If ∫$$cos$$⁡$$4x+1cot$$⁡x−$$tan$$⁡$$xdx=Acos$$⁡$$4x+B$$, then the value of A is

∫$$cos2$$⁡$$2xsin$$⁡$$2xdxcos$$⁡$$2x=12$$∫$$sin$$⁡$$4xdx=$$−$$18cos$$⁡$$4x+B$$

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Question

$If \int \frac{\cos 4 x + 1}{\cot x - \tan x} d x = A \cos 4 x + B, then the value of A is$

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Solution

$\begin{array}{l} \int \frac{\cos^{2} 2 x \sin 2 x d x}{\cos 2 x} \\ = \frac{1}{2} \int \sin 4 x d x = - \frac{1}{8} \cos 4 x + B \end{array}$

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