Download Math 141 Exam #2 Key (Summer 2013) 1a Let u(x) = x and v (z

Math 141 Exam #2 Key (Summer 2013)
1a Let u(x) = x and v 0 (z) = cos 8x. Then u0 (x) = 1 and v(x) = 18 sin 8x, and so
Z
Z
1
x
1
x
sin 8x dx = sin 8x +
cos 8x + c.
x cos 8x dx = sin 8x −
8
8
8
64
1b Let u(x) = e−x and v 0 (x) = sin(4x). Then u0 (x) = −e−x and v(x) = − 14 cos(4x), and so
Z
Z
1
1 −x
−x
e sin(4x) dx = − e cos(4x) −
e−x cos(4x) dx + c.
4
4
For the integral on the right, let u(x) = e−x and v 0 (x) = cos(4x). Then u0 (x) = −e−x and
v(x) = 14 sin(4x), so that
Z
Z
1 −x
1 1 −x
1
−x
−x
e sin(4x) dx = − e cos(4x) −
e sin(4x) +
e sin(4x) dx + c,
4
4 4
4
and hence
1c For z ∈
Z
√2 , 2
3
e−x sin(4x) dx = −
1
4 −x
e cos(4x) − e−x sin(4x) + c.
17
17
we have
1
(sec−1 z)0 = √
.
z z2 − 1
√
Let u(z) = sec−1 z and v 0 (z) = z. Then u0 (z) = 1/ z z 2 − 1 and v(z) = 21 z 2 , and so
Z 2
Z 2
1 2 −1 2
z
−1
sec (z)dz = 2 z sec (z) 2/√3 − √ √
dz.
√
2
2/ 3 2 z − 1
2/ 3
Let w = z 2 − 1 in the integral at right. We obtain
Z 2
√ 1 Z 3 −1/2
2
−1
−1
−1
w
dw
sec (z)dz = 2 sec (2) − sec
2/ 3 −
√
3
4 1/3
2/ 3
p 5π
2π π 1 √
1
=
− −
3 − 1/3 =
−√ .
3
9 4
9
3
2a We have
Z
7
3
sin (x) cos (x) dx =
Z
[1 − cos2 (x)]3 cos3 (x) sin(x) dx,
and so if we let u = cos(x), so that sin(x) dx is replaced by −du by the Substitution Rule, we
obtain
Z
Z
Z
7
3
2 3 3
sin (x) cos (x) dx = − (1 − u ) u du = (u3 − 3u5 + 3u7 − u9 ) du
1
1
3
1
= u4 − u6 + u8 − u10 + c
4
2
8
10
2
=
1
1
3
1
cos4 − cos6 + cos8 − cos10 +c.
4
2
8
10
Alternatively: write
Z
sin7 (x)[1 − sin2 (x)] cos(x) dx,
and let u = sin(x) to obtain
Z
1
1
1
1
sin10 (x) + c.
u7 (1 − u2 ) du = u8 − u10 + c = sin8 (x) −
8
10
8
10
2b We have
Z
csc4 t
dt =
cot2 t
Z
csc2 t ·
cot2 t + 1
dt.
cot2 t
Substitute u = cot t, so that formally csc2 t dt = −du, and we obtain
Z 2
Z
Z
u +1
1
cot2 t + 1
2
dt = −
du = − (1 + u−2 )du = −u + + c
csc t ·
2
2
cot t
u
u
= − cot t + tan t + c.
2c We have
Z
π/3
Z
p
2
sec (ϕ) − 1 dϕ =
−π/3
π/3
Z
q
2
tan (ϕ) dϕ =
−π/3
Z
π/3
| tan(ϕ)| dϕ
−π/3
0
Z
π/3
tan(ϕ) dϕ
[− tan(ϕ)] dϕ +
=
−π/3
0
0
π/3
= − ln | sec(ϕ)| −π/3 + ln | sec(ϕ)| 0
= 2 ln(2) = ln(4).
3a Let x − 13 tan θ. Formally we obtain dx = 31 sec2 θdθ, and also
through the usual trigonometric substitution process yields
√
Z 1/3
1
2
dx
=
.
(9x2 + 1)3/2
6
0
√
9x2 + 1 = sec θ. Running
3b Let x = 11 sin θ for θ ∈ [−π/2, π/2], so that dx is replaced with 11 cos θ dθ as part of the
substitution. Observe that −π/2 ≤ θ ≤ π/2 implies cos θ ≥ 0, so that
√
cos2 θ = | cos θ| = cos θ.
3
Now,
Z √
121 −
x2
Z p
Z
p
2
dx =
121 − 121 sin θ · 11 cos θ dθ = 121 cos θ 1 − sin2 θ dθ
Z
Z
√
2
= 121 cos θ cos θ dθ = 121 cos2 θ dθ,
R
and with the deft use of the given formula for cosn θ dθ we obtain
Z √
Z
cos θ sin θ 1
121
121
2
121 − x dx = 121
+
cos θ sin θ +
θ + c.
(1) dθ =
2
2
2
2
From x = 11 sin θ comes sin θ = x/11, so θ = sin−1 (x/11) and θ may be characterized as
an angle in the right triangle
11
|x|
θ
√
121 − x2
Note that
√ x ≥ 0 if θ ∈ [0, π/2], and x < 0 if θ ∈ [−π/2, 0). From this triangle we see that
cos θ = 121 − x2 /11, and therefore
√
Z √
121
121 −1 x 121 − x2 x
121 − x2 dx =
·
·
+
sin
+c
2
11
11
2
11
√
x 121 − x2 121 −1 x =
+
sin
+ c.
2
2
11
4a We have
y+1
A
B
C
y+1
=
=
+
+
,
y 3 + 3y 2 − 18y
y(y + 6)(y − 3)
y
y+6 y−3
whence we obtain
y + 1 = (A + B + C)y 2 + (3A − 3B + 6C)y − 18A,
which results in the system
A+ B+ C=0
3A − 3B + 6C = 1
−18A
=1
5 4
1
Solving the system yields (A, B, C) = − 18
, − 54
, 27 . Hence
Z
Z y+1
1
5
4
dy =
−
−
+
dy
y 3 + 3y 2 − 18y
18y 54(y + 6) 27(y − 3)
1
5
4
= − ln |y| −
ln |y + 6| +
ln |y − 3| + c.
18
24
27
(
4
4b Again start with a decomposition, noting that x2 + 2x + 6 is an irreducible quadratic:
Z
Z 1/15
2
−x/15 − 2/5
dx =
+ 2
dx
(x − 4)(x2 + 2x + 6)
x−4
x + 2x + 6
Z
1
1
x+6
=
ln |x − 4| −
dx.
(1)
15
15
(x + 1)2 + 5
For the remaining integral, let u = x + 1 to obtain
Z
Z
Z
Z
u+5
u
1
x+6
√
dx =
du =
du + 5
du
2
2
2
(x + 1) + 5
u +5
u +5
u2 + ( 5)2
(2)
Letting w = u2 + 5 in the first integral in (2), and using Formula (12) for the second, we next
get
Z
Z
1/2
u
x+6
1
−1
√ +c
dx =
dw + 5 · √ tan
(x + 1)2 + 5
w
5
5
√
√
1
u
u
1
−1
−1
2
√ + c = ln(u + 5) + 5 tan
√ +c
= ln |w| + 5 tan
2
2
5
5
√
1
x+1
= ln[(x + 1)2 + 5] + 5 tan−1 √
+c
2
5
Returning to (1),
Z
ln |x − 4|
1 ln[(x + 1)2 + 5] √
2
−1 x + 1
√
dx =
−
+ 5 tan
+c
(x − 4)(x2 + 2x + 6)
15
15
2
5
√
ln |x − 4| ln(x2 + 2x + 6)
5
−1 x + 1
√
−
−
tan
=
+ c.
15
30
15
5
5a Letting u = 2x − 3, we have
Z 1
Z 1
Z −1
1
1
1/2
dx = lim
dx = lim
du
2
2
a→−∞ a (2x − 3)
a→−∞ 2a−3 u2
−∞ (2x − 3)
−1
1
1
1
1
1
= lim
−
= lim
1+
= .
a→−∞ 2
a→−∞
u 2a−3
2
2a − 3
2
R1
R4
5b We must evaluate 0 1/(x − 1) dx and 1 1/(x − 1) dx, if possible. By definition,
Z 1
Z b
1
1
dx = lim−
dx = lim− [ln |x − 1|]b0 = lim− (ln |b − 1| − ln | − 1|)
b→1
b→1
b→1
0 x−1
0 x−1
= lim− ln(1 − b) = −∞.
b→1
So
R1
0
1/(x − 1) dx diverges, and therefore
R4
0
1/(x − 1) dx also diverges.
5
6 The volume of the solid is
Z ∞
Z
−2 2
π(x ) dx = π lim
V=
2
b→∞
π
= π · 13 (2−3 − 0) = .
24
2
b
b
1
dx = π lim − 13 x−3 2 = π lim 31 (2−3 − b−3 )
4
b→∞
b→∞
x