📝 题目
7． $\displaystyle{\int} x \arcsin \frac{x}{2} \mathrm{~d} x$ ．
💡 答案与解析
[AI解答]
[AI解答] 我们要求解不定积分： $$ \int x \arcsin \frac{x}{2} \, \mathrm{d}x $$
**步骤1：令变量替换** 设 $$ t = \frac{x}{2} \quad \Rightarrow \quad x = 2t,\quad \mathrm{d}x = 2\,\mathrm{d}t $$ 代入原积分： $$ \int x \arcsin \frac{x}{2} \, \mathrm{d}x = \int (2t) \arcsin t \cdot 2 \, \mathrm{d}t = 4 \int t \arcsin t \, \mathrm{d}t $$
**步骤2：分部积分** 令 $$ u = \arcsin t,\quad \mathrm{d}v = t \, \mathrm{d}t $$ 则 $$ \mathrm{d}u = \frac{1}{\sqrt{1-t^2}} \, \mathrm{d}t,\quad v = \frac{t^2}{2} $$ 由分部积分公式 $\int u \, \mathrm{d}v = uv - \int v \, \mathrm{d}u$，得 $$ \int t \arcsin t \, \mathrm{d}t = \frac{t^2}{2} \arcsin t - \int \frac{t^2}{2} \cdot \frac{1}{\sqrt{1-t^2}} \, \mathrm{d}t $$
**步骤3：处理剩余积分** 计算 $$ \int \frac{t^2}{\sqrt{1-t^2}} \, \mathrm{d}t $$ 令 $t = \sin \theta$，则 $\mathrm{d}t = \cos \theta \, \mathrm{d}\theta$，且 $\sqrt{1-t^2} = \cos \theta$，于是 $$ \int \frac{t^2}{\sqrt{1-t^2}} \, \mathrm{d}t = \int \frac{\sin^2 \theta}{\cos \theta} \cdot \cos \theta \, \mathrm{d}\theta = \int \sin^2 \theta \, \mathrm{d}\theta $$ 利用恒等式 $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$，得 $$ \int \sin^2 \theta \, \mathrm{d}\theta = \frac{1}{2} \int (1 - \cos 2\theta) \, \mathrm{d}\theta = \frac{1}{2} \left( \theta - \frac{1}{2} \sin 2\theta \right) + C $$ 又 $\sin 2\theta = 2 \sin \theta \cos \theta = 2 t \sqrt{1-t^2}$，且 $\theta = \arcsin t$，所以 $$ \int \frac{t^2}{\sqrt{1-t^2}} \, \mathrm{d}t = \frac{1}{2} \arcsin t - \frac{t}{2} \sqrt{1-t^2} + C $$
**步骤4：代回分部积分结果** $$ \int t \arcsin t \, \mathrm{d}t = \frac{t^2}{2} \arcsin t - \frac{1}{2} \left( \frac{1}{2} \arcsin t - \frac{t}{2} \sqrt{1-t^2} \right) + C $$ 化简得 $$ \int t \arcsin t \, \mathrm{d}t = \frac{t^2}{2} \arcsin t - \frac{1}{4} \arcsin t + \frac{t}{4} \sqrt{1-t^2} + C $$ 合并 $\arcsin t$ 项： $$ = \frac{2t^2 - 1}{4} \arcsin t + \frac{t}{4} \sqrt{1-t^2} + C $$
**步骤5：代回原变量** 由 $t = \frac{x}{2}$，得 $$ \int x \arcsin \frac{x}{2} \, \mathrm{d}x = 4 \left[ \frac{2\left(\frac{x}{2}\right)^2 - 1}{4} \arcsin \frac{x}{2} + \frac{\frac{x}{2}}{4} \sqrt{1 - \left(\frac{x}{2}\right)^2} \right] + C $$ 化简： $$ 2\left(\frac{x}{2}\right)^2 = \frac{x^2}{2},\quad \frac{x^2}{2} - 1 = \frac{x^2 - 2}{2} $$ 于是 $$ \frac{2t^2 - 1}{4} = \frac{x^2 - 2}{8} $$ 以及 $$ \frac{t}{4} \sqrt{1-t^2} = \frac{x}{8} \sqrt{1 - \frac{x^2}{4}} = \frac{x}{8} \cdot \frac{\sqrt{4 - x^2}}{2} = \frac{x \sqrt{4 - x^2}}{16} $$ 所以原积分乘以4后： $$ 4 \left( \frac{x^2 - 2}{8} \arcsin \frac{x}{2} + \frac{x \sqrt{4 - x^2}}{16} \right) + C = \frac{x^2 - 2}{2} \arcsin \frac{x}{2} + \frac{x \sqrt{4 - x^2}}{4} + C $$
**最终结果** $$ \boxed{\displaystyle \frac{x^2 - 2}{2} \arcsin \frac{x}{2} + \frac{x}{4} \sqrt{4 - x^2} + C} $$
难度：★★★☆☆