solve underroot tanx.
Asked by priyadarshini_selvam
| 22nd Mar, 2012,
03:53: PM
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
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
Answered by
| 22nd Mar, 2012,
04:07: PM
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