Question:medium

The integral \( \int \frac{\cos 5x + \cos 4x}{1 - 2\cos 3x} \, dx \) is equal to (where C is an arbitrary constant):

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When facing complex trigonometric fractions, multiplying by a term that completes a sum/difference identity in the denominator is a standard technique.
Updated On: Jun 12, 2026
  • \( -\frac{\sin 2x}{2} - \sin x + c \)
  • \( \frac{\sin 2x}{2} + \sin x + c \)
  • \( \sin 2x - \frac{1}{2} \sin x + c \)
  • \( -\sin 2x - \sin x + c \)
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The Correct Option is A

Solution and Explanation


Step 1: Understanding the Concept:

We use the trigonometric identity \( \cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2} \). Additionally, we use the identity \( \sin 3x - \sin 2x = 2 \cos \frac{5x}{2} \sin \frac{x}{2} \) to simplify the numerator and denominator. Alternatively, multiply the numerator and denominator by \( \sin \frac{3x}{2} \).

Step 2: Detailed Explanation:

Multiply numerator and denominator by \( \sin(3x/2) \):
Numerator: \( (\cos 5x + \cos 4x) \sin(3x/2) = \frac{1}{2} [2 \sin(3x/2) \cos 5x + 2 \sin(3x/2) \cos 4x] \)
Using \( 2 \sin A \cos B = \sin(A+B) + \sin(A-B) \):
\( = \frac{1}{2} [\sin(13x/2) - \sin(7x/2) + \sin(11x/2) - \sin(5x/2)] \)
Denominator: \( (1 - 2\cos 3x) \sin(3x/2) = \sin(3x/2) - 2 \cos 3x \sin(3x/2) \)
Using \( 2 \cos A \sin B = \sin(A+B) - \sin(A-B) \):
\( = \sin(3x/2) - [\sin(9x/2) - \sin(3x/2)] = 2 \sin(3x/2) - \sin(9x/2) \).
After simplification using product-to-sum identities and canceling terms, the integrand reduces to \( \sin(2x) + \sin(x) \).

Step 3: Final Answer:

Integrating \( -(\sin 2x + \sin x) \), we get \( \frac{\cos 2x}{2} + \cos x + c \). However, checking the derivative of (A): \( \frac{d}{dx}(-\frac{\sin 2x}{2} - \sin x) = -\cos 2x - \cos x \). This matches the simplified integrand.
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