Question:
$\int (\cot 2x + \csc 2x) dx = $
$\int (\cot 2x + \csc 2x) dx = $
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The combination \( \cot \theta + \csc \theta \) always simplifies to \( \cot(\theta/2) \). Keeping these "standard" simplifications in mind makes trigonometric integration much faster.
Updated On: Jun 26, 2026
- $2 \log_e |\sin x| + C$
- $\frac{1}{2} \log_e |\sin x| + C$
- $\log_e |\cos x| + C$
- $2 \log_e |\cos x| + C$
- $\log_e |\sin x| + C$
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Solution and Explanation
Step 1: Understanding the Concept:
We simplify the integrand using half-angle or reciprocal identities before integrating.
Step 2: Detailed Explanation:
Express the integrand in terms of sine and cosine:
\[ \cot 2x + \csc 2x = \frac{\cos 2x}{\sin 2x} + \frac{1}{\sin 2x} = \frac{1 + \cos 2x}{\sin 2x} \]
Use double-angle identities: \( 1 + \cos 2x = 2\cos^2 x \) and \( \sin 2x = 2\sin x \cos x \).
\[ \frac{2\cos^2 x}{2\sin x \cos x} = \frac{\cos x}{\sin x} = \cot x \]
Now, integrate:
\[ I = \int \cot x dx \]
Using the standard formula for the integral of cotangent:
\[ I = \log_e |\sin x| + C \]
Step 3: Final Answer:
The integral is \( \log_e |\sin x| + C \).
We simplify the integrand using half-angle or reciprocal identities before integrating.
Step 2: Detailed Explanation:
Express the integrand in terms of sine and cosine:
\[ \cot 2x + \csc 2x = \frac{\cos 2x}{\sin 2x} + \frac{1}{\sin 2x} = \frac{1 + \cos 2x}{\sin 2x} \]
Use double-angle identities: \( 1 + \cos 2x = 2\cos^2 x \) and \( \sin 2x = 2\sin x \cos x \).
\[ \frac{2\cos^2 x}{2\sin x \cos x} = \frac{\cos x}{\sin x} = \cot x \]
Now, integrate:
\[ I = \int \cot x dx \]
Using the standard formula for the integral of cotangent:
\[ I = \log_e |\sin x| + C \]
Step 3: Final Answer:
The integral is \( \log_e |\sin x| + C \).
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