$\int_{0}^{1} x \arcsin \sqrt{4x - 4x^2}dx$

求积分${\int }_0^{1}x\arcsin \sqrt{4x-4x^2}dx$

这个积分本身不难，但是有易错点，①非主值区间三角函数不可求反函数；②计算量大，不能善用技巧容易算错

1. 整理形式
   $$\begin{array}{rl}{\int }_0^{1}x\arcsin \sqrt{4x-4x^2}dx& ={\int }_0^{1}x\arcsin \sqrt{-(4x^2-4x)}dx\\ & ={\int }_0^{1}x\arcsin \sqrt{-4\left(x^2-x+(-\frac{1}{2}{)}^2-(-\frac{1}{2}{)}^2\right)}dx\\ & ={\int }_0^{1}x\arcsin \sqrt{1-(2x-1{)}^2}dx\end{array}$$

2. 换元

   令$2x-1=\cos (t)$，则有$x=\frac{1+\cos (t)}{2}$，且$\cos (t)\in [-1,1]$，故可反解$t=\arccos (2x-1)$。

   换元3换

   1. 被积函数换：
      $$\begin{array}{rl}x\arcsin \sqrt{1-(2x-1{)}^2}& =\frac{(1+\cos (t))}{2}\cdot \arcsin \sqrt{1-{\cos }^2(t)}\\ & =\frac{(1+\cos (t))}{2}\cdot \arcsin (\sin (t))\end{array}$$

   2. 上下限换：
      $$t=\{\begin{array}{l}\arccos (2\cdot 0-1)=\arccos (-1)=\pi ,x=0\\ \arccos (2\cdot 1-1)=\arccos (1)=0,x=1\end{array}$$

   3. 积分变量换
      $$dx=d\left(\frac{1+\cos (t)}{2}\right)=-\frac{\sin (t)}{2}dt$$

3. 积分拆解
   $$\begin{array}{rl}& {\int }_0^{1}x\arcsin \sqrt{1-(2x-1{)}^2}dx\\ & ={\int }_{\pi }^0\frac{(1+\cos (t))}{2}\cdot \arcsin (\sin (t))\cdot \left(-\frac{\sin (t)}{2}\right)dt\\ & =\underset{利用定积分性质交换上下限}{\underbrace{-\frac{1}{4}{\int }_{\pi }^0\sin (t)\cdot \left[1+\cos (t)\right]\cdot \arcsin (\sin (t))dt}}\\ & =\frac{1}{4}{\int }_0^{\pi }\sin (t)\cdot \left[1+\cos (t)\right]\cdot \underset{t\notin \left[-\frac{\pi }{2},\frac{\pi }{2}\right],\text{分区间讨论}}{\underbrace{\arcsin (\sin (t))}}dt\\ & =\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\sin (t)\cdot \left[1+\cos (t)\right]\cdot \arcsin (\sin (t))dt+\frac{1}{4}{\int }_{\frac{\pi }{2}}^{\pi }\sin (t)\cdot \left[1+\cos (t)\right]\cdot \underset{\text{在}\left[\frac{\pi }{2},\pi \right],t=\pi -\arcsin (y)}{\underbrace{\arcsin (\sin (t))}}dt\\ & =\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\sin (t)\cdot \left[1+\cos (t)\right]\cdot tdt+\frac{1}{4}\underset{\text{令}\pi -t=u进行换元，记得换元三换}{\underbrace{{\int }_{\frac{\pi }{2}}^{\pi }\sin (t)\cdot \left[1+\cos (t)\right]\cdot (\pi -t)dt}}\\ & =\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\sin (t)\cdot \left[1+\cos (t)\right]\cdot tdt+\frac{1}{4}\underset{\text{利用诱导公式化简}}{\underbrace{{\int }_{\frac{\pi }{2}}^0\sin (\pi -u)(1+\cos (\pi -u))u\cdot (-du)}}\\ & =\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\sin (t)\cdot \left[1+\cos (t)\right]\cdot tdt+\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\sin (u)(1-\cos (u))u\cdot du\\ & =\frac{1}{4}\left[{\int }_0^{\frac{\pi }{2}}t\cdot \sin (t)\left[1+\cos (t)+1-\cos (t)\right]dt\right]\\ & =\frac{1}{2}\underset{\text{表格法快速求积}}{\underbrace{{\int }_0^{\frac{\pi }{2}}t\cdot \sin (t)dt}}\\ & =\frac{1}{2}\left[\sin (t)-t\cdot \cos (t)\right]{|}_0^{\frac{\pi }{2}}\\ & =\frac{1}{2}\end{array}$$