On 2:14 we can substitute -1 as -tan π/4 in numinator and tanx as tan x * tan π/4 (as tan π/4 = 1) then we get (tan x - tan π/4)/ (1 + tan x *tan π/4)] By using tan(A - B) formula we condensed it to the integral of tan (x - π/4) which gives log|cos(x - π/4)| as the alternate final answer from 2:14
definitely easy
On 2:14 we can substitute -1 as -tan π/4 in numinator and tanx as tan x * tan π/4 (as tan π/4 = 1) then we get (tan x - tan π/4)/ (1 + tan x *tan π/4)] By using tan(A - B) formula we condensed it to the integral of tan (x - π/4) which gives log|cos(x - π/4)| as the alternate final answer from 2:14
This question is easy to solve..When we complete miscellaneous exercises, examples and with normal exercises sums of Integrals chapter 💯.
Very interesting. On 2:14 in the video, can we write (tanx-1)/(tanx+1) as -cot(x+90°) and proceed?