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