lista10-c1-2016-2

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M
Universidade Federal do Rio de Janeiro
INSTITUTO DE MATEMÁTICA
Departamento de Métodos Matemáticos
10a Lista de Exerćıcios de Cálculo I - Monica Merkle
1. Calcule as integrais abaixo:
(a)
∫
(
√
2x− 1√
2x
) dx
(b)
∫
x2(4− x2)3 dx
(c)
∫
cotg 2x dx
(d)
∫
eln( cossec
2x) dx
(e)
∫
(ex − e−x) dx
(f)
∫
(x+1)2√
x
dx
2. Explique o paradoxo aparente:∫
(x + 1) dx = x
2
2
+ x + c
Resolvendo-a pela substituição u = x + 1 ,∫
(x + 1) dx =
∫
u du = u
2
2
+ c = (x+1)
2
2
+ c.
3. Calcule as seguintes integrais por substituição:
(a)
∫
xe−x
2
dx
(b)
∫
ex
e2x+2ex+1
dx
(c)
∫
e
√
x√
x
dx
(d)
∫
x
√
1− x2 dx
(e)
∫
ln(cos x) tg x dx
(f)
∫
ee
x
ex dx
(g)
∫
sen x cos3 x dx
(h)
∫
e senh x cosh x dx
(i)
∫
cos(ln x)
x
dx
(j)
∫
x2
4+x3
dx
(k)
∫
sen 3x dx
(l)
∫
cossec x dx
(m)
∫
sen 2x dx
(n)
∫
1
cotg 3x
dx
(o)
∫
cotg (ex)ex dx
(p)
∫
tg x
cos2 x
dx
(q)
∫
arccotg x
1+x2
dx
(r)
∫
2x
2
x dx
(s)
∫
3xex dx
(t)
∫
1
1+2x2
dx
(u)
∫
arccos x−x√
1−x2 dx
(v)
∫ √
1+
√
x√
x
dx
(w)
∫
cos3 x
sen 4x
dx
(x)
∫
3
√
tg 2x
cos2 x
dx
4. Calcule as integrais abaixo, usando integração por partes:
(a)
∫
x2ex dx
(b)
∫
x3ex
2
dx
(c)
∫
x2 sen x dx
(d)
∫
ln(ln x)
x
dx
(e)
∫
ex cos x dx
(f)
∫ √
x ln x dx
(g)
∫
x
√
1 + x dx
(h)
∫
sen 2x dx
(i)
∫
ln3 x dx
(j)
∫
x cos(ax) dx
(k)
∫
x sec2 x dx
(l)
∫
cossec 3x dx
(m)
∫
ln(1− x) dx
(n)
∫
arcsen
√
x√
x
dx
(o)
∫
(ex + x2)2 dx
5. Calcule as seguintes integrais por substituição trigonométrica:
(a)
∫
x2
√
4− x2 dx
(b)
∫
1
x2
√
1+x2
dx
(c)
∫ √
x2−a2
x
dx
(d)
∫
1√
(a2+x2)3
dx
(e)
∫
1
x3
√
x2−9 dx
6. Resolva as integrais abaixo pelo método de frações parciais:
(a)
∫
1
(x−2)(x−3) dx
(b)
∫
2x−3
(x−1)(x−7) dx
(c)
∫
x+1
(x−1)2(x−2) dx
(d)
∫
x2+x+2
x2−1 dx
(e)
∫
x3
x2+5x−6 dx
(f)
∫
x5+x4−8
x3−4x dx
(g)
∫
x5
x3−1 dx
7. Resolva as integrais a seguir:
(a)
∫
(1 + cos x)5 sen x dx
(b)
∫
ex+1
e2x−1 dx
(c)
∫
cos5 x dx
(d)
∫
ln(x2)
x2
dx
(e)
∫
(
√
1 + 1
2x
) 1
x3
dx
(f)
∫
x2 sen 2x dx
(g)
∫
1
x ln x ln(ln x)
dx
(h)
∫
2+3 cos x
sen 2x
dx
(i)
∫
e
√
x dx
(j)
∫
x+3
(3−x)2/3 dx
(k)
∫
cos(ln x) dx
(l)
∫
sen (4x) cos(2x) dx
(m)
∫
1
1+cos x
dx
(n)
∫
(x + 1
x
)3/2(x
2−1
x2
) dx
Respostas
1.
(a) 2
√
2
3
x3/2 −√2√x + C
(b) −x
9
9
+ 12x
7
7
− 48x5
5
+ 64x
3
3
+ C
(c) −x− cotg x + C
(d) − cotg x + C
(e) ex + e−x + C
(f) 2
5
x5/2 + 2x1/2 + 4
3
x3/2 + C
2. (x+1)
2
2
+ C = x
2
2
+ x + 1
2
+ C, onde 1
2
+ C é constante.
3.
(a) −e
−x2
2
+ C
(b) −1
ex+1
+ C
(c) 2e
√
x + C
(d)
−
√
(1−x2)3
3
+ C
(e) −(ln(cos x))
2
2
+ C
(f) ee
x
+ C
(g) − cos4 x
4
+ C
(h) e senh x + C
(i) sen (ln x) + C
(j) 1
3
ln |4 + x3|+ C
(k) − cos x + cos3 x
3
+ C
(l) − ln | cossec x + cotg x|+ C
(m) x
2
− sen (2x)
4
+ C
(n) −1
3
ln | cos(3x)|+ C
(o) ln | sen ex|+ C
(p) tg
2x
2
+ C
(q) − arccotg
2x
2
+ C
(r) 2
x2
2 ln 2
+ C
(s) 3
xex
ln 3+1
+ C
(t) 1√
2
arctg (
√
2x) + C
(u) −1
2
(arccos x)2 +
√
1− x2 + C
(v) 4
3
√
(1 +
√
x)3 + C
(w) 1
sen x
− 1
3 sen 3x
+ C
(x) 3
5
tg 5/3x + C
4.
(a) (x2 − 2x + 2)ex + C
(b) x
2ex
2
2
− ex2
2
+ C
(c) −x2 cos x + 2x sen x + 2 cos x + C
(d) ln x(ln(ln x)− 1) + C
(e) ex (cos x+sen x)
2
+ C
(f) 2
3
x3/2 ln x− 4
9
x3/2 + C
(g) 2
3
x
√
(1 + x)3 − 4
15
√
(1 + x)5 + C
(h) x− sen x cos x
2
+ C
(i) x(ln x)3 − 3x(ln x)2 + 6x ln x− 6x + C
(j) x sen (ax)
a
+ cos(ax)
a2
+ C
(k) x tg x + ln | cos x|+ C
(l) − cotg x cossec x−ln | cossec x+cotg x|
2
+ C
(m) −x− (1− x) ln |1− x|+ C
(n) 2
√
x arcsen
√
x + 2
√
1− x + C
(o) e
2x
2
+ x
5
5
+ 2x2ex − 4xex + 4ex + C
2
5.
(a) 2 arcsen x
2
− 1
2
x
√
4− x2 + 1
4
x3
√
4− x2 + C
(b) −
√
1+x2
x
+ C
(c)
√
x2 − a2 − a arccos a
x
+ C
(d) x
a2
1√
a2+x2
+ C
(e) 1
54
arcsec (x
3
) +
√
x2−9
18x2
+ C
6.
(a) − ln |x− 2|+ ln |x− 3|+ C
(b) 11
6
ln |x− 7|+ 1
6
ln |x− 1|+ C
(c) −3 ln |x− 1|+ 3 ln |x− 2|+ 2
x−1 + C
(d) x + 2 ln |x− 1| − ln |x + 1|+ C
(e) x
2
2
− 5x + 31 ln |x + 6|+ 1
7
ln |x−1
x+6
|+ C
(f) x
3
3
+ x
2
2
+ 4x + ln |x2(x−2)5
(x+2)3
|+ C
(g) 1
3
(x3 + ln |x3 − 1|) + C
7.
(a) − (1+cos x)6
6
+ C
(b) ln |1− e−x|+ C
(c) sen x + sen
5x
5
− 2 sen 3x
3
+ C (por subst.)
1
5
( sen x cos4 x + 4 sen x cos2 x + 8 sen
3x
3
) + C
(por partes)
(d) − ln x2
x
− 2
x
+ C
(e) −8
5
(1 + 1
2x
)5/2 + 8
3
(1 + 1
2x
)3/2 + C
(f) −x2 sen x cos x
2
+ x
3
6
− x cos(2x)
4
+ sen (2x)
8
+ C
(g) ln(ln(ln x)) + C
(h) −2 cotg x− 3 cossec x + C
(i) 2e
√
x
√
x− 2e√x + C
(j) 3
4
(3− x)4/3 − 18(3− x)1/3 + C
(k) x(cos(ln x)+ sen (ln x))
2
+ C
(l) − cos3(2x)
3
+ C (por subst.)
−1
3
(cos(4x) cos(2x) + sen (4x) sen (2x)
2
)
(por partes)
(m) − cotg x + cossec x + C
(n) 2
5
(x + 1
x
)5/2 + C
3