∫1⁄cos3x-cosx dx

  dx/cos3x-cosx =dx/4cos^3x-4cosx   by using identity   cos3x =4cos^3x - 3cosx

=  dx/4.cosx(cos^2x-1)

=cosxdx/4cos^x(cos^x-1)

=-cosxdx/4sin^2x(1-sin^x)

 now putting sinx=t,cosxdx=dt

  =-dt/4(1-t^2)t^2

  =1/4 {-dt/t^2  -dt/1-t^2 }

  =1/4 .{1/t + log1-t/1+t } +c

 =1/4 .(cosecx + log1-sinx/1+sinx } +c