find integral :- root cos(2x)/sin x

put cos2x=1-tan^2x /1+tan^2x

  expression becomes root(1-tan^2x)/tanx

now multiply and divide by 1+tan^2x

now  expression  is root(1-tan^2x) .sec^2x /(1+tan^2x)tanx

  now put tanx=t  , we get sec^2xdx=dt 

now expression is root(1-t^2)/t.(1+t^2)

  now multiply divide by t

 expression is root (1-t^2)t/t^2 .(1+t^2)

 now put t^2=u, so 2tdt=du

 now expression is root1-u/2u(1+u)

  now again put 1-u=v^2 ,so -du=2vdv

 expression is -v^2/(1-v^2)(2-v^2)  

 =1/2-v^2  +  -1/(2-v^2)(1-v^2)  

now integrate using partial fraction and get the answer