$9\mathrm{.}16.$ Consider the definite integral ${\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}si{n}^n(x)dx,$ where $n2.$

$($a$)$ Integrate by parts using $u=si{n}^{n-1}$ and $dv=sin(x)dx$ to show that

${\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}si{n}^n(x)dx=\frac{n-1}{n}{\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}si{n}^{n-2}(x)dx$

This involves both the going$-$in$-$circles trick and the identity $co{s}^2(x)=1-si{n}^2(x)$

$($b$)$ Using the recursion from part $($a$),$ compute

${\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}si{n}^5(x)dx$ and ${\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}si{n}^6(x)dx$