Question

Evaluate the integral: \[ \int \frac{\sin \theta \sin 2\theta}{1 - \cos 2\theta} d\theta. \]

• A. \( 1 + \cos \theta + C \)
• B. \( 1 + \sin \theta + C \)
• C. \( \sin \theta + C \)
• D. \( 1 + \cos 2\theta + C \)
• E. \( 1 + \sin 2\theta + C \)

Hint:

Use trigonometric identities to simplify integrals before attempting direct integration.

Answer 1

The Correct Option is C

Step 1: Simplify the denominator Using the identity: \[ 1 - \cos 2\theta = 2 \sin^2 \theta. \] Step 2: Expand the numerator Using \( \sin 2\theta = 2 \sin \theta \cos \theta \), the numerator becomes: \[ \sin \theta \cdot 2 \sin \theta \cos \theta = 2 \sin^2 \theta \cos \theta. \] Thus, the integral simplifies to: \[ \int \frac{2 \sin^2 \theta \cos \theta}{2 \sin^2 \theta} d\theta. \] Canceling \( 2 \sin^2 \theta \): \[ \int \cos \theta d\theta. \] Step 3: Integrate \[ \int \cos \theta d\theta = \sin \theta + C. \] Thus, the correct answer is (C) \( \sin \theta + C \).