Question

The value of \( \int \sin^2 x \, dx \) is:

• A. \( \frac{x}{2} - \sin 2x \)
• B. \( \frac{x}{2} + \sin 2x \)
• C. \( \frac{x}{2} - \cos 2x \)
• D. \( \frac{x}{2} + \cos 2x \) \bigskip

Hint:

Use trigonometric identities to simplify integrals involving trigonometric functions.

Answer 1

The Correct Option is C

Step 1: Use the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \). Step 2: Substitute this identity into the integral: \[ \int \sin^2 x \, dx = \int \frac{1 - \cos 2x}{2} \, dx \] Step 3: Integrate each term separately: \[ \int \frac{1}{2} \, dx = \frac{x}{2}, \quad \int \frac{\cos 2x}{2} \, dx = \frac{\sin 2x}{4} \] Thus, the integral is: \[ \frac{x}{2} - \frac{\cos 2x}{2}. \] \bigskip