Question

\(\int \left(\frac{\sin 3x}{\sin x}-\frac{\cos 3x}{\cos x}\right)\,dx\) is equal to

• A. \(2\cos 2x+C\)
• B. \(-2\cos 2x+C\)
• C. \(2\sin 2x+C\)
• D. \(2x+C\)
• E. \(x+C\)

Hint:

Whenever \(\sin 3x\) or \(\cos 3x\) appears with division by \(\sin x\) or \(\cos x\), use triple-angle formulas first. The expression usually simplifies dramatically.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The problem involves integrating a trigonometric expression. The first step is to simplify the integrand using trigonometric identities before attempting to integrate.
Step 2: Key Formula or Approach:
1. Combine the two fractions into a single fraction.
2. Use the angle subtraction formula for sine: \( \sin(A-B) = \sin A \cos B - \cos A \sin B \).
3. Use the double angle formula for sine: \( \sin(2\theta) = 2 \sin \theta \cos \theta \).
Step 3: Detailed Explanation:
Let's start by simplifying the integrand:
\[ \frac{\sin 3x}{\sin x} - \frac{\cos 3x}{\cos x} \] Combine the terms by finding a common denominator, which is \( \sin x \cos x \):
\[ = \frac{\sin 3x \cos x - \cos 3x \sin x}{\sin x \cos x} \] The numerator matches the sine angle subtraction formula, \( \sin(A-B) \), with \( A = 3x \) and \( B = x \).
\[ \sin 3x \cos x - \cos 3x \sin x = \sin(3x - x) = \sin(2x) \] The denominator can be related to the sine double angle formula. We know \( \sin(2x) = 2 \sin x \cos x \), so \( \sin x \cos x = \frac{1}{2}\sin(2x) \).
Substituting these back into the expression:
\[ = \frac{\sin(2x)}{\frac{1}{2}\sin(2x)} \] Assuming \( \sin(2x) \neq 0 \), we can cancel the terms:
\[ = 2 \] So, the original integral simplifies to:
\[ \int 2 dx \] Integrating this simple expression gives:
\[ 2x + C \] Step 4: Final Answer:
The integral is equal to \( 2x + C \). This corresponds to option (D).