integrais_por_substitui__o_trigonometricas

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Calcular o valor, em função de x, das seguintes integrais aplicando o método de integração por
substituição trigonométrica:
1) I = d
⌠
⌡

1
x
3
 − x
2 9
x ; 
 Solução 
 intervalos: ≤ pi θ e < θ
3 pi
2
 , se x < -3 ; ≤ 0 θ < 
pi
2
 , se x > 3 
 considerando x = 3 ( )sec θ => dx = 3 ( )sec θ ( )tg θ dθ 
 
 temos que: x3 = ( )3 ( )sec θ 3 = 27 ( )sec θ 3 e 
 − x
2 9 = − ( )3 ( )sec θ 2 9 = − 9 ( )sec θ 2 9 = 3 − ( )sec θ 2 1 = 3 ( )tg θ 2 = 
3 ( )tg θ 
 
 substituindo em I , temos:
 
 I = d
⌠
⌡

3 ( )sec θ ( )tg θ
27 ( )sec θ 3 3 ( )tg θ
θ = 
1
27
d
⌠
⌡

1
( )sec θ 2
θ = 
1
27
d
⌠
⌡ ( )cos θ
2 θ = 
1
27
d
⌠
⌡

 + ( )cos 2 θ 1
2
θ = 
 
 = 
1
54
d⌠⌡ ( )cos 2 θ θ + 
1
54
d⌠⌡1 θ = 
1
54
 ( 1
2
 ( )sen 2 θ ) + 1
54
 θ + K = 
θ
54
 + 
1
54
 
( )sen θ ( )cos θ + K
 
 temos que: x = 3 ( )sec θ = 3( )cos θ => ( )cos θ = 
3
x
 
 temos também: − x2 9 = 3 ( )tg θ => ( )tg θ = − x
2 9
3
 => θ = 
Page 1
 − 





arctg − x
2 9
3
 
 fazendo: ( )sen θ = ( )cos θ ( )sen θ( )cos θ = ( )cos θ ( )tg θ => ( )sen θ = 
3
x
 
 − x
2 9
3
 = 
 − x
2 9
x
 
 substituindo estes valores em I , temos:
 I = 
1
54
 





arctg − x
2 9
3
 + 
1
54
 
 − x
2 9
x
 
3
x
 + K = 
 = 
1
54
 





arctg − x
2 9
3
 + 
1
18 x2
 − x
2 9 + K 
 ------------------------------------------
2) I = d
⌠
⌡

12 x3
 + 2 x2 7
x ; 
 Solução 
 intervalos: ≤ 0 θ e < θ
pi
2
 , se ≤ 0 x ; −
pi
2
 < θ < 0 , se x < 0 
 I = d
⌠
⌡

12 x3
 + 2 x2 7
x = d
⌠
⌡

12 x3
 + 
2 x2
7
1
x 
 considerando x = 
7
2
 ( )tg θ => dx = 7
2
 ( )sec θ 2 dθ 
 
 temos que: 12x3 = 12




7
2
( )tg θ
3
 = 42 
7
2
 ( )tg θ 3 e 
Page 2
 
 + 2 x2 7 = + 2




7
2
( )tg θ
2
7 = + 7 ( )tg θ 2 7 = 7 ( ) + ( )tg θ 2 1 = 
 = 7 + ( )tg θ 2 1 = 7 ( )sec θ 2 = 7 ( )sec θ 
 
 subtituindo em I , temos:
 I = d
⌠
⌡

42
7
2
( )tg θ 3 7
2
( )sec θ 2
7
2
( )sec θ
θ = 42 
7
2
7
 d
⌠
⌡ ( )tg θ
3 ( )sec θ θ = 21 7 
d
⌠
⌡ ( )tg θ
3 ( )sec θ θ = 
 = 21 7 d
⌠
⌡( ) − ( )sec θ
2 1 ( )tg θ ( )sec θ θ = 21 7 
d
⌠
⌡ − ( )sec θ
2 ( )tg θ ( )sec θ ( )tg θ ( )sec θ θ = 
 = 21 7 d
⌠
⌡ ( )sec θ
2 ( )sec θ ( )tg θ θ −21 7 d⌠⌡ ( )sec θ ( )tg θ θ 
 considerando u = ( )sec θ => du = ( )sec θ ( )tg θ dθ 
 sutstituindo , temos:
 
 I = 21 7 d
⌠
⌡u
2
u −21 7 d⌠⌡1 u = 21 7 
u
3
3
 −21 7 u + K 
 sustituindo u por seu valor inicial, temos:
 I = 21 7 
( )sec θ 3
3
 −21 7 ( )sec θ + K = 7 7 ( )sec θ 3 −21 7 ( )sec θ + K 
 
 como temos que + 2 x2 7 = 7 ( )sec θ => ( )sec θ = 7 + 2 x
2 7
7
 
 substituindo este valor em I, temos:
Page 3
 I = 7 7





7 + 2 x
2 7
7
3
 −
21 7 7 + 2 x2 7
7
 = 7 7
7
3
 + 2 x2 7
3
73
 
−21 + 2 x2 7 = 
 = − + 2 x2 7
3
21 + 2 x2 7 + K = − ( ) + 2 x2 7 + 2 x2 7 21 + 2 x2 7 + K = 
 
 = − 2 x2 + 2 x2 7 14 + 2 x2 7 + K 
 
 ------------------------------------------
3) I = d
⌠
⌡

 − − x
2 2 x 3
 + x 1
x ; 
 
 Solução 
 considerando x + 1 = u => x = u - 1 => dx = du 
 substituindo em I , temos:
 I = d
⌠
⌡

 − − ( ) − u 1 2 2 ( ) − u 1 3
u
u = d
⌠
⌡

 − + − + − u2 2 u 1 2 u 2 3
u
u = 
 = d
⌠
⌡

 − u
2 4 u
u
u = d
⌠
⌡

 − u
2 4 u
u − u
2 4 u
u = d
⌠
⌡

 − u 4
 − u
2 4 u
u = d
⌠
⌡

 − u 4
 − ( ) − u 2 2 4
u 
 no intervalo 0 < θ < 
pi
2
 
 podemos considerar 
 − u 2 = 2 ( )sec θ => u = 2 ( )sec θ + 2 => du = 
2 ( )sec θ ( )tg θ dθ 
 assim, temos: − u 4 = − 2 ( )sec θ 2 
 
Page 4
 e − ( ) − u 2 2 4 = − ( ) + − 2 ( )sec θ 2 2 2 4 = − 4 ( )sec θ 2 4 = 
 = 2 − ( )sec θ 2 1 = 2 ( )tg θ 2 = 2 ( )tg θ 
 subtituindo estes valores em I , temos:
 I = d
⌠
⌡

( ) − 2 ( )sec θ 2 2 ( )sec θ ( )tg θ
2 ( )tg θ θ = d
⌠
⌡ − 2 ( )sec θ
2 2 ( )sec θ θ = 
 − 2 d
⌠
⌡ ( )sec θ
2 θ 2 d⌠⌡ ( )sec θ θ = 
 
 = 2 ( )tg θ −2 ln( | sec(θ) + tg(θ) | ) + K
 mas temos que: − u 2 = 2 ( )sec θ => sec(θ) = − u 2
2
 
 e que: − ( ) − u 2 2 4 = 2 tg(θ) => tg(θ) = − ( ) − u 2
2 4
2
 
 substituindo em I , temos:
 I = 2 
 − ( ) − u 2 2 4
2
 −2 ln( | − u 2
2
 + 
 − ( ) − u 2 2 4
2
 | ) + K 
 mas temos que: u = x + 1
 substituindo em I , temos:
 I = 
 − ( ) + − x 1 2 2 4 −2 ln( | + − x 1 2
2
 + 
 − ( ) + − x 1 2 2 4
2
 | ) + K = 
 = − ( ) − x 1 2 4 −2 ln( | − x 1
2
 + 
 − ( ) − x 1 2 4
2
 | ) + K = 
 = − − x
2 2 x 3 −2 ln( | − + x 1 − − x
2 2 x 3
2
 | ) + K 
 
 ----------------------------------------
Page 5
4)
 I = d
⌠
⌡

1
( ) − − x2 2 x 3




3
2
x ; 
 Solução 
 I = d
⌠
⌡

1
( ) − − x2 2 x 3




3
2
x = d
⌠
⌡

1
[ ] − ( ) − x 1 2 4




3
2
x = d
⌠
⌡

1
[ ] − ( ) − x 1 2 4
3
x 
 
 no intervalo ≤ 0 θ e θ < 
pi
2
 
 podemos considerar 
 − x 1 = 2 sec(θ) => x = 2 sec(θ) + 1 => dx = 2 sec(θ) tg(θ) 
dθ 
 e [ ] − ( ) − x 1 2 4
3
 = − ( )2 ( )sec θ 2 4
3
 = 23 − ( )sec θ 2 1
3
 = 8 ( )tg θ 2
3
 = 
8 ( )tg θ 3 
 substituindo estes valores em I , temos:
 I = d
⌠
⌡

2 ( )sec θ ( )tg θ
8 ( )tg θ 3
θ = 
1
4
 d
⌠
⌡

( )sec θ
( )tg θ 2
θ = 
1
4
 d
⌠
⌡

( )cos θ
( )sen θ 2
θ 
 considerando u = sen(θ) => du = cos(θ) dθ 
 substituindo em I , temos:
 I = 
1
4
 d
⌠
⌡

1
u
2
u = −
1
4 u
 + K
 substituindo u por seu valor inicial, temos:
Page 6
 I = −
1
4 ( )sen θ + K
 como temos que: − x 1 = 2 sec(θ) = 2( )cos θ => cos(θ) = 
2
 − x 1
 
 e que: [ ] − ( ) − x 1 2 4
3
 = 8 ( )tg θ 3 = 8 


( )sen θ
( )cos θ
3
 => 8 
( )sen θ
( )cos θ = 
 − ( ) − x 1 2 4 => 
 => 8 sen(θ) = ( )cos θ − ( ) − x 1 2 4 = 2
 − x 1
 − ( ) − x 1 2 4 = 2 
 − ( ) − x 1 2 4
( )
 − x 1 2
 = 
 = 2 − 1




2
 − x 1
2
 => sen(θ) = 1
4
 − 1




2
 − x 1
2logo, I = 
1
16
 − 1




2
 − x 1
2
 + K 
 
 -----------------------------------------
5) I = d
⌠
⌡

1
( ) + x 2 + + x2 4 x 3
x ; 
 Solução 
 
 I = d
⌠
⌡

1
( )
 + x 2 + + x2 4 x 3
x = d
⌠
⌡

1
( )
 + x 2 − ( ) + x 2 2 1
x 
 
 no intervalo 0 < θ < 
pi
2
 
Page 7
 podemos considerar x + 2 = sec(θ) => x = sec(θ) −2 => dx = sec(θ) tg(θ) dθ 
 e − ( ) + x 2 2 1 = − ( )sec θ 2 1 = ( )tg θ 2 = tg(θ) 
 substituindo estes valores em I , temos:
 I = d
⌠
⌡

( )sec θ ( )tg θ
( )sec θ ( )tg θ θ = d
⌠
⌡1 θ = θ + K 
 como temos que: x + 2 = sec(θ) => θ = arcsec( x + 2 )
 
 logo, I = arcsec( x + 2 ) + K 
 ----------------------------------------
 
============================================
 Jailson Marinho Cardoso
Aluno do curso de Matemática
Universidade Federal da Paraíba
Campus I 
31/07/2000 
============================================
Page 8