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If $\int \frac{\cos 4 x + 1}{\cot x - \tan x} dx = Acos 4 x + B$ then

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a

A=−12

b

A=−18

c

A=−14

d

None of these

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detailed solution

Correct option is B

∫cos⁡4x+1cot⁡x−tan⁡xdx=Acos⁡4x+BLet,I=∫cos⁡4x+1cot⁡x−tan⁡xdx=∫2cos2⁡2xcos⁡xsin⁡x−sin⁡xcos⁡xdx=∫2cos2⁡2xcos⁡2xsin⁡xcos⁡xdx=∫sin⁡2xcos⁡2xdx=12∫2sin⁡2xcos⁡2xdx=12∫sin⁡4xdx=12−cos⁡4x4+B=−18cos⁡4x+Bbut I=Acos⁡4x+B( given )∴A=−18

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