If int frac{cos 4x + 1}{cot x - tan x} dx = k | vedclass

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(B) Given integral: $I = \int \frac{\cos 4x + 1}{\cot x - 	an x} dx$ Using the identity $\cos 4x + 1 = 2 \cos^2 2x$ and $\cot x - 	an x = \frac{\cos

Vedclass · Accepted Answer

$\frac{-1}{8}$

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(B) Given integral: $I = \int \frac{\cos 4x + 1}{\cot x - 	an x} dx$ Using the identity $\cos 4x + 1 = 2 \cos^2 2x$ and $\cot x - 	an x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} = \frac{\cos 2x}{\frac{1}{2} \sin 2x} = \frac{2 \cos 2x}{\sin 2x}$. Substituting these into the integral: $I = \int \frac{2 \cos^2 2x}{\frac{2 \cos 2x}{\sin 2x}} dx$ $I = \int \cos 2x \sin 2x dx$ $I = \frac{1}{2} \int \sin 4x dx$ $I = \frac{1}{2} \left( \frac{-\cos 4x}{4} ight) + c$ $I = \frac{-1}{8} \cos 4x + c$ Comparing with $k \cos 4x + c$,we get $k = \frac{-1}{8}$.

If $\int \frac{\cos 4x + 1}{\cot x - \tan x} dx = k \cos 4x + c$,then $k$ is equal to

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