Técnicas de integração (com gabarito)

Prévia do material em texto

2
a
Lista de Exercícios de Cálculo Diferencial e Integral II - Prof
a
Vanessa Munhoz
Técnicas de integração
RESPOSTAS
1. Resolva as seguintes integrais indefinidas utilizando o método de integração por partes.
(a)
∫
(x sin(5x)) dx = −x cos(5x)
5
+
sin(5x)
25
+ k
(b)
∫ (
t e4t
)
dt =
e4t
4
(
t− 1
4
)
+ k
(c)
∫
(x ln(3x)) dx =
x2
2
(
ln(3x)− 1
2
)
+ k
(d)
∫ (
x cossec2x
)
dx = −x cotx+ ln | sinx|+ k
(e)
∫ (
(x− 1) sec2 x) dx = (x− 1) tanx+ ln | cosx|+ k
(f)
∫ (
(x− 1)e−x) dx = −xe−x + k
(g)
∫ (
x2 ex
)
dx = ex(x2 − 2x+ 2) + k
(h)
∫ (
x3 sin(4x)
)
dx = −x
3
4
cos(4x) +
3x2
16
sin(4x) +
3x
32
cos(4x)− 3
128
sin(4x) + k
(i)
∫
((x+ 1) cos(2x)) dx =
(
x+ 1
2
)
sin(2x) +
1
4
cos(2x) + k
(j)
∫ (
x5 ex
2
)
dx = ex
2
(
x4
2
− x2 + 1
)
+ k
(k)
∫ (
(x+ 3)2ex
)
dx = ex(x2 + 4x+ 5) + k
(l)
∫ (
x2 lnx
)
dx =
x3
3
(
lnx− 1
3
)
+ k
2. Utilize o método de integração por partes para calcular as seguintes integrais definidas.
(a)
∫ pi
2
0
(x sinx) dx = 1
(c)
∫ 2
1
(x lnx) dx = 2 ln 2− 3
4
(e)
∫ 1
0
(
x cossec2x
)
dx = −cotg1 + ln | sin 1|
(g)
∫ 0
−2
(x ex) dx =
3
e2
− 1
(b)
∫ 1
−1
(
t e4t
)
dt =
3e4
16
+
5
16e4
(d)
∫ 1
0
(ex cosx) dx =
1
2
[e(sin 1 + cos 1)− 1]
(f)
∫ 4
1
((x− 1)ex) dx = 2e4 + e
(h)
∫ 3pi
2
pi
(
(x− 1) sec2 x) dx = 0
1
(i)
∫ pi
pi
2
((x+ 1) cos(2x)) dx =
1
2
3. Avalie as seguintes integrais trigonométricas:
(a)
∫
(sin 7x cos 5x) dx = −1
2
(
cos(12x)
12
+
cos(2x)
2
)
+ k
(b)
∫
(cos 5x cos 3x) dx =
1
2
(
sin(2x)
2
+
sin(8x)
8
)
+ k
(c)
∫
(sin 5x sin 2x) dx =
1
2
(
sin(3x)
3
− sin(7x)
7
)
+ k
(d)
∫ (
cos2 θ
)
dθ =
1
2
(
θ +
sin(2θ)
2
)
+ k
(e)
∫ (
sin4 θ
)
dθ =
1
4
(
3
2
θ − sin(2θ) + sin(4θ)
8
)
+ k
(f)
∫ (
sin3 x cos5 x
)
dx = −cos
6 x
6
+
cos8 x
8
+ k ou
sin4 x
4
− sin
6 x
3
+
sin7 x
7
+ k
(g)
∫ (
cos5 x
)
dx = sinx− 2sin
3 x
3
+
sin5 x
5
+ k
(h)
∫ (
sin3 3θ
)
dθ = −1
3
(
cos(3θ)− cos
3(3θ)
3
)
+ k
(i)
∫ (
sin7 x cos2 x
)
dx =
cos9 x
9
− 3cos
7 x
7
+ 3
cos5 x
5
− cos
3 x
3
+ k
(j)
∫ (
sin2 x cos4 x
)
dx =
1
8
(
x− x
2
− sin(4x)
8
+
sin3(2x)
6
)
+ k
(k)
∫ (
sin3 2θ cos4 2θ
)
dθ = −1
2
(
cos5(2θ)
5
− cos
7(2θ)
7
)
+ k
(l)
∫ (
15 sin2 x cos3 x
)
dx = 15
(
sin3 x
3
− sin
5 x
5
)
+ k
(m)
∫ (
tan3 x secx
)
dx =
sec3 x
3
− secx+ k
(n)
∫ (
tan5 x sec2 x
)
dx =
tan6 x
6
+ k
(o)
∫ (
tan6 x
)
dx =
tan5 x
5
− tan
3 x
3
+ tanx− x+ k
2
4. Calcule as integrais aplicando a substituição trigonométrica adequada.
(a)
∫ (
1
x2
√
x2 + 4
)
dx = −
√
x2 + 4
4x
+ k
(b)
∫ (
1
x2
√
x2 − 9
)
dx =
√
x2 − 9
9x
+ k
(c)
∫ (√
1− 4x2
)
dx =
1
4
arcsin(2x) +
1
2
x
√
1− 4x2 + k
(d)
∫ (
x3√
x2 + 9
)
dx =
1
3
(x2 − 18)√x2 + 9 + k
(e)
∫ (
1
x2
√
25− x2
)
dx = −
√
25− x2
25x
+ k
(f)
∫ 2
√
2
(
1
t3
√
t2 − 1
)
dt =
pi
24
+
√
3− 2
8
5. Usando o método das frações parciais, calcule:
(a)
∫ (
2x3 + 3x2 − 2x
x+ 2
)
dx =
2x3
3
+
x2
2
+ k
(b)
∫ (
x3 − 3x2 − x+ 3
x2 − 2x− 3
)
dx =
x2
2
− x+ k
(c)
∫ (
1
x2 − 4
)
dx =
1
4
ln |x− 2| − 1
4
ln |x+ 2|+ k
(d)
∫ (
x4 + 2x+ 1
x3 − x2 − 2x
)
dx =
x2
2
+ x− 1
2
ln |x|+ 7
2
ln |x− 2|+ k
(e)
∫ (
x+ 3
x2 − 3x+ 2
)
dx = −4 ln |x− 1|+ 5 ln |x− 2|+ k
(f)
∫ (
2x+ 1
x2 − 1
)
dx =
3
2
ln |x− 1|+ 1
2
ln |x+ 1|+ k
(g)
∫ (
x+ 1
x3 + x2 − 6x
)
dx = −1
6
ln |x|+ 3
10
ln |x− 2| − 2
15
ln |x+ 3|+ k
(h)
∫ (
x3 + 3x− 1
x4 − 4x2
)
dx = −3
4
ln |x| − 1
4x
+
13
16
ln |x− 2|+ 15
16
ln |x+ 2|+ k
(i)
∫ (
x+ 1
x3 − x
)
dx = − ln |x|+ 1
2
ln |x− 1|+ k
(j)
∫ (
x3 − 2x2 − 3
x+ 1
)
dx =
x3
3
− 3x
2
2
+ 3x− 6 ln |x+ 1|+ k
3
(k)
∫ (
x5 + 3
x3 − 4x
)
dx =
x3
3
+ 4x− 3
4
ln |x|+ 35
8
ln |x− 2| − 29
8
ln |x+ 2|+ k
(l)
∫ (
5x2 + 1
x− 1
)
dx =
5x2
2
+ 5x+ 6 ln |x− 1|+ k
6. Calcule as seguintes integrais:
(a)
∫ [√
3 + x(x+ 1)2
]
dx =
2
7
(x+ 3)
7
2 − 8
5
(3 + x)
5
2 +
8
3
(x+ 3)
3
2 + k
(b)
∫ (
x2
√
1 + x
)
dx =
2
7
(x+ 1)
7
2 − 4
5
(x+ 1)
5
2 +
2
3
(x+ 1)
3
2 + k
(c)
∫ (
cosx
5 + sin2 x
)
dx =
1√
5
arctan
(
sinx√
5
)
+ k
(d)
∫ (
cosx
2 sin2 x+ 3 cos2 x
)
dx =
1
2
√
3
ln
∣∣∣∣∣
√
3 + sin x√
3− sinx
∣∣∣∣∣+ k
(e)
∫ (
sinx cosx√
cos2 x− sin2 x
)
dx = −1
2
√
cos 2x+ k ou −1
2
√
2 cos2 x− 1
(f)
∫ [
x sin
(x
2
)]
dx = −2x cos pi
2
+ 4 sin
pi
2
+ k
(g)
∫ (
2x+ 1
x2 + 2x+ 2
)
dx = ln |(x+ 1)2 + 1| − arctan(x+ 1) + k
(h)
∫ (
8x2
(x3 + 2)3
)
dx = − 4
3(x3 + 2)2
+ k
(i)
∫ [(
1 +
1
x
)3
1
x2
]
dx = −1
4
(
1 +
1
x
)4
+ k
(j)
∫ 12
7
dx = 5
(k)
∫ 2
−3
|x+ 1|dx = 13
2
(l)
∫ 0
1
[
t2 (t
1
3 −√t)
]
dt = − 1
70
(m)
∫ 2
3
(
x2 − 1
x− 1
)
dx = −7
2
(n)
∫ (
2x2 + 4
x3 − 8
)
dx = ln |x− 2|+ 1
2
ln |x2 + 2x+ 4| −
√
3
3
arctan
(
x+ 1√
3
)
+ k
4
(o)
∫ 1
0
(
x2 ex
)
dx = e− 2
(p)
∫ pi
pi
2
(
sin2 x
)
dx =
pi
4
(q)
∫ 0
−1
[
x (x+ 1)100
]
dx =
1
102
− 1
101
= − 1
102× 101
(r)
∫ pi
2
0
(ex cosx) dx =
1
2
(
e
pi
2 − 1)
(s)
∫ 1
0
(
sinx e1+cosx
)
dx = e2 − e1+cos 1
(t)
∫ 2
1
(lnx) dx = 2 ln 2− 1
(u)
∫ 5
5
(√
x2 +
√
x5 + 1
)
= 0
(v)
∫ √2
1
(
x 3−x
2
)
dx =
1
9 ln 3
(x)
∫ (
tan3 x sec4 x
)
dx =
sec6 x
6
− sec
4 x
4
+ k
(w)
∫ (
x
3− 2x− x2
)
dx = −1
4
ln |x− 1| − 3
4
ln |x+ 3|+ k
(y)
∫ (
cosx
4− sin2 x
)
dx =
1
4
ln
∣∣∣∣2 + sin x2− sinx
∣∣∣∣+ k
(z)
∫ (
sin 2x
1 + cos x
)
dx = −2 cosx+ 2 ln |1 + cos x|+ k
(α)
∫ (
sin2 x
)
dx =
1
2
(
x− sin(2x)
2
)
+ k
(β)
∫ [
sin5(5x) cos3(5x)
]
dx =
1
5
(
sin6(5x)
6
− sin
8(5x)
8
)
+ k
(γ)
∫ (
sin2 θ cos5 θ
)
dθ =
sin3 θ
3
− 2 sin
5 θ
5
+
sin7 θ
7
+ k
(δ)
∫ (
9 sin3 x cos2 x
)
dx =
9 cos5 x
5
− 3 cos3 x+ k
(θ)
∫ (
x
1 + x4
)
dx =
1
2
arctan (x2) + k
5
(φ)
∫ (
2x− 3
(x− 1)3
)
dx = − 2
x− 1 +
1
2(x− 1)2 + k
(ψ)
∫ (
x+ 2
x3 + 2x2 + 5x
)
dx =
2
5
ln |x| − 1
5
ln |x2 + 2x+ 2|+ 3
10
arctan
(
x+ 1
2
)
+ k
(ε)
∫ (
x2
3
√
x2 + 4
)
dx =
1
6
x
√
x2 + 4− 4
3
ln
∣∣∣∣∣
√
x2 + 4 + x
2
∣∣∣∣∣+ k
(σ)
∫ 0
−4
(√
t+ 4
t
)
dt = 4 + 2 ln 0− 2 ln 1 = 4
(ρ)
∫ (
x√
x+ 9
)
dx = 2
√
x+ 9
(
x− 18
3
)
+ k
(ξ)
∫ (√
x+ 9
x
)
dx = 2
√
x+ 9 + 3 ln
∣∣∣∣√x+ 9− 3√x+ 9 + 3
∣∣∣∣+ k
(µ)
∫ (
t√
t2 − 7
)
dt =
√
t2 − 7 + k
(ϕ)
∫ 2
3
0
(
x3
√
4− 9x2
)
dx =
64
1215
6