The integral ∫$$cos$$⁡$$4x$$−$$1cot$$⁡x−$$tan$$⁡x is equal to

Correct option is 4none of these

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

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−12cos⁡4x+C

−14cos⁡4x+C

−12sin⁡2x+C

none of these

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detailed solution

Correct option is D

Put cos⁡2x=t Then −2sin⁡2xdx=dt and theintegral becomes∫−2sin2⁡2xcos2⁡x−sin2⁡xcos⁡xsin⁡xdx=−∫1−cos2⁡2xcos⁡2xsin⁡2xdx=12∫1−t2tdt=12log⁡|t|−t22+C=12log⁡|cos⁡2x|−14cos2⁡2x+C.

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The integral ∫cos⁡4x−1cot⁡x−tan⁡x is equal to

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