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If cos4x+1cotxtanx=Kcos4x+C, then

 

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a
K=−1/2
b
K=−1/8
c
K=−1/5
d
none of these

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detailed solution

Correct option is B

∫cos⁡4x+1cot⁡x−tan⁡xdx=∫2cos2⁡2xcos2⁡x−sin2⁡x⋅sin⁡xcos⁡xdx=∫cos⁡2xsin⁡2xdx=12∫sin⁡4xdx=−18cos⁡4x+CHence K=−1/8


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