Question

$\int (\cot 2x + \csc 2x) dx = $

• A. $2 \log_e |\sin x| + C$
• B. $\frac{1}{2} \log_e |\sin x| + C$
• C. $\log_e |\cos x| + C$
• D. $2 \log_e |\cos x| + C$
• E. $\log_e |\sin x| + C$

Hint:

The combination \( \cot \theta + \csc \theta \) always simplifies to \( \cot(\theta/2) \). Keeping these "standard" simplifications in mind makes trigonometric integration much faster.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
We are dealing with the integral of a trigonometric sum. We can convert the terms to sine and cosine to simplify the expression into a single, easily integrable trigonometric function.
Step 2: Key Formula or Approach:
Convert using \(\cot 2x = \frac{\cos 2x}{\sin 2x}\) and \(\csc 2x = \frac{1}{\sin 2x}\).
Combine and use half-angle formulas: \(1 + \cos 2x = 2\cos^2 x\) and \(\sin 2x = 2\sin x \cos x\).
Integrate the simplified function.
Step 3: Detailed Explanation:
Simplify the integrand:
\[ \cot 2x + \csc 2x = \frac{\cos 2x}{\sin 2x} + \frac{1}{\sin 2x} = \frac{1 + \cos 2x}{\sin 2x} \] Apply the double angle identities:
\[ \frac{1 + \cos 2x}{\sin 2x} = \frac{2\cos^2 x}{2\sin x \cos x} \] Cancel \(2\) and \(\cos x\):
\[ = \frac{\cos x}{\sin x} = \cot x \] Now, evaluate the integral:
\[ \int (\cot 2x + \csc 2x) dx = \int \cot x \, dx \] The standard integral of \(\cot x\) is:
\[ = \log_e|\sin x| + C \] Step 4: Final Answer:
The result is \(\log_e | \sin x | + C\).