∫dz/(1+z4) yeilds
12√2 tan-1(z2 - 1√2 z) - 14√2 log|z2 + 1 - √2 z z2 + 1 + √2 z| + c
but from tan x = z2 how are we getting this integral???? plz xplain... didnt get it....
put x-a=x+a-2a
now u have to do ∫√cotx
take tanx=z^2
then u will get
∫dz/(1+z^4)
write numerator as (1/2) * ( z^2+1 -(z^-1))
now proceed
∫dz/(1+z4) yeilds
12√2 tan-1(z2 - 1√2 z) - 14√2 log|z2 + 1 - √2 z z2 + 1 + √2 z| + c
but from tan x = z2 how are we getting this integral???? plz xplain... didnt get it....