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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
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