B.1. SOLVING OF 2 sin 1

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APPENDIX B – INTEGRALS SOLVING

B.1. SOLVING OF ² sin ¹

b

a

x a x

x

−     ∂

∫  

Making the variable substitution a y ≜ : x

( )

( ) ( ^{( )} )

( )

1 2

2 1 1

2 2

/ 1

3 4 1

/

1 1

3 3

3 1 3

2 / /

sin sin

sin

sin 1

3 3 1

b

a a b

a b

a b a b

a a a

x x y y

x y y

a y y y

y y

a y a y

y

− −

− −

− −

−

 

  ∂ =  −  ∂

     

= − ⋅ ∂

 

= −  −  − ∂

−

 

∫ ∫

∫

∫

(1)

where the last step has been made integrating by parts. The last integral can be solved defining z ≜ 1 − y ² :

( )

( )

1 /

1 3

2 2 2

/ 0

1 1 1

3 1 3 1

a b

a b

y y z

y z

− −

∂ = − ∂

− −

∫ ∫ (2)

Since:

( ^{1 1 z 2} ) ^{2 1 4 ( z 1 1 ) ( 2 z 1 1 ) 2 1 2 z 2}

 

=  + + 

− + −

 

−  

the integral at the second member of the (2) becomes:

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( )

( )

( ) ( )

( )

2

1 /

1 /

1 2 2

0 0

1 1 1 1 1

4 1 1 2

1

a b a b

z tgh

z z z

z

− −

 −   

∂ =  − − − + +     

−  

∫

where:

( )

1 1 1

2 ln 1 tgh x x

x

− +

≜ −

Substituting in the (1):

( )

1 /

3 3

2 1 1 1

2

0

sin sin 1

3 6 1

a b b b

a a

a x a a z

x x tgh

x x z z

−

−    −      −    

∂ = +  + 

       −   

           

∫

B.2. SOLVING OF ∫ x ² x ² − ∂ a ² x The integral can be solved by parts:

( ) ( )

( ) ( )

( )

2 2 2 2 2

3/ 2 3/ 2

2 2 2 2

2 2 3/ 2

2 2 2 2 2 2

3 3

1 1

3 3 3

x x a x x x x a x

x x a x a

x x x a

x x a x a x a x

− ∂ = ⋅ − ∂

− −

= − ∂

= − − − ∂ + − ∂

∫ ∫

∫

∫ ∫

The first integral at the second member is the same of the one

we are trying to solve. Bringing it to the first member:

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( )

( ) ( )

2 2 2 2 2 3/ 2 2 2 2

2 4

2 2 3/ 2 2 2 2 2

1 1

4 4

1 ln

4 8 8

x x a x x x a a x a x

a a

x x a x x a x x a

− ∂ = − + − ∂

= − + − − + −

∫ ∫

B.3. SOLVING OF

2 2

2 1

sin x 2 a

x x

x

− − ∂

∫

Integrating by parts:

( )

2 2 2 2

2 1 2 1

2 2

3 2 2 2

1

2 2 2

sin sin

3 sin 3

x a x a

x x x x

x x

x x a a x

x x a x

− −

−

 

− ∂ = ⋅   −   ∂

 

= − − ∂

−

∫ ∫

∫

(1)

Even the integral at the second member can be solved integrating by parts:

( )

( )

2

2 2 2 2

2 2 2 2

2 2 2

2 2

2 2 ln

x x

x x x

x a x a

x x a x a x

x x a a

x x a

 

∂ =   ∂

−  − 

= − − − ∂

= − + + −

∫ ∫

∫

Substituting, the (1) becomes:

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( )

2 2

2 1

2

3 2 2 3

1 2 2 2 2

2

sin

sin ln

3 6 6

x a

x x

x

x x a a a

x x a x x a

x

−

−

− ∂ =

= − − − + + −

∫