrm{d}x = -\frac{1}{2} \cos 2x $$ 所以 $$ \int x \cos 2x \, \mathrm{d}x = \frac{x}{2} \sin 2x - \frac{1}{2} \left( -\frac{1}{2} \cos 2x \right) = \frac{x}{2} \sin 2x + \frac{1}{4} \cos 2x $$
**步骤3：代回原式** 原积分 $$ = -\frac{1}{2}(x^2 - 1) \cos 2x + \left( \frac{x}{2} \sin 2x + \frac{1}{4} \cos 2x \right) + C $$ 合并含 $\cos 2x$ 的项： $$ -\frac{1}{2}x^2 \cos 2x + \frac{1}{2} \cos 2x + \frac{1}{4} \cos 2x = -\frac{1}{2}x^2 \cos 2x + \frac{3}{4} \cos 2x $$ 因此结果为： $$ \boxed{-\frac{1}{2}x^{2}\cos 2x + \frac{x}{2}\sin 2x + \frac{3}{4}\cos 2x + C} $$
难度：★★☆☆☆