substitute sinx and cosx and get the answer.please tell me whether you had got it or not?

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

please answer tis question.

which question??