First slide

Introduction to definite integration

Question

If $θ_{1}$ and $θ_{2}$ be respectively the smallest and the largest values of $θ \ln (0, 2 π) - \{π\}$ which satisfy the equation $2 \cot^{2} θ - \frac{5}{\sin θ} + 4 = 0$ then $\int_{θ_{1}}^{θ_{2}} \cos^{2} 3 θ d θ$ is equal to:

Difficult

A

$\frac{2 π}{3}$
B

$\frac{π}{3} + \frac{1}{6}$
C

$\frac{π}{3}$
D

$\frac{π}{9}$

Solution

$2 \cot^{2} θ - \frac{5}{\sin θ} + 4 = 0$
$\Rightarrow \frac{2 \cos^{2} θ}{\sin^{2} θ} - \frac{5}{\sin θ} + 4 = 0$
$\Rightarrow 2 \cos^{2} θ - 5 \sin θ + 4 \sin^{2} θ = 0, \sin θ \neq 0$
$\Rightarrow 2 \sin^{2} θ - 5 \sin θ + 2 = 0$
$\Rightarrow (2 \sin θ - 1) (\sin θ - 2) = 0$
$\Rightarrow \sin θ = \frac{1}{2}$
$\Rightarrow θ = \frac{π}{6}, \frac{5 π}{6}$
$∴ \int_{π / 6}^{5 π / 6} \cos^{2} 3 θ d θ = \int_{π / 6}^{5 π / 6} \frac{1 + \cos 6 θ}{2} d θ$
$= \frac{1}{2} {[θ + \frac{\sin 6 θ}{6}]}_{π / 6}^{5 π / 6} = \frac{1}{2} [\frac{5 π}{6} - \frac{π}{6} + \frac{1}{6} (0 - 0)] = \frac{1}{2} . \frac{4 π}{6} = \frac{π}{3}$

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