Mathematics - Definite Integration Question with Solution | TestHub

$2{\cot }^2\mathrm{\theta }-\frac{5}{\sin \mathrm{\theta }}+4=0$
$\frac{2{\cos }^2\mathrm{\theta }}{{\sin }^2\mathrm{\theta }}-\frac{5}{\sin \mathrm{\theta }}+4=0$
$2{\cos }^2\mathrm{\theta }-5\sin \mathrm{\theta }+4{\sin }^2\mathrm{\theta }=0, \sin \mathrm{\theta }\ne 0$
$2{\sin }^2\mathrm{\theta }-5\sin \mathrm{\theta }+2=0$
$\left(2\sin \mathrm{\theta }-1\right)\left(\sin \mathrm{\theta }-2\right)=0$
$\sin \mathrm{\theta }=\frac{1}{2}$
$\mathrm{\theta }=\frac{\pi }{6},\frac{5\pi }{6};$$\theta \in \left(0,2\pi \right)-\left\{\pi \right\}$
$\therefore \underset{\pi /6}{\overset{5\mathrm{\pi }/6}{\int }}{\cos }^{2}3\mathrm{\theta }\mathrm{d}\mathrm{\theta }=\underset{\pi /6}{\overset{5\mathrm{\pi }/6}{\int }}\frac{1+\cos 6\mathrm{\theta }}{2}\mathrm{d}\mathrm{\theta }$
$=\frac{1}{2}{\left[\theta +\frac{\sin 6\mathrm{\theta }}{6}\right]}_{x/6}^{5\mathrm{\pi }/6}=\frac{1}{2}\left[\frac{5\pi }{6}-\frac{\pi }{6}+\frac{1}{6}\left(0-0\right)\right]=\frac{1}{2}.\frac{4\pi }{6}=\frac{\pi }{3}$