(sin x+sec x)^-1 dx

Re:solvethe given eqn can be written as

cosx / (sinxcosx+1) dx

=2cosx/ sin2x+2 dx

=cosx+sinx / 2+sin2x dx   +cosx-sinx/ 2+sin2x dx

     let that be equal to    

    for   put sinx-cosx= t   ............................(1)

    (sinx+cosx)dx=dt

     squaring (1)

    1-sin2x=t^2

     sin2x=1-t^2

similarly for   take sinx+cosx= p and proceed

finally by solving we get

dt/3-t^2  +dt/1+t^2

=1/2  log[root3+2/ root3-2] +tan^-1 (p)

where t=sinx-cosx and p=sinx+cosx

correct if i am wrong!!