Question

The value of \[ \int \cos^2 x \, dx \] will be

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

Hint:

To solve integrals involving \( \cos^2 x \), always use the identity \( \cos^2 x = \frac{1 + \cos(2x)}{2} \) to simplify the expression.

Answer 1

The Correct Option is A

Step 1: Using the Trigonometric Identity:
We use the identity for \( \cos^2 x \) to simplify the integral: \[ \cos^2 x = \frac{1 + \cos(2x)}{2} \] Thus, the integral becomes: \[ \int \cos^2 x \, dx = \int \frac{1 + \cos(2x)}{2} \, dx \]
Step 2: Splitting the Integral:
Now we split the integral into two parts: \[ = \frac{1}{2} \int 1 \, dx + \frac{1}{2} \int \cos(2x) \, dx \]
Step 3: Solving the Integrals:
The first integral is straightforward: \[ \int 1 \, dx = x \] For the second integral: \[ \int \cos(2x) \, dx = \frac{1}{2} \sin(2x) \]
Step 4: Final Solution:
Putting it all together: \[ \int \cos^2 x \, dx = \frac{x}{2} + \frac{1}{4} \sin(2x) + c \]
Step 5: Conclusion:
Thus, the correct answer is \( \frac{x}{2} + \frac{1}{4} \sin(2x) + c \).