If θ1 and θ2 be respectively the smallest and the largest values of θ in (0,2π)−{π} which satisfy the equation 2cot2θ−sinθ5+4=0, then ∫θ1θ2cos23θdθ is equal to:
Solutions · 1
Official solution
\[2\cot^2\theta-\frac5{\sin\theta}+4=0\Rightarrow\frac{2\cos^2\theta}{\sin^2\theta}-\frac5{\sin\theta}+4=0\]
Multiplying by sin2θ: 2cos2θ−5sinθ+4sin2θ=0⇒2sin2θ−5sinθ+2=0 (using cos2θ=1−sin2θ)
\[\Rightarrow(2\sin\theta-1)(\sin\theta-2)=0\Rightarrow\sin\theta=\frac12\Rightarrow\theta=\frac\pi6,\frac{5\pi}6\]
So θ1=6π,θ2=65π.
\[\int_{\pi/6}^{5\pi/6}\cos^23\theta\,d\theta=\int_{\pi/6}^{5\pi/6}\frac{1+\cos6\theta}{2}d\theta=\frac12\left[\theta+\frac{\sin6\theta}6\right]_{\pi/6}^{5\pi/6}=\frac12\left(\frac{5\pi}6-\frac\pi6\right)=\frac12\cdot\frac{4\pi}6=\frac\pi3\]