Question

Compute \( \int (\cot 2x + \cos 2x) \, dx \)

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

Hint:

For integrals involving trigonometric functions, look for standard identities to simplify the expression. For example, \( \cot x = \frac{\cos x}{\sin x} \) and use substitution to make the integral easier to handle.

Answer 1

The Correct Option is B

Step 1: Break the integral.
We can split the integral into two parts: \[ I = \int \cot 2x \, dx + \int \cos 2x \, dx \]
Step 2: Solve \( \int \cot 2x \, dx \).
Using the identity \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), we can rewrite: \[ \int \cot 2x \, dx = \frac{1}{2} \ln |\sin 2x| \]
Step 3: Solve \( \int \cos 2x \, dx \).
For the integral of \( \cos 2x \), we have: \[ \int \cos 2x \, dx = \frac{1}{2} \sin 2x \]
Step 4: Combine the results.
Thus, the total integral is: \[ I = \frac{1}{2} \ln |\sin 2x| + \frac{1}{2} \sin 2x + C \]
Final Answer: \( \frac{1}{2} \ln |\sin 2x| + \frac{1}{2} \sin 2x + C \).