If the real part of the complex number $(1-\cos \theta+2 i \sin \theta)^{-1}$ is $\frac{1}{5}$ for $\theta \in(0, \pi)$, then the value of the integral $\int_{0}^{\theta} \sin x d x$ is equal to :

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Complex Numbers

If the real part of the complex number

(1 - cos θ + 2 i sin θ)^{- 1}

\frac{1}{5}

for

θ \in (0, π)

, then the value of the integral

\int_{0}^{θ} sin x d x

is equal to :

Answer: (D)

Rationalising, the real part of

(1 - cos θ + 2 i sin θ)^{- 1}

\frac{1 - cos θ}{( 1 - cos θ ) ^{2} + 4 sin ^{2} θ} = \frac{1}{5}

Writing

c = cos θ

and using

sin^{2} θ = (1 - c) (1 + c)

5 (1 - c) = (1 - c)^{2} + 4 (1 - c) (1 + c) = (1 - c) (5 + 3 c)

For

θ \in (0, π)

the factor

1 - c

is non-zero, so

5 = 5 + 3 c

, giving

c = 0

and

θ = \frac{π}{2}

. Then

\int_{0}^{π /2} sin x d x = [- cos x]_{0}^{π /2} = 1

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option