Calculo Aplicado - Lista 1

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1 
 
 
 
 
 
 
FACULDADE DE CIÊNCIA E TECNOLOGIA 
 
CURSO ________________________________________________________________ 
DISCIPLINA: CÁLCULO APLICADO 
ALUNO (A): ______________________________________________________________ 
PROFESSOR(A)__________________________________________________________ 
 
 
Caro aluno, 
 
Pratique o máximo possível e não deixe de consultar o seu Professor nas possíveis 
dúvidas. 
 
 
LISTA DE EXERCÍCIOS I 
 
 
1) Use o conceito de primitiva e verifique se as seguintes integrais indefinidas estão 
corretas: 
 
(a) 
      cxcoslndx xtg
= ln|sec(x)| +C (b) 
  c)x7sen(dx )x7cos(
 
(c)
  cedx
e
x x
x


 
3
3
2
2
2
6
1
 (d)   
 

 cd
6
4sen2
 4sen2)4(cos
3 
(e)
 
   c|tln|lndttln.t
3
 (f) 
  

cxarctg2dx
x1
2
2
 
(g) 
  cedy
y
e y
y (h) sen(3 ) 1
ln |1 cos(3 ) |
1 cos(3 ) 3
t
t C
t

  

 
 
2) Determine: 
 
 a) Uma função f(x) tal que f ‘ (x) + 6 sen(3x) = 0 e f (0) = 5 
 b) A primitiva F(x) da função f (x) = 
3
22
x
 1)-(2x que passa pelo ponto P=(1, 3/2) 
 c) A imagem f 
 
4

, sabendo-se que 
  Cxxxdxx
2x
2
1
cos.sen)f(
 
2 
 
3) Resolva as integrais abaixo usando substituição de variável: 
 
 
a)
dx2 x5
 
C
)2ln(5
2
:.spRe
x5

 
b) 
0a com,dx)axsen( 
 
C
a
)axcos(
:.spRe 
 
c)

 )1x3(sen
dx
2
 
 
 
C
3
)1x3(gcot
:.spRe 


 
d)
 dx)x5cos(
 
 
 
C
5
)x5sen(
:.spRe 
 
e)
  7x3
dx
 
 
 
C7x3ln
3
1
:.spRe 
 
f)
 dx)x2(tg
 
 
 
Cx2cosln
2
1
:.spRe 
 
g)
dxe)e(g(cot xx
 
 
 
C)esen(ln:.spRe x 
 
h)
  xdx.1x
2
 
 
 
C)1x(
3
1
:.spRe 32 
 
i)

 3x2
xdx
2
 
 
 
C3x2
2
1
:.spRe 2 
 
j)
dx
xsen
)x(gcot
2
 
 
 
C
2
xgcot
:.spRe
2

 
l) 

1tgxxcos
dx
2
 
 
 
C1tgx2:.spRe 
 
m)
dx
1x
)1xln(
 

 
 
 
C
2
)1x(ln
:.spRe
2

 
n)

1xsen2
xdxcos
 
 
 
C1xsen2:.spRe 
 
n)

 xsen1
dx)x2sen(
2
 
 
 
Cxsen12:.spRe 2 
 
3 
 
o)

 2x1
xdxarcsen
 
 
 
C
2
xarcsen
:.spRe
2

 
p)

 2
2
x1
xdxarctg
 
 
 
C
3
xarctg
:.spRe
3

 
q)
dx
3x2x
1x
2 

 
 
 
C3x2xln
2
1
:.spRe 2 
 
r)
 xlnx
dx
 
 
 
Cxlnln:.spRe 
 
s)
dx)2x(3 3x4x
2


 
 
 
C
)3ln(.2
3
.spRe
3x4x2

 
t)

 2x21
dx
 
 
 
C)x.2(arctg
2
1
.spRe 
 
u)

 2x916
dx
 
 
 
C
4
x3
arcsen
3
1
.spRe 
 
 
 
 
4) Use integração por partes para resolver as integrais: 
 
 
a)
dxe)x2x( x2 
 
 
 Resp.: x2 ex + C 
b)
  dx)xln()1x4x16(
3
 
 
 Resp.: ln(x).(4x4+2x2+x) - (x4+x2 + x) + C 
 
c)
  xdxsen)1x(
2
 
 
 Resp.: - (x2 –1) cos(x) +2xsen(x) + C 
 
d) 
dx)x3(arctg
 
 
 
C)1x9ln(
6
1
)x3(arctg.x:.spRe 2 
 
e) arcsen(x 2)dx 
 
 
 
C3x4x)2xarcsen()2x(:.spRe 2 
 
4 
 
x
f) dx 
2sen x
 
 
 
C|)xsen(|ln)x(gcotx:.spRe 
 
8 3g) 3x .cos(x )dx
 
 
 
C)xsen(2)xcos(x2)xsen(x:.spRe 33336 
 
35 xh) x (1 4e )dx
 
 
 
C
6
x
3
4x4
e:.spRe
63
x3 







  
2x 1 i) e .dx
 
 
 
Ce)11x2(:.spRe 1x2  
 
 
 
 
 
5) Resolva as integrais contendo um trinômio ax2 + bx + c: 
 
 
2
)
2 5
dx
a
x x 
 
 
 
C
2
1x
arctg
2
1
:.spRe 

 
2
)
6 5
dx
b
x x 
 
 
C
1x
5x
ln
4
1
:.spRe 


 
 
(x 5)dx
 c)
22x 4x 3

 
 
 
 
C)]1x(2[arctg.22|3x4x2|ln
4
1
:.spRe 2 
 
d) 
dx
x4x43
3x
2 

 
 
 
C
2
1x2
arcsen
4
7
x4x43
4
1
:.spRe 2 


 
 
 
 
 
 
 
 
 
 
 
 
 
 
5 
 
6) Resolva as integrais de funções racionais: 
 
 
1
)
2 1
x
a dx
x


 
 
 
C1x2ln
4
1
x
2
1
.spRe 
 
)
( 1)( 3)( 5)
xdx
b
x x x  
 
 
 
C
)1x()5x(
)3x(
ln
8
1
.spRe
5
6


 
2
)
( 1) ( 2)
dx
c
x x 
 
 
 
C
1x
2x
ln
1x
1
.spRe 




 
3 2
8
)
4 4
x
d dx
x x x

 
 
 
 
C
x
2x
ln
2x
3
.spRe
2





 


 
3x 1
e) dx 
34x x


 
 
C|]1x2|ln7|1x2|ln9[
16
1
|x|ln
4
x
:.spRe 
 
2
2
2 3 3
)
( 1)( 2 5)
x x
f dx
x x x
 
  
 
 
 
C
2
1x
arctg
2
1
1x
)5x2x(
ln:.spRe
2
3
2





 


 
3
4 2
6
)
6 8
x
g dx
x x

 
 
 
 
C
2
x
arctg
2
3
2
x
artg
2
3
2x
4x
ln:.spRe
2
2















 
3 2
3 7
)
4 4
x
h dx
x x x

  
 
 
 
C)2/x(arctg
2
1
)1x(
4x
ln:.spRe
2
2


 
4
8x 16
i) dx 
16 x


 
 
 
C
2
x
arctg|x2|lnx4ln:.spRe 2 






 
2
2 2
(x 2x 3)dx
j) 
(x 1)(x 1)
 
 
 
 
 
C
x1
1
|1x|ln1xlnarctgx:spRe 2 


 
3
3 2
(5x 12)dx
 l) 
x 5x 4x

 
 
 
 
C|4|xln
3
83
|1|xln
3
17
xln3x5sp.:Re 
 
2 2
dx
 m) 
(x 9)
 
 
 
C
3
x
arctg
54
1
)9x(18
x
.spRe
2








 
6 
 
2 2
(x 1)dx
 n) 
(x 9)


 
 
 
C
3
x
arctg
54
1
)9x(18
9x
.spRe
2









 
2 2
(2x 3)dx
 o) 
(x 2 10)x

 
 
 
C
3
1x
arctg
54
1
)10x2x(18
17x
.spRe
2





 



 
 
 p) 
24 9
dx
x Cx32
x32
ln
12
1
.spRe 


 
 
 
 
7) Resolva as integrais de funções irracionais: 
 
 
 
( 3)
a) 
2 5
x dx
x x x

 
 
 
 
C
2
1x
arctg22xln2:.spRe 







 

 
3 3
4
)
6
x x
b dx
x


 
 
 
Cx
13
2
x
27
2
:.spRe
12 134 9 
 
 5 26 3
)
( 2) ( 2) 1
dx
c
x x  
 
 
 
C2xarctg3
12x
12x
ln
2
3
:.spRe 6
6
6


 
2
3) .(1 )d x x dx
 
 
 
C)x1(
5
3
)x1(
8
3
:.spRe 3
5
3
8

 
3
)
2
dx
e
x x
 
 
 
C)x2ln(48x24x6x2:.spRe 663 
 
2
1
)
1
x dx
f
x x

C
x
x
xx
xx
sp 



 21
11
11
ln:.Re
 
1
)
1
x dx
g
x x


 
 
 
C
x1x1
x1x1
ln
x1
x1
arctg2:.spRe 




 
 
 
7 
 
8) Resolva as integrais de funções trigonométricas: 
 
 
3) sen ( )a x dx
 
 
 
C)xcos()x(cos
3
1
:spRe 3 
 
2 3) sen ( )cos ( )b x x dx
 
 
 
C)x(sen
5
1
)x(sen
3
1
:spRe 53 
 
3
4
cos ( )
)
sen ( )
x
c dx
x
 
 
 
C)x(csc
3
1
)x(csc:spRe 3 
 
) sec(2 )d x dx
 
 
 
C
)x2sen(1
)x2sen(1
ln
4
1
:spRe 


 
3
3 4
sen
)
cos
xdx
e
x
 
 
 
C
)xcos(
3
)x(cos
5
3
:.spRe
3
3 5 
 
2) sen (3 )f x dx
 
 
 
C
12
)x6(sen
2
x
:.spRe 
 
2 2) sen ( ).cos ( )g x x dx
 
 
 
C
32
)x4(sen
8
x
:.spRe 
 
3)h tg x dx
 
 
 
 
C)xcos(ln
2
xtg
:.spRe
2

 
)
( ) 1
dx
i
tg x 
 
 
 
C
2
x
4
)1)x(tgln(
2
|1)x(tg|ln
:.spRe
2



 
2 2
)
sen
dx
j
x tg x
 
 
 
C
2
tgx
arctg
2
1
)x(gcot
2
1
:.spRe 












 
8 
 
2
2
sen ( )
)
1 cos ( )
x dx
l
x
 
 
 
Cx
2
tgx
arctg2:.spRe 





 
 
 
( )
)
1 ( )
sen x dx
m
sen x
 
 
n) 
1 cos
dx
senx x 
 
 
o) 
(5 ) (3 )sen x sen x dx
 
 
 
 
p) 
( )cos(5 )sen x x dx
 
 
 
Resp.: 
2
1
2
x C
x
tg
 
 
  
 
 
Resp.: 
 
ln 1
2
x
tg C
 
   
  
Resp.: 
 
1 (8 )
(2 )
4 4
sen x
sen x C
 
  
  
Resp.: 
 
cos(6 ) cos(4 )
12 8
x x
C  
 
 
 
 
 
9) Resolva as integrais usando substituição trigonométrica: 
 
 
2 2
2
)
a x
a dx
x


 
 
 
C
a
x
arcsen
x
xa
:.spRe
22



 
2 2) 4b x x dx
 
 
 
Cx4x
4
1
x4x
2
1
2
x
arcsen2:.spRe 232 
 
2 2
)
1
dx
c
x x
 
 
 
C
x
x1
:.spRe
2



 
2 2
)
x a
d dx
x


 
 
 
C
x
a
arccos.aax:.spRe 22 






 
2 5
dx
 e) 
(4 x )
 
 
 
C
x4)x4(3
x
x4
x
16
1
:.spRe
22
3
2












 
9 
 
4 2
dx
f) 
(x+1) . x 2x 10 
 
 
 
C
)1x(3
])1x(9[
)1x(3
)1x(9
:.spRe
35
32
4
2





 
2) 4 x g dx
 
 
 
Cx4
2
x
xx4ln2:.spRe 22 




 
 
h) dx
 
2 2(x 1) x 2x 2  
 
 
i) 
x.arctg(x)
dx
21+x

 
 
 
j) 
2
( 5)
2 4 3
x dx
x x

 
 
 
 
l) 
3 5
(2 1)
x
dx
x x


 
 
 
 
 
C
1x
2x2x
:.spRe
2




 
 
Cx1xln)x(arctgx1:.spRe 22 
 
 
 
2 21Re .: 2 4 3 2 2 ln | 2 4 3 2( 1) |
2
sp x x x x x C       
 
 
 
 
C)xx2(81x4ln
24
23
xx2
2
3
:.spRe 22 