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

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Question

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

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$\frac{-1}{8}$

Vedclass · Answer

(D) Let $I = \int \frac{1+\cos (4 x)}{\cot x-	an x} d x$. Using the identity $1+\cos(2	heta) = 2\cos^2(	heta)$,we have $1+\cos(4x) = 2\cos^2(2x)$. Also,$\cot x - 	an x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} = \frac{\cos(2x)}{\frac{1}{2}\sin(2x)} = 2\frac{\cos(2x)}{\sin(2x)}$. Substituting these into the integral: $I = \int \frac{2\cos^2(2x)}{2\frac{\cos(2x)}{\sin(2x)}} dx = \int \cos(2x) \sin(2x) dx$. Using the identity $\sin(2	heta) = 2\sin	heta \cos	heta$,we have $\sin(2x)\cos(2x) = \frac{1}{2}\sin(4x)$. $I = \int \frac{1}{2}\sin(4x) dx = \frac{1}{2} \left( \frac{-\cos(4x)}{4} ight) + C = -\frac{1}{8}\cos(4x) + C$. Comparing this with $A\cos(4x)+B$,we get $A = -\frac{1}{8}$.

If $\int \frac{1+\cos (4 x)}{\cot (x)-\tan (x)} d x=A \cos (4 x)+B$,then $A$ is equal to

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