solve underroot tanx.

I=??tan x dx

Let tan x = t

So, sec^2 x dx = 2 t dt

So, (1+tan^2 x) dx = 2 t dt

 

So, dx = (2 t dt) / (1+ t^4)

Substituting in I, eventually I becomes

? {(t^2+1 ) + (t^2-1)} / (t^4 +1)

 

On further simplifying,

 

I= ? (t^2+!) / (t^4+1) dt + ? (t^2-1) / (t^4 +1) dt

I=?(1+ 1/t^2) / (t^2 + !/t^2) dt + ? (1- 1/t^2) / (t^2+ 1/t^2) dt

 

For da 1st integral sub t- 1/t =u

 

So,da 1st integral becums ?du / (u^2 + 2)

 

Further simplifyn, it becums 1/?2 tan?1{(t – 1/t) / ?2}

 

 

For 2nd integral, sub t+ 1/t =v

So,da 2nd interal becums ?dv / (v^2 -2)

 

Now solv both da integrals

Ans is

(1/?2) tan?1 {(?tan x - ?cot x +?2 ) / ?2} + (-1/ 2?2) log {(?tan x +?cot x +?2) / (?tan x + ?cot x - ?2)} + c