$($b$){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{x(4-x)}dx=$

$8pts$

Solution: Since $x(4-x)=4-(x-2{)}^2,$ setting $x-2=2sin$:$in[-\frac{}{2},\frac{}{2}],$ we obtain

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{x(4-x)}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{4-(x-2{)}^2}dx=2{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{4-4si{n}^2}cosd$

$=4{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}|cos|cosd={?}^{(***)}4{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}co{s}^{2}d=2{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(cos(2)+1)d$

$=sin(2)+2+C=2sincos+2+C$

$=(x-2)\frac{\sqrt[\phantom{2}]{x(4-x)}}{2}+2si{n}^{-1}(\frac{x-2}{2})+C$