If  θ1and θ2 be respectively the smallest and the largest values of  θln0,2π−πwhich satisfy the equation  2cot2θ−5sinθ+4=0then ∫θ1θ2cos23θdθis equal to:

2cot2θ−5sinθ+4=0⇒2cos2θsin2θ−5sinθ+4=0⇒2cos2θ−5sinθ+4sin2θ=0,sinθ≠0⇒2sin2θ−5sinθ+2=0⇒2sinθ−1sinθ−2=0⇒sinθ=12⇒θ=π6,5π6∴∫π/65π/6cos23θdθ=∫π/65π/61+cos6θ2dθ                              =12θ+sin6θ6π/65π/6 =125π6−π6+160−0 =12.4π6=π3