Resolução de exercícios do capítulo 6 do livro Cálculo A PDF

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Resolução de exercícios do capítulo 6, seção 6.4 do livro Cálculo A, sexta edição...

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6.4 – EXERCÍCIOS – pg. 250 Calcular as integrais seguintes usando o método da substituição.

1.

 ( 2x

2

+ 2x − 3)10 (2 x + 1) dx

Fazendo − se : u = 2 x 2 + 2x − 3 du = (4 x + 2)dx = 2(2x + 1)dx Temos : 2 10  (2 x + 2 x − 3) (2 x + 1) dx =

2.

 (x

3

1 (2 x 2 + 2 x − 3) 11 + c. 2 11

− 2)1 / 7 x 2 dx

Fazendo − se : u = x3 − 2 du = 3x 2 dx Temos :

 ( x − 2) 3

3.



1/ 7

(

1 x3 − 2 x dx = 8 3 7 2

8 7

)

+c =

7 3 x −2 24

(

8 7

)

+ c.

x dx 5

x2 − 1

 (x

2

−1

− 1) 5 x dx

Fazendo − se : u = x 2 −1 du = 2 x dx Temos :

 4.

5

 5x

(

)

1 x2 −1 = 4 x2 − 1 2 5

x dx

4/5

+c =

4 5 2 (x − 1) 5 + c 8

4 − 3x2 dx

451

(

1

)

1

=  5x (4− 3x2 )2 dx =  5x 4 − 3x2 2 dx Fazendo − se : u = 4 − 3 x2 du = −6 x dx Temos :

(

− 1 4 − 3x 2 5 4 3 5 . x x dx − =  3 6 2 2

5.



x2 + 2 x4 dx

=

 x (1 + 2 x ) 2

(

1 1 + 2x 2 = 3 4 2 1 = 1 + 2x 2 6

(

6.

 (e

2 t

1

3 2

+c=

+c

3 2

3 −5 4 − 3 x 2 2 + c. 9

(

)

dx

3 2

) )

1 2

)

Fazendo:

u = 1 + 2x 2 du = 4x dx

+c

t

+ 2) 3 e2 dt

Fazendo − se : u = e 2t + 2 du = 2 e2 t dt Temos :

(

1 e2 t + 2 3 e e dt + = ( 2 )  4 2 3 2t

7.

2t

4 3

)

+ c=

4 3 2t e + 2 3 + c. 8

(

)

e tdt  et + 4 =

8.

1

du

u

t

t

t

= ln e + 4 + c , sendo que u = e + 4 e du = e dt.

e1 / x + 2  x 2 dx

452

1 1 x −1 1 2 −2 x dx x dx e + = − + + c = −e x − + c. 2 2 . 2  x x −1 Considerando - se : 1

= e x

1

u = ex

1

du = ex . 9.

tg x sec =

10.

2

x dx

tg 2x + c . considerando-se: 2

 sen =

11.

−1 . x2

u = tg x du = sec 2 x dx

4

x cos x dx

sen 5x +c 5

considerando-se:

u = sen x du = cos x dx

sen x dx 5 x

 cos

=  cos −5 x . sen x dx =−

cos −4 x 1 = +c −4 4 cos4 x

utilizando:

u = cos x du = − senxdx

1 = sec 4 x + c 4 12.



2 sen x− 5 cos x dx cos x sen x − 5  dx cos x = − 2 ln | cos x |− 5x + c = 2

13.

e

x

utilizando:

u = cos x du = − senxdx

x cos 2 e dx

1 = sen 2e x + c . 2 Considerando - se : u = 2 ex du = 2e x dx.

453

14.

x

 2 cos x

2

dx

11 1 sen x 2 + c = sen x 2 + c 22 4 Considerando - se : =

u = x2 du = 2 x dx. 15.

 sen (5θ − π )dθ 1 = − cos (5θ − π )+ c. 5 Considerando - se : u = 5θ − π du = 5dθ .

16.

arc sen y

2

1−y 2

dy

1 (arc sen y )2 1 2 + c = (arc sen y ) + c. 2 2 4 Considerando - se : u = arc sen y =

du =

17.

1 1− y 2

dy .

2 sec 2 θ

 a + b tg θ dθ 1 = 2. ln | a + b tg θ | +C b Considerando-se: u = a + b tgθ du = b. sec 2 θ d θ

18.

dx

 16 + x

2

454

1 =  16

19.

20.

y

dy −4y + 4

=

u = y− 2 ( y − 2)− 1 1 dy − 2 y dy = − = +c= ( 2 ) + c , utilizando:  (y − 2)2  du = dy −1 2− y



sen θ cos θ dθ

3

=

21.

2

x u= 1 1 x x 4 = 4 arc tg + c = arc tg + c , utilizando: 2 16 4 4 4 1 x du = dx 1+   4 4 dx

1/ 3  (senθ ) cos θ

dθ =

4 ( senθ ) 4 / 3 3 + c = sen 3θ + c. 4 4 3

ln x 2  x dx

(

)

2

1 ln x 2 1 1 2 2 + c = (ln x2 )2 + c = 4 (ln x ) + c = (ln x ) + c. 2 2 4 4 Considerando - se : u = ln x 2 2x 2 du = 2 = dx. x x

22.

 (e

ax

+e

− ax 2

) dx

(

1 1 2 ax e + 2 x − e −2 ax + c 2a 2a sen h 2 ax + 2x + c = + 2 x + c. a

)

2

=  e2 ax + 2 + e−2 ax dx =

23.

=

1 2 ax − 2 ax e −e 2a



3t 4 + t 2 dt

(

)

455

(

)

3

1 3t 2 + 1 2 =  t 3 t + 1 dt =  t 3t + 1 dt = +c 3 6 2 3 3 1 3 1 = . . 3t 2 + 1 2 + c = . 3t 2 + 1 2 + c. 6 2 9 2

(

)

2

(

)

(

2

(

)

1 2

)

Considerando-se: u = 3t 2 + 1 . du = 6t dt 24.

 4x

2

4 dx + 20x + 34

5 5   2x+  x +  1 2 2 2 + c. arc tg  = + c = arc tg  2 2 = 3 3 3 3 5  3   x+  +   2 2 2  2   dx

25.

x

2

3 dx − 4x+ 1

dx dx dx 3 1 3 = = −3  2 2 = −3  3 (x − 2 ) 3 (x − 2 ) − 3 (x − 2)2 − 1− 3 3 3 x−2 1 x+ 3−2 3 +c = − 3 = − 3 ln + c. ln x − 2 2 1− 2 3+ 2−x 3 1+

Considerando-se: u2 = u=

(x − 2 )2

3 x−2

du =

3 1 3

dx

Resposta alternativa:

456

arc tg h

x−2 3

arc cot g h

26.

se

x −2 1. 3

e x dx  e2x +16 1 ex = arc tg + c 4 4 Considerando-se: u 2 = e 2x u = ex du = e2 xdx

27.



x +3 dx x −1

2 u2 4  du  u   u du = . 2  u 2 −3 −1  u2 − 4 du = 2  1 + u2 − 4  du = 2 u + 4  u 2 − 4 u − du 1+ du 1 2 +c u = 2u + 8  4 2 = 2u − 2  ln 2 = 2 − 2 .2 u 2 4 u u   1− − 1−  2 4 4 2  

=

= 2u − 2 ln

2+u 2+ + c = 2 x + 3 − 2 ln 2−u 2−

x+ 3 x+ 3

+ c.

Considerando-se:

u2 = x +3 x = u2 − 3 dx = 2u du

28.

3 dx 2 3x

 x ln

457

=  ( ln 3x)

−2

3dx (ln 3x ) + c = − 3 + c. =3 x −1 ln 3 x

Considerando-se: u = ln 3x 3 du = dx 3x 29.

(sen 4 x + cos 2π ) dx =  sen 4 x dx + cos 2π  dx =

30.

2

x 2 +1

1 (− cos 4x )+ x cos 2π + c . 4

x dx 2

2

1 2 x +1 2x = + c= + c. 2 ln 2 ln 2 Considerando-se: u = x 2 +1 du = 2 x dx 31.

x e

3 x2

=

dx

1 3x 2 e +c 6

Considerando-se: u = 3x 2 du = 6 x dx 32.

dt

 (2 + t )

2

=  (2 + t ) 2 = −

(2 + t) −1

−1

+c =

−1 +c . 2 +t

Considerando-se: u = 2 +t du = dt

458

33.

dt

 t ln t = ln ln t + c. u = ln t Considerando-se:

34.

 8x

du =

dt t

1 − 2 x2 dx 3

(

)

− 1 1 − 2 x2 2 −4 +c= =8 1 − 2x2 3 4 3 2

(

)

3 2

+ c.

Considerando-se: u = 1 − 2 x2 du = −4 x dx 35.

 (e

2x

x

+ 2)5 e2 dx

(

)

6

1 e2 x + 2 1 2x 6 e + 2 +c. = +c = 2 6 12

(

)

Considerando-se: u = e 2x + 2 du = 2e 2 x dx 36.

 =

4t dt 4t 2 + 5

(

−

)

4t2 − 5

(

1 2

4t dt

1

)

1 4 t2 + 5 2 = + c = 4t 2 +5 + c. 1 2 2 Considerando-se: u = 4t 2 + 5 du = 8t dt

459

37.

cos x

 3 − sen x dx = − ln | 3 − sen x | + c Considerando-se: u = 3 − sen x du = − cos x dx

38.



dv v (1 + v )5

(1 + v )

−4

=2

+c

−4 1

=−

(

2 1+ v

)

4

+c

Considerando-se: u = 1+ v 1 du = dv 2 v 39.

x

2

1 + x dx

Considerando-se:

1 + x = u2 x = u2 − 1 

x

2

1+ x dx =

dx = 2u du

 (u

2

)

2

− 1 u 2 u du =

 (u

4

)

− 2 u2 + 1 2 u2 du

u7 u5 u3 −4 +2 +c 7 5 3 2 = (1 + x )7 − 4 (1 + x )5 + 2 (1 + x )3 + c 3 5 7 2 4 2 = (1 + x )3 1 + x − (1 + x )2 1 + x + (1 + x) 1 + x + c. 7 5 3

(

)

=  2u 6 − 4u 4 + 2u 2 du = 2

40.

x e

4 − x5

dx

460

=

− 1 − x5 e +c 5

Considerando-se: u = −x 5 du = −5x 4 41.

 t cos t dt 2

=

42.

1 u = t2 sen t 2 + c , utilizando: 2 du = 2tdt

8 x

2

6 x 3 + 5 dx 3

(

)

3 3 1 6 x3 + 5 2 4 2 8 =8 6x 3 + 5 2 + c = 6x 3 + 5 2 + c . +c = 3 18 9 3 27 2

(

)

(

)

Considerando-se: u = 6 x3 + 5 du = 18 x 2 dx 43.

 sen

1/ 2

2θ cos 2θ dθ 3

1 1 (sen 2θ )2 = + c = (sen2θ )3 / 2 + c . 3 3 2 2 Considerando-se: u = sen 2θ du = 2 cos 2θ dθ 44.

sec (5 x + 3)dx 2

1 = tg (5x + 3) + c . 5 Considerando-se:

461

u = 5x + 3 du = 5dx 45.

sen θ d θ

 (5 − cosθ )

3

− 5 − cos θ) 2 ( = +c.

−2

Considerando-se: u = 5 − cos θ du = sen θ dθ 46.

 cot g u du =

cos u du = ln | sen u | +c sen u u = sen u

Considerando-se: 47.

 (1 + e

− at 3 / 2

)

e− at dt,

(

− 1 1 + e−at = 5 a 2

du = cos u du a>0

5 2

)

+c= −

2 −at 1+ e 5a

(

5 2

) + c.

Considerando-se:

u = 1 + e − at du = e− at (− a ) dt 48.



cos x dx x

= 2 sen x + c . Considerando-se: u= x du =

1 2 x

dx

462

49.

t

t − 4 dt

(

)

(

)

=  u 2 + 4 .u 2u du =  2u 4 + 8u 2 du = 2

u5 u3 +8 +c 5 3

2 (t − 4)5 + 8 (t − 4)3 + c 5 3 2 8 2 = ( t − 4) t − 4 + (t − 4) t − 4 + c 5 3 =

Considerando-se:

t − 4 = u2 t = u2 + 4  50.

x

2

dt = 2 u du

(sen 2 x3 + 4 x) dx

=  x 2sen 2 x 3dx +  4x 3dx =

x4 −1 −1 cos 2 x 3 + 4 + c = cos 2 x 3 + x 4 + c , x 6 6

sendo que na primeira integral usamos: u = 2 x3 du = 6 x2 dx

463...