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

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A=−1/2

A=−1/8

A=−1/4

None of these

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detailed solution

Correct option is B

∫cos⁡4x+1cot⁡x−tan⁡xdx=∫2cos2⁡2xcos2⁡x−sin2⁡xsin⁡xcos⁡xdx=∫cos⁡2xsin⁡2xdx=14∫sin⁡4xdx=−18cos⁡4x+C Hence, A=−1/8 and B∈R

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If ∫cos⁡4x+1cot⁡x−tan⁡xdx=Acos⁡4x+B, then

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