Integeration help..... substituiton or triginometric identites method
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Integeration help..... substituiton or triginometric identites method

[From: ] [author: ] [Date: 11-06-22] [Hit: ]
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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.

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1) sin(4x) = 2sin(2x)cos(2x). Simplify

2) Use the power rule


3) What is the integrand? Use parentheses (PEMDAS!!) Use Weierrstrauss substitution, u = tan(x/2)
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keywords: triginometric,method,or,help,Integeration,substituiton,identites,Integeration help..... substituiton or triginometric identites method
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