int frac{1+cos 4 x}{cot x-tan x} d x is equal | vedclass

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(D) Let $I = \int \frac{1+\cos 4 x}{\cot x-	an x} dx$. Using the identity $1+\cos 4x = 2\cos^2 2x$ and $\cot x - 	an x = \frac{\cos x}{\sin x} - \fr

Vedclass · Accepted Answer

$-\frac{1}{8} \cos 4 x+C$

Vedclass · Answer

(D) Let $I = \int \frac{1+\cos 4 x}{\cot x-	an x} dx$. Using the identity $1+\cos 4x = 2\cos^2 2x$ and $\cot x - 	an x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} = \frac{\cos 2x}{\frac{1}{2}\sin 2x} = 2\cot 2x$. Substituting these into the integral: $I = \int \frac{2\cos^2 2x}{2\cot 2x} dx = \int \frac{\cos^2 2x}{\frac{\cos 2x}{\sin 2x}} dx = \int \sin 2x \cos 2x dx$. Using the identity $\sin 2x \cos 2x = \frac{1}{2} \sin 4x$: $I = \int \frac{1}{2} \sin 4x dx = \frac{1}{2} \left( -\frac{\cos 4x}{4} ight) + C = -\frac{1}{8} \cos 4x + C$.

$\int \frac{1+\cos 4 x}{\cot x-\tan x} d x$ is equal to

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