Question:

If $\int \frac{\cos 8x + 1}{\cot 2x - \tan 2x} dx = A\cos 8x + c$, then $A =$

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Convert products like $\sin x \cos x$ into single-angle identities for faster integration.
Updated On: Jun 10, 2026
  • $-\frac{1}{16}$
  • $\frac{1}{16}$
  • $-\frac{1}{8}$
  • $\frac{1}{8}$
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The Correct Option is A

Solution and Explanation

Step 1: Simplify denominator \[ \cot 2x - \tan 2x = \frac{\cos 2x}{\sin 2x} - \frac{\sin 2x}{\cos 2x} \] \[ = \frac{\cos^2 2x - \sin^2 2x}{\sin 2x \cos 2x} = \frac{\cos 4x}{\frac{1}{2}\sin 4x} = 2\cot 4x \]

Step 2: Simplify numerator \[ \cos 8x + 1 = 2\cos^2 4x \]

Step 3: Integral simplification \[ I = \int \frac{2\cos^2 4x}{2\cot 4x} dx = \int \cos 4x \sin 4x dx \] \[ = \frac{1}{2}\int \sin 8x dx \]

Step 4: Integration \[ I = -\frac{1}{16}\cos 8x + c \] Thus, \(A = -\frac{1}{16}\).
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