If  ${\theta }_1$and ${\theta }_2$ be respectively the smallest and the largest values of  $\theta \ln \left(0,2\pi \right)-\left\{\pi \right\}$which satisfy the equation  $2{\cot }^2\theta -\frac{5}{\sin \theta }+4=0$then $\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}{\cos }^{2}3\theta d\theta$is equal to:

$2{\cot }^2\theta -\frac{5}{\sin \theta }+4=0$

$\Rightarrow \frac{2{\cos }^2\theta }{{\sin }^2\theta }-\frac{5}{\sin \theta }+4=0$

$\Rightarrow 2{\cos }^2\theta -5\sin \theta +4{\sin }^2\theta =0,\sin \theta \ne 0$

$\Rightarrow 2{\sin }^2\theta -5\sin \theta +2=0$

$\Rightarrow \left(2\sin \theta -1\right)\left(\sin \theta -2\right)=0$

$\Rightarrow \sin \theta =\frac{1}{2}$

$\Rightarrow \theta =\frac{\pi }{6},\frac{5\pi }{6}$

$\therefore \underset{\pi /6}{\overset{5\pi /6}{\int }}{\cos }^{2}3\theta d\theta =\underset{\pi /6}{\overset{5\pi /6}{\int }}\frac{1+\cos 6\theta }{2}d\theta$

                              $$\begin{gathered} =\frac{1}{2}{\left[\theta +\frac{\sin 6\theta }{6}\right]}_{\pi /6}^{5\pi /6} \\ =\frac{1}{2}\left[\frac{5\pi }{6}-\frac{\pi }{6}+\frac{1}{6}\left(0-0\right)\right] \\ =\frac{1}{2}.\frac{4\pi }{6}=\frac{\pi }{3} \end{gathered}$$