Question

Evaluate \( \int \sin x \cdot \sin 2x \, dx \)

• A. \( -\frac{1}{4} \cos 3x + \frac{1}{4} \cos x \)
• B. \( -\frac{1}{4} \cos 3x - \frac{1}{4} \cos x \)
• C. \( \frac{1}{4} \cos 3x + \frac{1}{4} \cos x \)
• D. \( \frac{1}{4} \cos 3x - \frac{1}{4} \cos x \)

Hint:

For trigonometric integrals, use the product-to-sum identities to simplify the expression before integrating.

Answer 1

The Correct Option is A

We are given the integral \( \int \sin x \cdot \sin 2x \, dx \). To solve this, we use the product-to-sum identity: \[ \sin x \cdot \sin 2x = \frac{1}{2} [ \cos(x - 2x) - \cos(x + 2x) ] = \frac{1}{2} [ \cos(-x) - \cos(3x) ] \] Since \( \cos(-x) = \cos x \), this simplifies to: \[ \sin x \cdot \sin 2x = \frac{1}{2} [ \cos x - \cos 3x ] \] Thus, the integral becomes: \[ \int \sin x \cdot \sin 2x \, dx = \frac{1}{2} \int [ \cos x - \cos 3x ] \, dx \] Now, integrate each term: \[ \frac{1}{2} \left[ \sin x - \frac{1}{3} \sin 3x \right] = \frac{1}{4} \cos 3x - \frac{1}{4} \cos x \] Thus, the correct answer is \( -\frac{1}{4} \cos 3x + \frac{1}{4} \cos x \).