Question

If the integral $\int \frac{cos 8x+1}{cot 2x-tan 2x} dx=A cos 8x+k,$ where $k$ is an arbitrary constant, then $A$ is equal to:

• A. $-\frac{1}{16}$
• B. $\frac{1}{16}$
• C. $\frac{1}{8}$
• D. $-\frac{1}{8}$

Answer 1

The Correct Option is A

To solve the integral $\int \frac{\cos 8x + 1}{\cot 2x - \tan 2x} \, dx$, we need to simplify the integrand.

1. First, consider the expression for the denominator: $\cot 2x - \tan 2x$. We know that: $\cot 2x = \frac{\cos 2x}{\sin 2x}$ and $\tan 2x = \frac{\sin 2x}{\cos 2x}$.
2. Combine the terms: \[ \cot 2x - \tan 2x = \frac{\cos 2x}{\sin 2x} - \frac{\sin 2x}{\cos 2x} = \frac{\cos^2 2x - \sin^2 2x}{\sin 2x \cdot \cos 2x} \] Note that $\cos^2 2x - \sin^2 2x = \cos 4x$.
3. Thus, \[ \cot 2x - \tan 2x = \frac{\cos 4x}{\sin 2x \cdot \cos 2x} = \frac{\cos 4x}{\frac{1}{2}\sin 4x} = \frac{2\cos 4x}{\sin 4x} \] which simplifies to $2\cot 4x$.
4. Substitute this back into the integral: \[ \int \frac{\cos 8x + 1}{2\cot 4x} \, dx = \frac{1}{2} \int (\cos 8x + 1) \tan 4x \, dx \]
5. Let us solve $\int \cos 8x \tan 4x \, dx$. Use the substitution method: Let $u = \sin 4x$, thus $du = 4\cos 4x\, dx$.
6. Then, the integral becomes: \[ \frac{1}{2} \int \frac{\cos 8x \cdot 4 \cos 4x}{4 \cos^2 4x} \, du = \frac{1}{8} \int (2 \cos 4x \cdot \cos 8x) \, du \] This simplifies to: \[ \frac{1}{8} \int (2\cos 4x \cdot \cos 8x) \, du = -\frac{1}{16}\cos 8x \]
7. Combine this with the $+1 \tan 4x$ component, and by the linearity of integration, we find: \[ \int \frac{\cos 8x + 1}{\cot 2x - \tan 2x} \, dx = A \cos 8x + k \] \] where $A = -\frac{1}{16}$.

Therefore, the correct answer is $-\frac{1}{16}$.