integral of root tanx divided by only secx

multiply up and down by secxtanx then subsitute secx=t
after that put root(t)=m
and then the eq. becomes,,..
2.int.[1/(1+m^4)]
which can easily be found.
regards

hey man you can find int.[1/1+m^4]...
put 1=1/2[1-m^2/1+m^4  + 1+m^2/1+m^4]..!!!
its an algebric twin

how u r getting 2integral of 1 divided by 1 plus mpower4

multiply both num and deno with tanx.secx

then sub    secx=t

then divide the quad in denominator to get (1-1\t^2)in the root

then solve by taking √quad as m and find

multiply secx tanx 

subsitute secx=t

and then the eq. becomes,,..
2.int.[1/(1+m^4)]
which can easily be found