If θ1 and θ2 be respectively the smallest and the largest values of θ in (0, 2π) – {π} which satisfy the equation, 2 \left(cot\right)^{2} \theta - \frac{5}{sinθ} + 4 = 0 , then \int_{\left(\theta\right)_{1}}^{\left(\theta\right)_{2}} \left(cos\right)^{2} 3 \theta \textrm{ }\textrm{ } dθ is equal to:

Engineering

Mathematics

Introduction to Definite Integral

Question

If θ₁ and θ₂ be respectively the smallest and the largest values of θ in (0, 2π) – {π} which satisfy the equation, $2 \cot^{2} θ - \frac{5}{sinθ} + 4 = 0, then \int_{θ_{1}}^{θ_{2}} \cos^{2} 3 θ dθ$ is equal to:

$\frac{π}{9}$

$\frac{2 π}{3}$

$\frac{π}{3} + \frac{1}{6}$

$\frac{π}{3}$

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$2 \cot^{2} θ - \frac{5}{sinθ} + 4 = 0$

$\frac{2 \cos^{2} θ}{\sin^{2} θ} - \frac{5}{sinθ} + 4 = 0$

2 cos² θ – 5 sin + 4 sin² θ = 0, sinθ $\neq$ 0

2 sin² θ – 5 sin + 2 = 0

(2sinθ – 1)(sinθ – 2) = 0

$sinθ = \frac{1}{2}$

$θ = \frac{π}{6}, \frac{5 π}{6}$

$∴ \int_{π / 6}^{5 π / + 6} \cos^{2} 3 θ dθ = \int_{π / 6}^{5 π / 6} \frac{1 + \cos 6 θ}{2} dθ$

$= \frac{1}{2} {[θ + \frac{\sin 6 θ}{6}]}_{π / 6}^{5 π / 6} = \frac{1}{2} [\frac{5 π}{6} - \frac{π}{6} + \frac{1}{6} (0 - 0)] = \frac{1}{2} . \frac{4 π}{6} = \frac{π}{3}$