Apply : sin(x) = [ 2*tan(x/2) ] / [1+{tan(x/2)}^2]
Also Apply : cos(x) = [ 1- {tan(x/2)}^2 ] / [1+{tan(x/2)}^2]
On Apply the Formulas in (1) :
On simplification we get :
=> I = ∫{ [1+{tan(x/2)}^2] / [2*(2*tan(x/2))+(1- {tan(x/2)}^2) +3*(1+{tan(x/2)}^2)] } dx
=> I = ∫{ [1+{tan(x/2)}^2] / [2*{tan(x/2)}^2 +4*tan(x/2)+4 ] } dx
Apply : 1+{tan(x/2)}^2 = {sec(x/2)}^2
Therefore : ∫{ [{sec(x/2)}^2] / [2*{tan(x/2)}^2 +4*tan(x/2)+4 ] } dx
Substitution : u = tan(x/2)
=> du = { sec(x/2)}^2 * (1/2) dx
=> 2*du = {sec(x/2)}^2 dx -------(2)
Apply in (2) :
Therefore : I = ∫{ 1 / [2*{u}^2 +4*u+4 ] } (2*du)
=> I = 2* (1/2) ∫{ 1 / [{u}^2 +2*u+2 ] } du
=> I = ∫{ 1 / [u^2 +2*u+2 ] } du ---------(3)
The Quadratic Polynomial : [u^2 +2*u+2 ]
Take Coefficient of u : 2
Perform the Operation : [2] / 2 = 1 ; then : {1}^2
Apply in Quadratic Polynomial :
=> u^2 +2*u+2
=> u^2 +2*u+2 + [{1}^2-{1}^2]
=>(u^2 +2*u+{1}^2 ) + [2-{1}^2]
=> (u+1)^2 + 1
Therefore : [u^2 +2*u+2 ] = (u+1)^2 + 1 ---------(4)
Apply (4) in (3) :
=> I = ∫{ 1 / [(u+1)^2 + 1 ] } du
Apply : ∫{ 1 / [(x)^2 +(a)^2 ] } dx = (1/a) * (tan^(-1){x/a})
Therefore : I = ∫{ 1 / [(u+1)^2 + 1 ] } du
=> I = (1/1) * (tan^(-1){[u+1]/1}) +C
=> I = tan^(-1){[u+1]} +C
As known : u = tan(x/2)
Therefore : I = tan^(-1){[tan(x/2)+1]} +C
I wish you will appreciate my answer.