∫sin2xa2sin2x+b2cos2xdx
1a2+b2loga2sin2x+b2cos2x+C
1a2−b2loga2sin2x+b2cos2x+C
1a2−b2loga2sin2x−b2cos2x+C
1a2+b2loga2sin2x−b2cos2x+C
ddxa2sin2x+b2cos2x=a2−b2sin2x Now, ∫sin2xa2sin2x+b2cos2xdx=1a2−b2∫a2−b2sin2xa2sin2x+b2cos2xdx=1a2−b2∫ddxa2sin2x+b2cos2xa2sin2x+b2cos2xdx=1a2−b2loga2sin2x+b2 cos2x+C