If [ ]1 and [ ]2 be respectively the smallest and the largest values of [ ] in (0, 2[ ]) -…

Question

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 :

Accepted Answer

2cot2$$\theta $$ - $${5 \over {\sin \theta }}$$ + 4 = 0

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

$$ \Rightarrow $$ 2sin2
$$\theta $$ -- 5sin$$\theta $$ + 2 = 0

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

$$ \therefore $$ sin$$\theta $$ = $${1 \over 2}$$ [ sin$$\theta $$ = 2 not possible]

$$ \therefore $$ $$\theta $$1 = $${\pi  \over 6}$$ and $$\theta $$2 = $${{5\pi } \over 6}$$ as $$\theta $$1 $<$ $$\theta $$2

$$ \therefore $$ I = $$\int\limits_{{\pi  \over 6}}^{{{5\pi } \over 6}} {{{\cos }^2}3\theta d\theta } $$

= $$\int\limits_{{\pi  \over 6}}^{{{5\pi } \over 6}} {{{1 + \cos 6\theta } \over 2}d\theta } $$

= $${1 \over 2}\left[ {\theta  + {{\sin 6\theta } \over 2}} \right]_{{\pi  \over 6}}^{{{5\pi } \over 6}}$$

= $${{\pi  \over 3}}$$

Solutions · 1