∫cos⁡4x−1cot⁡x−tan⁡xdx is equal to

We can write I=∫−2sin2⁡2xcos⁡xsin⁡xcos2⁡x−sin2⁡xdx=−∫sin3⁡2xcos⁡2xdx=−∫sin3⁡2x1−sin2⁡2xcos⁡2xdxPut sin⁡2x=t to obtain I=−12∫t31−t2dt=12∫t1−t2−t1−t2dt=14t2+14log⁡1−t2+C=14sin2⁡2x+12log⁡|cos⁡2x|+C.