If int frac{cos 8 x+1}{cot 2 x-tan 2 x} ,d x= | vedclass

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(D) Given integral $I = \int \frac{\cos 8 x+1}{\cot 2 x-	an 2 x} \,d x$.Using the identity $1+\cos 	heta = 2 \cos^2 \frac{	heta}{2}$, we have $\cos

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$\frac{-1}{16}$

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(D) Given integral $I = \int \frac{\cos 8 x+1}{\cot 2 x-	an 2 x} \,d x$.Using the identity $1+\cos 	heta = 2 \cos^2 \frac{	heta}{2}$, we have $\cos 8x + 1 = 2 \cos^2 4x$.The denominator is $\frac{\cos 2x}{\sin 2x} - \frac{\sin 2x}{\cos 2x} = \frac{\cos^2 2x - \sin^2 2x}{\sin 2x \cos 2x} = \frac{\cos 4x}{\frac{1}{2} \sin 4x} = \frac{2 \cos 4x}{\sin 4x}$.Substituting these into the integral:$I = \int \frac{2 \cos^2 4x}{\frac{2 \cos 4x}{\sin 4x}} \,d x = \int \cos 4x \sin 4x \,d x$.Multiply and divide by $2$:$I = \frac{1}{2} \int 2 \sin 4x \cos 4x \,d x = \frac{1}{2} \int \sin 8x \,d x$.Integrating $\sin 8x$ gives $\frac{-\cos 8x}{8}$.So, $I = \frac{1}{2} \left( \frac{-\cos 8x}{8} ight) + c = \frac{-\cos 8x}{16} + c$.Comparing with $A \cos 8x + c$, we get $A = \frac{-1}{16}$.

If $\int \frac{\cos 8 x+1}{\cot 2 x-\tan 2 x} \,d x=A \cos 8 x+c$, where $c$ is an arbitrary constant, then the value of $A$ is

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