multiply and divide by root(cos2x) now write seperately  - integrate seperately for first part  take sin^2x common now take cot x =z dx=dz/-cosec^2x therefore its integral is [-log(cotx+root(cot^2x-1))] +c1  now integrate second one put cosx=z dx=dz/-sinz take -1/root2 common and form quad. final integral- (-1/root 2 log(cosx+root[cos^2x-1/2]) +c2 ans.[-log(cotx+root(cot^2x-1))]+(root 2 log(cosx+root[cos^2x-1/2]) +C