IN integration of (tanx)^1/2...  

CAnt we take whole function as t forsimplification?

 also plz tell me its ans by this method..am confused !!

U have to take whole as t...

let (tan x)^1/2  =  t

tan x = t^2

sec^2  x  dx  =  2t dt    [sec^2 x = tan^2  x  + 1]

dx = 2t /  *( t^4   +  1)    dt

so,

(int) t^2     *  2 t /  ( t^4   +  1)    dt

(int)  [2 t^3  /   ( t^4   +  1)   ] dt

2( t^4   +  1)  /  3       + c (constant of integration)

2 (tan^2  x  +  1)  / 3    +   c..........!!!

Putting

We get

The question now becomes

Writing 2t² as (1+t²) - (1-t²), we get 2 integrals.

Divide numerator and denominator by t² in both the integrals to get the 2 integrals as

In first integral, put t - 1/t = u

In second integral, put t + 1/t = v

Integrate....hope you can do the rest.........