Question

1+ COSX
dx =
... +
sinx coSX​

Answer 1

Answer:

∫1−(1−2sin22x)x−sinxdx=∫2sin22xx−sinxdx

=21∫sin22xx−21∫sin22x2sin2xcos2xdx

=21∫x⋅cosec22x−∫cot2xdx

=21[x⋅∫cosec22x−∫{dxd<