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

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

Vedclass · Accepted Answer

$k=\frac{-1}{8}$

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

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

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