Folha de exercícios nº 2

Departamento de Matemática e Engenharias
MATEMÁTICA I
Licenciaturas em Bioquímica e Química
1o Semestre 2005/2006
Resoluções da folha de exercícios no 2 - Primitivas e Integrais
Nas resoluções seguintes, c e k representam constantes reais arbitrárias.
1. Uma primitiva de f (x) é uma função F (x) tal que F ′ (x) =
f (x). O integral de f (x) é o
conjunto de todas as primitivas de f (x) e representa-se por f (x) dx. Assim, f (x) dx =
F (x) + c, onde F (x) é uma primitiva de f (x) e c ∈ R.
2. Integrais imediatos:
(a) 5e3x dx = 35 3e3x dx = 35 e3x + c
(b)
(c)
(d)
√
1
3
x dx = x 3 dx =
=
(e)
(f)
(g)
(h)
(i)
(j)
4
1
4
x3 ex dx =
cos(2x)
dx
(1+sin(2x))2
+c=
ex
4
1
4+9x2 dx
x
dx
1+4x4
=
=
=
2e2x +ex
1+e2x dx
√ 3x
dx
4x2 +3
+c=
3
4
√
3
x4 + c
cos (2x) (1 + sin (2x))−2 dx =
1
2
1
+ c = − 2(1+sin(2x))
+c
1
x
4
=
1
4
3(1+ln x) 3
4
2 cos (2x) (1 + sin (2x))−2 =
1
2
1+( 32 x)
+c
dx =
1
4
2
1
4x
+c
2 dx = 4 arctg 2x
2
1+(2x )
2e2x
1+e2x dx +
4
1
(1 + ln x) 3 dx =
1
dx
4(1+ 94 x2 )
1
=
4
3
+c
2
ex dx = e x dx = e x2 + c
√
3
1+ln x
dx
x
4
x3
4
4
4x3 ex dx =
=
−1
1 (1+sin(2x))
2
−1
1
x 3 +1
1
+1
3
ex
dx
1+(ex )2
− 1
= 3 x 4x2 + 3 2 dx =
1
3
8
×
2
3
1+(
3
2
2
3
x
2
)
dx = 16 arctg
= ln 1 + e2x + arctg (ex ) + c
3 2x + c
√
− 1
8x 4x2 + 3 2 dx = 34 4x2 + 3 + c
(k)
(l)
x
ae ex dx =
1
ln a
sin x cos x dx =
x
ln a ae ex dx =
sin2 x
2
x
ae
ln a
+c
+c
2
sin x cos x dx = − − sin x cos x dx = − cos2 x + +l, l ∈ R
Ou sin x cos x dx = 21 2 sin x cos x dx = 12 sin (2x) dx = − 14 −2 sin (2x) dx =
+k
= cos(2x)
4
x
(m) cotg x dx = cos
sin x dx = ln |sin x| + c
Ou
(n)
(o)
(p)
sin2 x dx =
20
(1+x2 )arctgx
1−cos(2x)
dx
2
dx = 20
=
1
1+x2
1
1
2 dx − 2
arctgx dx
cos (2x) dx =
x
2
−
sin(2x)
4
+c
= 20 ln |arctgx| + c
1+(arcsin
3x)2
3x)2
1
√
√
dx = √1−9x
dx + (arcsin
dx =
2
1−9x2
1−9x2
3x)3
2
3
= 31 √ 3 2 dx + 13 √1−9x
dx = 13 arcsin 3x + (arcsin
2 (arcsin 3x)
9
1−(3x)
3x3 +14x2 +7x−3
x+4
+c
1
3x2 + 2x − 1 + x+4
dx = x3 + x2 − x + ln |x + 4| + c
√ √2
1
1
3
2 dx =
dx = 43 · 23
(r) x2 +x+1
dx = 3 1 1 2 dx = 43
1 2
4
1
2
x+
+
x+
1+
)
(
)
(
√
√
x+
1+
4
2
3
2
3
3
= √23 arctg √23 x + √13 + c
(q)
(s)
(t)
dx =
sin x cos (cos x) dx = − − sin x cos (cos x) dx = − sin (cos x) + c
earctgx +x ln(1+x2 )+1
1+x2
= earctgx +
ln2 (1+x2 )
4
dx =
earctgx
1+x2
dx +
+ arctgx + c
1
2
2x ln(1+x2 )
1+x2
dx +
1
1+x2
dx =
3. Integração por partes:
x
x x4x dx = ln4 4 x − ln14 4x dx = ln4 4 x − ln14 + c
(b) ex x2 − 4 dx = ex x2 − 4 − 2 xex dx = ex x2 − 4 − 2 xex − ex dx =
= ex x2 − 2x − 2 + c
(c) x sec2 x dx = x tan x − tan x dx = x tan x + ln |cos x| + c
(a)
(d)
(e)
−
x cosh (3x) dx = x sinh(3x)
3
ln x dx =
1
3
1 · ln x dx = x ln x −
sinh (3x) dx = x sinh(3x)
− 19 cosh (3x) + c
3
x x1 dx = x (ln x − 1) + c
2
(f)
=
(g)
(h)
sin4 x cos3 x dx =
sin5
x
5
cos2 x −
2
35
sin4 x cos x cos2 x dx =
sin7
x
7
+c
cos2 x −
2
5
sin6 x cos x dx =
2x
e2x cos (3x) dx = e2 cos (3x) + 23 e2x sin (3x) dx=
2x
2x
= e2 cos (3x) + 32 e2 sin (3x) − 23 e2x cos (3x) dx =
2x
= e2 cos (3x) + 34 e2x sin (3x) − 49 e2x cos (3x) dx
2x
2x
Então, 1 + 49
e cos (3x) dx = e2 cos (3x) + 34 e2x sin (3x) ⇔
2x
⇔ 13
e2x cos (3x) dx = e2 cos (3x) + 43 e2x sin (3x) ⇔
4
2x
⇔ e2x cos (3x) dx = e13 (2 cos (3x) + 3 sin (3x)) + c
3
√
2
x+
x 2 − 5x dx = − 15 (2−5x)
3
2
3
2
2
− 15
4
375
2
15
5
3
(2 − 5x) 2 dx =
(2 − 5x) x −
(2 − 5x) 2 + c
√
√
x
2
1
arcsin x dx = − 12 1−x
1 − x2 √1−x
(i) √
arcsin x +
1
2 dx =
2
2
1−x
√
= − 1 − x2 arcsin x + x + c
x
1
2 +c
(j) arctg x dx = 1 · arctg x dx = xarctg x − 1+x
2 dx = xarctg x − 2 ln 1 + x
=
4.
sin5 x
5
ax
eax sin (bx) dx = ea sin (bx) − ab eax cos (bx) dx =
ax
ax
= ea sin (bx) − ab ea cos (bx) + ab eax sin (bx) dx =
2 ax
ax
= ea sin (bx) − bea2 cos (bx) − ab 2 eax sin (bx) dx
2 ax
ax
Assim, eax sin (bx) dx + ab 2 eax sin (bx) dx = ea sin (bx) − bea2 cos (bx) ⇔
ax
ax
2
2 eax sin (bx) dx = ea sin (bx) − bea2 cos (bx) ⇔
⇔ a a+b
2
ax
beax
eax
⇔ eax sin (bx) dx = aae
2 +b2 sin (bx) − a2 +b2 cos (bx) + c = a2 +b2 (a sin (bx) − b cos (bx)) + c
5. Método de substituição:
(a)
t
Sustituição: t = ln x ⇔ x = et ⇒ dx
dt = e
t
√1
√ 1 dt = arcsin t + c = arcsin (ln x) + c
√ 1 2 dx =
t
2 e dt =
2
x
(b)
1
dx
1−ln2 (x)
√
x
e
1−ln (x)
√1+√x
dx
obtemos:
√1+√x
dx =
√
x
1−t
Vamos utilizar a substituição t =
√
x
√
1+t
t 2t
√ 3
= 34 (1 + x) 2 + c
1−t
√
x ⇔ x = t2 ⇒
dx
dt
= 2t. Com esta substituição,
3
1
2
dt = 2 (1 + t) 2 dt = 2 (1+t)
+c=
3
2
3
4
3
3
(1 + t) 2 + c =
(c)
tan4 (x) dx
Substituição: t = tan x ⇔ x = arctg t ⇒
t4
1
dt =
tan4 (x) dx = t4 1+t
2 dt =
1+t2
dx
dt
=
1
1+t2
4
t
1
2
(efectuando a divisão dos polinómios, obtemos 1+t
2 = t − 1 + 1+t2 )
2
1
t3
1
t dt − 1dt + 1+t
= t2 − 1 + 1+t
2 dt =
2 dt = 3 − t + arctg t + c =
(d)
=
tan3 (x)
3
√ 1
4−x2
− tan x + x + c
dx
Substituição: x = 2 sin t ⇒
√ 1
4−x2
dx =
2 cos t
4−(2 sin t)2
√
= 1 dt = t + c = arcsin
e2x +2e3x
(e)
1−ex dx
x
2
dx
dt
= 2 cos t
dt = √ 2 cos t
2
4−(2 sin t)
dt =
+c
√ 2 cos t 2 dt =
4−4 sin t
1
Utilizamos a substituição: ex = t ⇔ x = ln t ⇒ dx
dt = t
(ex )2 +2(ex )3
2 +2t3 1
t2 (1+2t) 1
e2x +2e3x
dx = t 1−t
1−ex dx =
1−ex
t dt =
1−t t dt =
√2 cos t 2 dt =
2
2t2 +t
1−t dt
1−sin t
=
2
+t
3
(efectuando a divisão dos polinómios, obtemos 2t1−t
= −2t − 3 + 1−t
)
3
1
−1
= −2t − 3 + 1−t
dt = − 2t dt − 3 dt + 3 1−t
dt = −t2 − 3t − 3 1−t
dt =
= −t2 − 3t − 3 ln |1 − t| + c = −e2x − 3ex − 3 ln |1 − ex | + c
1
dx
(f) √1−x2 arcsin(x)
Substituição: arcsin (x) = t ⇔ x = sin t ⇒ dx
dt = cos t
1
cos(t)
1
√
√ 12
dt
=
dx
=
cos
(t)
dt
=
2
t dt = ln |t| + c =
cos(t)
t
1−x arcsin(x)
1−sin (t) t
= ln |arcsin (x)| + c
(g)
(h)
(1 + sin (x))9 cos (x) dx
√ 1
Substituição: sin (x) = t ⇔ x = arcsin t ⇒ dx
dt = 1−t2
√
1
dt =
(1 + sin (x))9 cos (x) dx = (1 + sin (x))9 1 − sin2 (x)dx = (1 + t)9 1 − t2 √1−t
2
10
10
= (1 + t)9 dt = (1+t)
+ c = (1+sin(x))
+c
10
10
1
1+sin x
dx
Substituição: t = tan x2 ⇔ x = 2 arctan t ⇒
dx
dt
=
2
1+t2
Com esta substituição sin x = 2 sin x2 cos x2 = 2 tan x2 cos2
1
1+sin x
dx =
= − tan 2x +1 + c
2
1
1+ 2t 2
1+t
2
dt
1+t2
=
2
dt
1+t2 +2t
4
x
2
= 2 tan x2 sec12 x =
2
2
= 2 (t + 1)−2 dt = − t+1
+c =
2t
1+t2
(i)
√
x2 1 + x dx
√
Substituição: 1 + x = t ⇔ 1 + x = t2 ⇔ x = t2 − 1 ⇒ dx
dt = 2t
2√
4
2
2
x 1 + xdx =
t − 1 t 2t dt = 2 t − 2t2 + 1 t2 dt = 2 t6 − 2t4 + t2 dt =
√
√
√
7
7
5
3
( 1+x)
( 1+x)
( 1+x)
t5
t3
t
−2
+
=2 7 −25 + 3 +c=2
+c=
7
5
3
5
2
7
2
3
2
− 4 (1+x)
+ 2 (1+x)
+c
= 2 (1+x)
7
5
3
(j)
arcsin x dx
Substituição: t = arcsin x ⇔ x = sin t ⇒
dx
dt
= cos t
arcsin x dx = t cos t dt = t sin t − sin t dt = t sin t + cos t + c =
√
= t sin t + 1 − sin2 t + c = x arcsin x + 1 − x2 + c
√
3
tan2 (x)
(k) f (x) = cos2 (x)
Começamos por recordar que cos21(x) = sec2 (x) = 1 + tan2 (x) e assim podemos
escrever:√
3
tan2 (x)
f (x) = cos2 (x) =
1
cos2 (x)
2
3
tan2 (x) = 1 + tan2 (x) (tan (x)) 3
1
Podemos, assim, utilizar a substituição: t = tan (x) ⇔ x = arctan t ⇒ dx
dt = 1+t2
3
2 1
√
2
tan2 (x)
1 + tan2 (x) (tan (x)) 3 dx =
1 + t2 t 3 1+t
dx =
f (x) dx =
2 dt =
cos2 (x)
=
5
2
t 3 dt =
√
t3
5
3
5
+ c = 35 t 3 + c =
3
5
5
(tan (x)) 3 + c =
3
5
3
tan5 (x) + c
2
(l) f (x) = xx+3
√
√
Substituição: t = x2 + 3 ⇔ x2 + 3 = t2 ⇔ x = t2 − 3 ⇒
2
√x2 +3
dx = √t2t−3 √t2t−3 dt = t2t−3 dt =
f (x) dx =
x
dx
dt
=
√ t
t2 −3
2
(efectuando a divisão dos polinómios, obtemos t2t−3 = 1 + t23−3 )
1
= 1 + t23−3 dt = 1dt + t23−3 dt = t − 3 3−t
(*)
2 dt
1
Calculemos agora, separadamente, a primitiva, 3−t2 dt.
1
1
dt =
dt. Podemos, assim, utilizar a substituição:
√ 2
3−t2
( 3) −t2
√
3 cos u, obtendo-se:
√ √ √
1
1
1
3 cos u du = 33 cos12 u cos u du = 33 cos1 u du =
√
2u
2 dt = 3
1−sin
3−( 3 sin u)
√ √ √ 3
u−tan u
sec2 u−sec u tan u
du =
= 33 sec u du = 33 sec u sec
sec u−tan u du = 3
sec u−tan u
√ √
√
u tan u−sec2 u
u
=
= − 33 sec sec
du = − 33 ln |sec u − tan u| = − 33 ln 1−sin
u−tan u
cos u
t=
√
3 sin u ⇒
dt
du
=
5
=
√
− 33
t
√
1−sin u √
1−
3
3
√
=
−
ln ln
3
2
1−sin2 u 1− √t3 Voltamos agora a (*) e colocamos lá este resultado. Fica:


t
t
√
1− √3 1− √3 √
1
3


(*) = t − 3 3−t2 dt = t − 3 × − 3 ln 2 + c = t + 3 ln 2 + c
1− √t3 1− √t3 √
Resta substituir t por x2 + 3 e obtemos:
√
2 +3
x
t
1− √3 √
√
√
√
1−
f (x) dx = t + 3 ln 2 + c = x2 + 3 + 3 ln √ 3 2 + c =
t
2
1− √3 1− x√3+3 1− x2 +3 √
√
3
= x2 + 3 + 3 ln x2 +3 + c
1− 3 6. f (x) =
c =?
f ′ (x) dx =
lim f (x) = e ⇔ lim
x→∞
x→∞
1
x2
sin x1 ecos( x ) dx = ecos( x ) + c
1
1
1
ecos( x ) + c = e ⇔ e + c = e ⇔ c = 0
Então f (x) = ecos( x ) .
1
7. f ′ (x) =
c =?
−2
f ′′ (x) dx = − 2x x2 + 1
dx =
f ′ (0) = 1 ⇔
1
1+0
Então f ′ (x) =
f (x) =
k =?
+c
+c=1⇔c=0
1
1+x2
f ′ (x) dx =
lim f (x) =
x→+∞
1
1+x2
1
dx
1+x2
= arctg x + k
3π
3π
⇔ lim (arctg x + k) =
⇔
x→+∞
4
4
Então f (x) = arctg x +
π
.
4
6
π
2
+k =
3π
π
⇔k=
4
4
8. (a)
(b)
2
x3 −2x
√ +3 dx
x
=
x3
√
dx − 2
x
x2
√
dx + 3
x
√1 dx
x
7
5
1
= 72 x 2 − 45 x 2 + 6x 2
4
sin x + 5 cos3 x dx = sin4 x cos3 x dx + 5 cos3 x dx =
= sin4 x cos x 1 − sin2 x dx + cos x 1 − sin2 x dx =
= sin4 x cos x dx − sin6 x cos x dx + 5 cos x dx − 5 sin2 x cos x dx =
=
1
5
sin5 x − 71 sin7 x + 5 sin x − 53 sin3 x
9. Integrais definidos:
(a)
π
π
sin2 (3x) cos3 (3x) dx = 0 sin2 (3x) cos (3x) 1 − sin2 (3x) dx =
π 1 5
π
π
π
sin (3x) 0 =
= 0 sin2 (3x) cos (3x) dx− 0 sin4 (3x) cos (3x) dx = 19 sin3 (3x) 0 − 15
0
=
(b)
1
9
2π
π
2
(0 − 0) −
1
15
|sin x| dx =
(0 − 0) = 0
π
π
2
sin x dx +
2π
π
− sin x dx = − [cos x]ππ + [cos x]2π
π =
2
= − (−1 − 0) + 1 − (−1) = 1 + 2 = 3
 2
 x
(c) Sendo f (x) =
3x + 1
 1
se −∞ < x ≤ 1
se 1 < x < 2
,
se
2
≤
x
<
+∞
x
3 1 2
2
3
1 2
2
3 1
x
x
+
3
+ [ln |x|]32 =
dx
=
+
x
f
(x)
dx
=
x
dx
+
3x
+
1
dx
+
3
2
0
0
1
2 x
=
1
3
0
+ 6 + 2 − 32 + 1 + (ln 3 − ln 2) =


1
3
+
11
2
+ ln 32 =
1
35
6
+ ln 32
x2
x+1
se −∞ < x ≤ 1
,
se 1 < x ≤ 2
 ex
se 2 < x < +∞
1+ex
3
1 x2
2
3 ex
0 f (x) dx = 0 x+1 dx + 1 x ln x dx + 2 1+ex dx =
2
2
1
2
1
= 0 x − 1 + x+1
dx + x2 ln x − 12 1 x dx + [ln (1 + ex )]32 =
(d) Sendo f (x) =
x ln x
1
=
x2
2
− x + ln |x + 1|
1
0
= − 21 + ln 2 + 2 ln 2 −
3
1
2
+ 2 ln 2 −
1
2
2
x2
2 1
+ ln 1 + e3 − ln 1 + e2 =
2 − 21 + ln 1 + e3 − ln 1 + e2 =
3
= − 54 + ln 8+8e
= − 45 + 3 ln 2 + ln 1+e
1+e2
1+e2
7
10. Primitivação de funções racionais:
(a)
2x+1
−x+3 dx
=−
2x+1
x−3 dx
= (∗)
Efectuando a divisão dos polinómios, obtemos
7
2x+1
x−3 = 2 + x−3 .
Assim,
(b)
(c)
(∗) = − 2 +
7
x−3 dx
= −2x − 7 ln |x − 3| + c
1
1
1
dx
=
3
(x−2)
(x+1)
(x−2) − (x+1) dx =
1
x−2 3
= 31 (ln |x − 2| − ln |x + 1|) + c = 13 ln x+1
+ c = ln x−2
x+1 + c
1
(x−2)(x+1) dx
=
3x+1
x(x−1)(x+2) dx
1
3
=
1
3
−
4
3
1
− x2 +
−
(x−1)
5
6
(x+2) dx
=
4
= − 21 ln |x| + 43 ln |x − 1| − 56 ln |x + 2| + c = ln
(d)
x3 +x+1
dx
x4 −2x3 +x2
1
x
= 3 ln |x| −
(e)
(f)
9
49
x3 −1
3
x−1
9
49
(x−2)
5 1
7 x−2
−
(x−2)2
3
x
+
1
x2
−
−
9
49
(x+5) dx
2
x−1
3
x−1
+
+c
3
dx
(x−1)2
=
3
|x|
+ ln (x−1)
2 +c
=
9
49
5
ln |x + 5| + c = − 7x−14
+ ln x−2
+c
x+5 1
dx
(x−1)(x2 +x+1)
1
1
x−1 dx − 3
+ c = − x1 −
5
7
+
9
49
=
5
1
|x| 2 |x+2| 6
x+2
dx
x2 +x+1
=
=
1
3
1
x2
dx
x4 −1
=
1
3
x−1
−
dx = 31
2x+4
11
1
x+ 23
3
2
x +x+1
1
x−1
−
x+2
dx
x2 +x+1
dx =
x2 +x+1
1
2x+1
= 13 ln |x − 1| − 6
dx + 3 x2 +x+1
dx = (ver 1. (r))
x2 +x+1
= 31 ln |x − 1| − 16 ln x2 + x + 1 − 12 √23 arctg √23 x + √13 + c
√
= 31 ln |x − 1| − 16 ln x2 + x + 1 − 33 arctg √23 x + √13 + c
=
(g)
1
3
dx =
=
ln |x − 2| −
1
x3 +x+1
dx
x2 (x2 −2x+1)
− 2 ln |x − 1| −
2x+1
dx
(x−2)2 (x+5)
=
=
|x−1| 3
1
4
x−1
=
−
x2
dx
(x2 −1)(x2 +1)
1
4
x+1
+
1
2
x2 +1
dx =
=
1
4
ln |x − 1| −
32
x2
dx
(x−1)(x+1)(x2 +1)
=
ln |x − 1| − 14 ln |x + 1| + 21 arctg x =
1
x−1 4 1
= ln x+1 + 2 arctg x + c
8
=
11. Área de regiões planas:
(a) y = ex , y = e−x , x = 2
Cálculo
área:
2 da
2
x
A = 0 e − e−x dx = [ex + e−x ]0 = e2 + e−2 − (1 + 1) = e2 + e−2 − 2
(b) y = sin x, y = cos x, x = 0, x = π
Cálculo
π
π da área:
π
A = 04 cos x − sin x dx + π sin x − cos x dx = [sin x + cos x]04 + [− cos x − sin x]ππ =
4
4
π
π
π
π
= sin 4 + cos 4 − (sin 0 + cos 0) + (− cos π − sin π) − − cos 4 − sin 4 =
√
√
√
√ √
= 22 + 22 − (0 + 1) + (1 − 0) − − 22 − 22 = 2 2
(c) y 2 = 4 + x, x + 2y = 4
9
Cálculo
2
2 dos pontosde 2intersecção:
y=2
y = −4
y + 2y − 8 = 0
y = 4 + 4 − 2y
y =4+x
∨
⇔
⇔
⇔
x=0
x = 12
x = 4 − 2y
x + 2y = 4
Cálculo
√
0 da
12
√ área: √
A = −4 4 + x − − 4 + x dx + 0 − x2 + 2 − − 4 + x dx =
12
12 √
0 √
4 + x dx =
= 2 −4 4 + x dx + 0 − x2 + 2 dx + 0
2
12 3 0
3 12
2
(4+x) 2
x
2
= 2 (4+x)
+
−
= 32
+
2x
+
3
3
4
3 − 36 + 24 + 3 (64 − 8) =
=
32
3
OU
A=
2
−4
− 12 +
112
3
0
=
108
3
2
0
= 36
2
2
2 dy = 8y − y2 −
4
−
2y
−
y
−
4
dy
=
8
−
2y
−
y
−4
−4
2
= 16 − 4 −
(d) y = x2 , y =
8
3
− −32 − 16 +
x2
2 ,
64
3
y = 2x
= 12 −
8
3
+ 48 −
64
3
= 60 −
72
3
2
y3
3 −4
=
= 60 − 24 = 36
Cálculo
dos pontos
de intersecção:
y=4
y=0
y = x2
∨
⇔
⇔
⇔
2
x=2
x=0
x (x − 2) = 0
x = 2x
y = 2x
⇔
Cálculo da área:
4
2
2
A = 0 x2 − x2 dx + 2 2x −
x2
2
=
2
y = x2
⇔
y = 2x
2
x3
6 0
+ 16 −
x2
2
64
6
= 2x
− 4 − 86 =
x
x
2
dx =
8
6
⇔
−2 =0
2
x2
0 2
+ 12 −
10
56
6
y=0
∨
x=0
dx + x2 −
4
x3
6 2
= 12 − 8 = 4
=
y = 16
x=4