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

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(C) Given the integral $I = \int \frac{\cos 4x + 1}{\cot x - 	an x} dx$. Using the identity $\cos 4x + 1 = 2 \cos^2 2x$,we have: $I = \int \frac{2 \c

Vedclass · Accepted Answer

$-\frac{1}{8}$

Vedclass · Answer

(C) Given the integral $I = \int \frac{\cos 4x + 1}{\cot x - 	an x} dx$. Using the identity $\cos 4x + 1 = 2 \cos^2 2x$,we have: $I = \int \frac{2 \cos^2 2x}{\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}} dx$ $I = \int \frac{2 \cos^2 2x}{\frac{\cos^2 x - \sin^2 x}{\sin x \cos x}} dx$ Since $\cos^2 x - \sin^2 x = \cos 2x$ and $\sin x \cos x = \frac{1}{2} \sin 2x$,we get: $I = \int \frac{2 \cos^2 2x}{\frac{\cos 2x}{\frac{1}{2} \sin 2x}} dx = \int \frac{2 \cos^2 2x \cdot \frac{1}{2} \sin 2x}{\cos 2x} dx$ $I = \int \cos 2x \sin 2x dx = \frac{1}{2} \int \sin 4x dx$ $I = \frac{1}{2} \left( -\frac{\cos 4x}{4} ight) + c = -\frac{1}{8} \cos 4x + c$. Comparing with $k \cos 4x + c$,we find $k = -\frac{1}{8}$. Thus,option $(C)$ is correct.

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

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