If the integral int frac{cos 8x + 1}{cot 2x - | vedclass

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(A) Let $I = \int \frac{\cos 8x + 1}{\cot 2x - 	an 2x} dx.$First,simplify the denominator:$\cot 2x - 	an 2x = \frac{\cos 2x}{\sin 2x} - \frac{\sin 2

Vedclass · Accepted Answer

$-\frac{1}{16}$

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(A) Let $I = \int \frac{\cos 8x + 1}{\cot 2x - 	an 2x} dx.$First,simplify the denominator:$\cot 2x - 	an 2x = \frac{\cos 2x}{\sin 2x} - \frac{\sin 2x}{\cos 2x} = \frac{\cos^2 2x - \sin^2 2x}{\sin 2x \cos 2x} = \frac{\cos 4x}{\frac{1}{2} \sin 4x} = 2 \cot 4x.$Using the identity $\cos 8x + 1 = 2 \cos^2 4x$,the integral becomes:$I = \int \frac{2 \cos^2 4x}{2 \cot 4x} dx = \int \frac{\cos^2 4x}{\frac{\cos 4x}{\sin 4x}} dx = \int \cos 4x \sin 4x dx.$Multiply and divide by $2$:$I = \frac{1}{2} \int 2 \sin 4x \cos 4x dx = \frac{1}{2} \int \sin 8x dx.$Integrating $\sin 8x$ gives:$I = \frac{1}{2} \left( -\frac{\cos 8x}{8} ight) + k = -\frac{1}{16} \cos 8x + k.$Comparing this with $A \cos 8x + k$,we get $A = -\frac{1}{16}$.

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

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