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

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

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$k = - 1/8$

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

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

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