If the real part of the complex number 1 - cos θ + 2 i sin θ - 1 is 1 5 for θ ∈ ( 0 , π ) , then the value of the integral ∫ 0 θ sin x d x is equal to:

Step-by-step solution

Let z = 1 1 - cos θ + 2 i sin θ ⇒ z = 1 - cos θ - 2 i sin θ 1 - cos θ + 2 i sin θ 1 - cos θ - 2 i sin θ ⇒ z = 2 sin 2 θ 2 - 2 i sin θ ( 1 - cos θ ) 2 + 4 sin 2 θ Put, sin θ = 2 sin θ 2 cos θ 2 ,   1 - cos θ = 2 sin 2 θ 2 ⇒ z = 2 sin 2 θ 2 - 4 i sin θ 2 cos θ 2 4 sin 4 θ 2 + 16 sin 2 θ 2 cos 2 θ 2 Hence, R e z = 2 sin 2 θ 2 4 sin 4 θ 2 + 16 sin 2 θ 2 cos 2 θ 2