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Wednesday, September 2, 2015
Page 334
Problem 1
Z
Problem. Find the indefinite integral
5
dx.
x
Solution.
Z
5
dx = 5 ln |x| + C.
x
Problem 3
Z
1
dx.
x+1
Solution. Let u = x + 1 and du = dx. Then we get
Z
Z
1
1
dx =
du
x+1
u
Problem. Find the indefinite integral
= ln |u| + C
= ln |x + 1| + C.
Problem 7
Z
x
dx.
−3
Solution. Let u = x2 − 3 and du = 2x dx. Then we get
Z
Z
2x
x
1
dx =
dx
2
2
x −3
2
x −3
Z
1
1
=
du
2
u
1
= ln |u| + C
2
1
= ln |x2 − 3| + C.
2
Problem. Find the indefinite integral
x2
Problem 12
Z
Problem. Find the indefinite integral
x3 − 8x
dx.
x2
1
Solution. First, divide x2 into each term of the numerator. Then integrate termwise.
Z 3
Z x − 8x
8
x−
dx
dx =
x2
x
1
= x2 − 8 ln |x| + C.
2
Problem 13
x2 + 2x + 3
dx.
x3 + 3x2 + 9x
Solution. Let u = x3 + 3x2 + 9x and du = (3x2 + 6x + 9) dx, which happens to be 3
times the numerator. Then we get
Z
Z
1
3x2 + 6x + 9
x2 + 2x + 3
dx
=
dx
x3 + 3x2 + 9x
3
x3 + 3x2 + 9x
Z
1
1
=
du
3
u
1
= ln |u| + C
3
1
= ln |x3 + 3x2 + 9x| + C.
3
Z
Problem. Find the indefinite integral
Problem 17
x3 − 3x2 + 5
Problem. Find the indefinite integral
dx.
x−3
Solution. Begin by using long division to divide x − 3 into x3 − 3x2 + 5. We get x2
with a remainder of 5, so
Z
x3 − 3x2 + 5
5
= x2 +
.
x−3
x−3
Now we can integrate.
Z
x3 − 3x2 + 5
dx =
x−3
Z x2 +
5
x−3
dx
1
= x3 + 5 ln |x − 3| + C.
3
Problem 27
Z
Problem. Find the indefinite integral
1
√ dx by u-substitution.
1 + 2x
2
√
Solution. Let u = 1 + 2x and du = √12x dx. Note that du =
that as dx = u du and use that for the substitution.
Z
Z 1
1
√ dx =
(u du)
1+u
1 + 2x
Z
u
=
du
1+u
Z 1
1−
=
du
1+u
1
u
dx, so we may rewrite
= u − ln |1 + u| + C
√
√
= 2x − ln |1 + 2x| + C.
Problem 29
Z
Problem. Find the indefinite integral
√
x
√
dx by u-substitution.
x−3
√
1
Solution. Let u = x and du = 2√1 x dx. Note that du = 2u
dx, so dx = 2u du. Now
substitute and integrate.
√
Z
Z x
u
√
dx =
· 2u du
u−3
x−3
Z 2u2
=
du
u−3
Z 18
=
du
2u + 6 +
u−3
= u2 + 6u + 18 ln |u − 3| + C
√
√
= x + 6 x + 18 ln | x − 3| + C.
Or, you could let u =
√
x − 3 and du =
1
√
2 x
3
dx. Then
√
x = u + 3 and du =
1
2(u+3)
dx,
so dx = 2(u + 3) du. Now substitute and integrate.
√
Z Z
u+3
x
√
dx =
· 2(u + 3) du
u
x−3
Z (u + 3)2
=2
du)
u
Z 2
u + 6u + 9
=2
du)
u
Z 9
u+6+
du)
=2
u
= u2 + 12u + 18 ln |u| + C
√
√
√
= ( x − 3)2 + 12( x − 3) + 18 ln | x − 3| + C.
Problem 32
Z
Problem. Find the indefinite integral
tan 5θ dθ.
Solution. Let u = 5θ and du = 5 dθ. Then
Z
Z
1
tan 5θ dθ =
5 tan 5θ dθ
5
Z
1
=
tan u du
5
1
= ln | sec u| + C
5
1
= ln | sec 5θ| + C.
5
Problem 34
Z
Problem. Find the indefinite integral
Solution. Let u =
x
2
sec
x
dx.
2
and du = 12 dx. Then
Z
Z
x
1
x
sec dx = 2
sec dx
2
2
2
Z
= 2 sec u du
= 2 ln | sec u + tan u| + C
x
x = 2 ln sec + tan + C.
2
2
4
Problem 39
Z
sec x tan x
dx.
sec x − 1
Solution. Let u = sec x − 1 and du = sec x tan x dx. Then
Z
Z
du
sec x tan x
dx =
sec x − 1
u
Problem. Find the indefinite integral
= ln |u| + C
= ln | sec x − 1| + C.
Problem 70
sin x
from x =
1 + cos x
Solution. Let u = 1 + cos x and du = − sin x dx. Then u π4 = 1 +
1 − √12 .
Problem. Find the area of the region under y =
Z
3π/4
π/4
sin x
dx = −
1 + cos x
Z
3π/4
1− √1
2
=−
1+ √1
to x =
√1
2
and u
3π
.
4
3π
4
=
− sin x
dx
1 + cos x
π/4
Z
π
4
du
u
2
1− √1
= − [ln |u|]1+ √12
2
1 1
= − ln 1 − √ + ln 1 + √ 2
2
1
1 + √ 2
= ln 1
1 − √2 √
2 + 1
= ln √
.
2 − 1
Problem 96
πx
over the interval [0, 2].
6
Solution. The formula for the average value of a function f (x) over an interval [a, b]
is
Z b
1
f (x) dx.
b−a a
Problem. Find the average value of f (x) = sec
5
So we need to find the value of
1
2
Let u =
πx
6
Z
2
sec
0
πx
dx.
6
π
6
and du = dx. Then u()0) = 0 and u(2) = π3 .
Z
Z
1 2
6 1 2π
πx
πx
dx = ·
sec
dx
sec
2 0
6
π 2 0 6
6
Z
3 π/3
sec u du
=
π 0
3
= [ln | sec u + tan u|]π/3
0
π
3
π
π =
ln sec + tan − ln |sec 0 + tan 0|
π
3
3
√ 3 ln 2 + 3 − ln 1
=
π
√ 3 = ln 2 + 3 .
π
Problem 97
Problem. A population of bacteria P is changing at a rate of
dP
3000
=
,
dt
1 + 0.25t
where t is the time in days. The initial population (when t = 0) is 1000. Write an
equation that gives the population at any time t. Then find the population when
t = 3.
Solution. The population P (t) is given by
Z
3000
P (t) =
dt.
1 + 0.25t
Let u = 1 + 0.25t and du = 0.25 dt. Then
Z
3000
P (t) =
dt
1 + 0.25t
Z
0.25
= 12000
dt
1 + 0.25t
Z
du
= 12000
u
= 12000 ln |u| + C
= 12000 ln |1 + 0.25t| + C.
6
To find the value of C, let t = 0.
P (0) = 12000 ln 1 + C
= C.
Therefore, C = 1000 and the population is
P (t) = 12000 ln (1 + 0.25t) + 1000.
When t = 3, the population is
P (3) = 12000 ln 1.75 + 1000
= 12000(0.5596) + 1000
= 7715.
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