If ${\displaystyle \underset{a}{\overset{b}{\int }}\left|\sin x\right|dx=8}$ and  ${\displaystyle {\int }_0^{a+b}\left|\cos x\right|dx=9}$ , then

${\displaystyle \underset{a}{\overset{b}{\int }}\left|\sin x\right|dx=8}$  and ${\displaystyle {\int }_0^{a+b}\left|\cos x\right|dx=9}$ 

${\displaystyle \underset{a}{\overset{b}{\int }}\left(\sin x\right)dx=8}$  area of curve  of side  function

${\displaystyle \underset{\theta }{\overset{\frac{\pi }{2}}{\int }}\left(\sin x\right)dx=1}$

$\Rightarrow \frac{\pi }{2}-0=1\left(\frac{\pi }{2}\right)$

${\displaystyle \underset{a}{\overset{b}{\int }}\left(\sin x\right)dx=8}$

$b-a=\frac{8\pi }{2}=4\pi$

$b-a=4\pi \cdots \cdots \left(1\right)$

${\displaystyle \underset{0}{\overset{a+b}{\int }}\left|\cos x\right|.dx=9}$

$\Rightarrow {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(\cos x\right).dx=1}$

$a+b-0=\frac{9\pi }{2}$

$a+b=\frac{9\pi }{2} \cdots \cdots \left(2\right)$

$b-a=4\pi$

$a+b=\frac{9\pi }{2}$  

$2b=\frac{17\pi }{2}\Rightarrow b=\frac{17\pi }{4}$

$a+b=\frac{9\pi }{2}$

$a=\frac{9\pi }{2}-\frac{17\pi }{4}=\frac{18\pi -17\pi }{4}=\frac{\pi }{4}$

$\frac{a}{b}=\frac{1}{17}$

${\displaystyle \underset{a}{\overset{b}{\int }}{\sec }^{2}xdx={\left(\tan x\right)}_a^b=-\tan \frac{\pi }{4}+\tan \frac{17\pi }{4}}$

$=\tan \left(4\pi +\frac{\pi }{4}\right)-\tan \frac{\pi }{4}$

$=\tan \frac{\pi }{4}-\tan \frac{\pi }{4}=0$

${\displaystyle \underset{a}{\overset{b}{\int }}{\sec }^{2}xdx=0}$