If $\theta$₁ and $\theta$₂ be respectively the smallest and the largest values of $\theta$ in (0, 2$\pi$) - {$\pi$} which satisfy the equation,
2cot²$\theta$ - ${5 \over {\sin \theta }}$ + 4 = 0, then
$\int\limits_{{\theta _1}}^{{\theta _2}} {{{\cos }^2}3\theta d\theta }$ is equal to :

2cot²$\theta$ - ${5 \over {\sin \theta }}$ + 4 = 0 <br><br>$\Rightarrow$ $$2{{{{\cos }^2}\theta } \over {{{\sin }^2}\theta }} - {5 \over {\sin \theta }} + 4$$ = 0 <br><br>$\Rightarrow$ 2sin² $\theta$ – 5sin$\theta$ + 2 = 0 <br><br>$\Rightarrow$ (2sin$\theta$ – 1)(sin$\theta$ – 2) = 0 <br><br>$\therefore$ sin$\theta$ = ${1 \over 2}$ [ sin$\theta$ = 2 not possible] <br><br>$\therefore$ $\theta$₁ = ${\pi \over 6}$ and $\theta$₂ = ${{5\pi } \over 6}$ as $\theta$₁ $&lt;$ $\theta$₂ <br><br>$\therefore$ I = $\int\limits_{{\pi \over 6}}^{{{5\pi } \over 6}} {{{\cos }^2}3\theta d\theta }$ <br><br>= $$\int\limits_{{\pi \over 6}}^{{{5\pi } \over 6}} {{{1 + \cos 6\theta } \over 2}d\theta } $$ <br><br>= $${1 \over 2}\left[ {\theta + {{\sin 6\theta } \over 2}} \right]_{{\pi \over 6}}^{{{5\pi } \over 6}}$$ <br><br>= ${{\pi \over 3}}$