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

∫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