Question

The value of \( \int \cos^2 x \, dx \) will be:

• A. \( -\frac{1}{4}\sin 2x + \frac{x}{4} + C \)
• B. \( -\frac{1}{2}\sin 2x + \frac{x}{4} + C \)
• C. \( \cos^2 x - \sin^2 x + C \)
• D. \( \frac{1}{4}\sin 2x + \frac{x}{2} + C \)

Hint:

To integrate even powers of sine and cosine like \( \sin^2 x \) or \( \cos^4 x \), always use the power-reducing formulas derived from the double-angle identities: \( \cos^2 x = (1+\cos 2x)/2 \) and \( \sin^2 x = (1-\cos 2x)/2 \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We need to find the indefinite integral of \( \cos^2 x \). Since there is no direct standard formula for integrating \( \cos^2 x \), we must first rewrite it using a trigonometric identity to reduce the power.
Step 2: Key Formula or Approach:
The half-angle identity (or power-reducing formula) for cosine is:
\[ \cos^2 x = \frac{1 + \cos(2x)}{2} \] We will substitute this into the integral and then integrate term by term.
Step 3: Detailed Explanation or Calculation:
\[ \int \cos^2 x \, dx = \int \frac{1 + \cos(2x)}{2} \, dx \] \[ = \int \left(\frac{1}{2} + \frac{1}{2}\cos(2x)\right) dx \] Split the integral into two parts:
\[ = \int \frac{1}{2} \, dx + \int \frac{1}{2}\cos(2x) \, dx \] \[ = \frac{1}{2}x + \frac{1}{2} \int \cos(2x) \, dx \] To integrate \( \cos(2x) \), we get \( \frac{\sin(2x)}{2} \).
\[ = \frac{1}{2}x + \frac{1}{2} \left(\frac{\sin(2x)}{2}\right) + C \] \[ = \frac{x}{2} + \frac{\sin(2x)}{4} + C \] Step 4: Final Answer:
The value of the integral is \( \frac{1}{4}\sin 2x + \frac{x}{2} + C \).