Primitive of (cos⁡x−sin⁡x)15−sin⁡2x is equal to

I=∫cosx−sinxdx15−sin2x Put sin⁡x+cos⁡x=t(cos⁡x−sin⁡x)dx=dtsin⁡x+cos⁡x=t1+2sin⁡xcos⁡x=t2sin⁡2x=t2−1I=∫dt15−t2−1I=∫dt15−t2−1=sin−1⁡t4+c=sin−1⁡sin⁡x+cos⁡x4+c