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Chapter 1
Techniques of Integration
In this chapter we learn the most common integration techniques. These tech-
niques will be explained throughout examples and remarks.
1.1 Integration by substitution
We solve the following problems:
1.∫
x(x2 + 4)6dx
Solution: Since the derivative of x2+4 is 2x which may be considered as
x, we do the substitution x2 +4 = t ⇒ 2xdx = dt ⇒ dx = dt2x
. Therefore,∫x(x2 + 4)6dx =
∫xt6
dt
2x=
1
2
∫t6dt =
1
14t7 + C =
1
14(x2 + 4)7 + C.
Thus, the idea of integration by substitution is that some quantity and
its derivative both exist. In such a case we let the quantity (not its
derivative) equal t.
2.∫
x+1x2+2x+7
dx
1
Solution: Let x2 +2x+7 = t ⇒ 2(x+1)dx = dt ⇒ dx = dt2(x+1)
. Hence,∫x + 1
x2 + 2x + 7dx =
∫x + 1
t
dt
2(x + 1)
=1
2
∫1
tdt
=1
2ln |t|+ C
=1
2ln |x2 + 2x + 7|+ C.
3.∫
sin√
x√x
dx
Solution: Let√
x = t ⇒ x = t2 ⇒ dx = 2tdt. Whence,∫sin√
x
xdx =
∫sin t
t2tdt = 2
∫sin t
dt
= −2 cos t + C = −2 cos√
x + C.
4.∫
1(2x−1)5
dx
Solution 2x− 1 = t ⇒ 2dx = dt ⇒ dx = dt2. Therefore,∫
1
(2x− 1)5dx =
∫1
t5dt
2=
1
2
∫t−5dt =
1
2
t−4
−4+ C
=−1
8
1
(2x− 1)4+ C.
5.∫
esin θ cos θdθ
Solution: sin θ = t ⇒ cos θdθ = dt ⇒ dθ = dtcos θ
. Whence,∫esin θ cos θdθ =
∫et cos θ
dt
cos θ=
∫etdt == et + C
= esin θ + C.
6.∫ √
5− tdt
Solution: 5− t = u ⇒ −dt = du ⇒ dt = −du. Thus,∫ √u(−du) = −
∫u1/2du = −2
3u3/2 + C
= −2
3(5− t)3/2 + C.
2
7.∫
x2(x3 + 5)9dx
Solution: x3 + 5 = t ⇒ 3x2dx = dt ⇒ dx = dt3x2 . Thus,∫
x2(x3 + 5)9dx =
∫x2t9
dt
3x2=
1
3
∫t9dt =
1
3
t10
10+ C
=1
30(x3 + 5)10 + C.
8.∫
sin πtdt
Solution: πt = x ⇒ dt = dxπ
,∫sin πtdt =
∫sin x
dx
π=
1
π(− cos x) + C
= − 1
πcos πt + C.
9.∫ (ln x)2
xdx
Solution: Let ln x = t ⇒ 1xdx = dt ⇒ dx = xdt. Thus∫
(ln x)2
xdx =
∫t2
xxdt =
∫t2dt =
t3
3+ C
=1
3(ln x)3 + C.
10.∫
cos θ sin6 θdθ
Solution: sin θ = t ⇒ cos θdθ = dt ⇒ dθ = dtcos θ
. Consequently∫cos θ sin6 θdθ =
∫cos θt6
dt
cos θ=
∫t6dt =
t7
7+ C
=sin7 θ
7+ C.
11.∫
ex√
1 + exdx
Solution: 1 + ex = t ⇒ exdx = dt ⇒ dx = dtex . Hence,∫
ex√
1 + exdx =
∫ex√
tdt
ex=
∫ √tdt =
2
3t3/2 + c
=2
3(1 + ex)3/2 + C.
3
12.∫
1x ln x
dx
Solution: ln x = t ⇒ 1xdx = dt ⇒ dx = xdt. Thus,∫
1
x ln xdx =
∫1
xtxdt =
∫1
tdt = ln |t|+ C
= ln | ln x|+ C.
13.∫ √
cot x csc2 xdx
Solution: cot x = t ⇒ − csc2 xdx = dt ⇒ dx = dt− csc2 x
. Therefore∫ √cot x csc2 xdx =
∫ √t csc2 x
dt
− csc2 x= −
∫ √tdt = −2
3t3/2 + C
= −2
3(cot x)3/2 + C.
14.∫
cot xdx
Solution: We observe that∫cot xdx =
∫cos xsin x
dx =∫
1tdt = ln |t|+ C
= ln | sin x|+ C.
15.∫
sec3 x tan xdx
Solution: sec x = t ⇒ sec x tan xdx = dt ⇒ dx = dtsec x tan x
. Therefore∫sec3 x tan xdx =
∫t3 tan x
dt
t tan x=
∫t2dt =
t3
3+ C =
sec3 x
3+ C.
16.∫
xa√
b + cxa+1dx
Solution: We see that the derivative of b+cxa+1 is (a+1)cxa. Thus the
substitution b + cxa+1 = t ⇒ (a + 1)cxadx = dt ⇒ dx = dt(a+1)cxa leads to∫
xa√
b + cxa+1dx =
∫xa√
tdt
(a + 1)cxa=
1
c(a + 1)
∫ √tdt
=1
c(a + 1)
2
3t3/2 + C =
2
3c(a + 1)(b + cxa+1)3/2 + C.
4
17.∫
sin t sec2(cos t)dt
Solution: cos t = u ⇒ − sin tdt = du ⇒ dt = − dusin t
. Thus,∫sin t sec2(cos t)dt =
∫sin t sec2(u)
−du
sin t= −
∫sec2 udu
= − tan u + C = − tan(cos t) + C.
18.∫
1+x1+x2 dx
Solution: An immediate substitution will not simplify the problem. A
deep look suggests that we split the top as follows:∫1 + x
1 + x2dx =
∫1
1 + x2dx +
∫x
1 + x2dx
= arctan(x) +1
2ln(x2 + 1) + C,
where we have used the fact that ddx
(arctan x) = 1x2+1
for the first integral
and the substitution x2+1 = t for the second one. The details are simple
and shall be left to the student.
19.∫
x4√x+2
dx Solution: Let 4√
x + 2 = t ⇒ x + 2 = t4 ⇒ dx = 4t3dt. Thus,∫x
4√
x + 2dx =
∫t4 − 2
t4t3dt = 4
∫t2(t4 − 2)dt
= 4
∫(t6 − 2t2)dt = 4(
t7
7− 2t3
3) + C
= 4
((x + 2)7/4
7− 2(x + 2)3/4
3
)+ C.
20.∫
arctan x1+x2 dx
Solution: Let arctan x = t ⇒ 1x2+1
dx = dt ⇒ dx = (x2 + 1)dt. Hence,∫arctan x
1 + x2dx =
∫t
1 + x2(x2 + 1)dt =
∫tdt
=t2
2+ C =
(arctan x)2
2+ C.
21.∫ √
x sin(1 + x3/2)dx
5
Solution: 1 + x3/2 = t ⇒ 32x1/2dx = dt ⇒ dx = 2
3√
xdt. Hence,∫ √
x sin(1 + x3/2)dx =
∫ √x sin(t)
2
3√
xdt =
2
3
∫sin tdt
= −2
3cos t + C = −2
3cos(1 + x3/2) + C.
22.∫
(1 + tan θ)5 sec2 θdθ
Solution: 1 + tan θ = t ⇒ sec2 θdθ = dt ⇒ dθ = dtsec2 θ
. Thus∫(1 + tan θ)5 sec2 θdθ =
∫t5 sec2 θ
dt
sec2 θ=
∫t5dt
=t6
6+ C =
(1 + tan θ)6
6+ C.
23.∫
ex
ex+1dx
Solution: ex + 1 = t ⇒ exdx = dt ⇒ dx = dtex . Therefore,∫
ex
ex + 1dx =
∫ex
t
dt
ex=
∫1
tdt = ln |t|+ C
= ln(ex + 1) + C.
24.∫ cos(π/x)
x2 dx
Solution: Let π/x = t ⇒ − πx2 dx = dt ⇒ dx = −x2dt
π. Therefore,∫
cos(π/x)
x2dx = −
∫cos t
x2
x2dt
π= − 1
π
∫cos tdt
= − 1
πsin t + C = − 1
πsin(π/x) + C.
25.∫
sin x1+cos2 x
dx
Solution: Let cos x = t ⇒ − sin xdx = dt ⇒ dx = − dtsin x
. Whence,∫sin x
1 + cos2 xdx = −
∫sin x
1 + t2dt
sin x=
∫−1
1 + t2dt
= − arctan t + C = − arctan(cos x) + C.
26.∫
x5 3√
x3 + 1dx
6
Solution: Let x3 + 1 = t ⇒ 3x2dx = dt ⇒ dx = dt3x2 . Thus,∫
x5 3√
x3 + 1dx =
∫x5 3√
tdt
3x2=
1
3
∫x3 3√
tdt
=1
3
∫(t− 1)t1/3dt =
1
3
∫(t4/3 − t1/3)dt
=1
3
(t7/3
7/3− t4/3
4/3
)+ C
=1
3
(3
7(x3 + 1)7/3 − 3
4(x3 + 1)4/3
)+ C.
27.∫
xx4+1
dx
Solution: Let x2 = t ⇒ 2xdx = dt ⇒ dx = dt2x
. Thus,∫x
x4 + 1=
∫x
t2 + 1
dt
2x=
1
2
∫1
t2 + 1dt
=1
2arctan t + C =
1
2arctan(x2) + C.
28.∫
x2√
1−xdx
Solution: Let√
1− x = t ⇒ 1− x = t2 ⇒ dx = −2tdt. Consequently,∫x2
√1− x
dx =
∫x2
t(−2tdt) = −2
∫(1− t2)2dt
= −2
∫(1− 2t2 + t4)dt = −2
(t− 2
3t3 +
1
5t5)
+ C
= −2
(√1− x− 2
3
√(1− x)3 +
1
5
√(1− x)5
)+ C.
29.∫ 2
0(x− 1)25dx
Solution: Let x− 1 = t ⇒ dx = dt. Hence∫(x− 1)25dx =
∫t25dt =
t26
26+ C =
(x− 1)26
26+ C.
Therefore, ∫ 2
0
(x− 1)25dx =(x− 1)26
26
]2
0
=1
26− 1
26= 0.
7
30.∫ π
0sec2(t/4)dt ∫
sec2(t/4)dt = 4 tan(t/4) + C.
Therefore,∫ π
0
sec2(t/4)dt = 4 tan(t/4)]π0 = 4 tan 0− 4 tan π/4 = 0− 1 = −1.
31.∫ 1
0xe−x2
dx
Solution: Let −x2 = t ⇒ dx = −dt2x
. Therefore,∫xe−x2
dx = −∫
xet dt
2x= −1
2
∫etdt = −1
2e−x2
+ C.
Consequently,∫ 1
0
xe−x2
dx = −1
2e−x2
]1
0
= −1
2e−1 +
1
2=
1
2(1− e−1).
32.∫ 2
−2(x + 3)
√4− x2dx
Solution:∫ 2
−2
(x + 3)√
4− x2dx =
∫ 2
−2
x√
4− x2dx + 3
∫ 2
−2
√4− x2dx
= −1
3
√(4− x2)3
]2
−2
+ 3× the area of a semi circle
whose radius is 2,
where the first integral has been solved using the substitution 4−x2 = t.
For the second integral: The equation of the circle centered at the origin
with radius 2 is x2 + y2 = 4 ⇒ y = ±√
4− x2. Therefore, the area of
the upper semicircle is given by the area bounded below by the x−axis
and above by the curve y =√
4− x2. This area may be evaluated by the
second integral. The area of this semicircle is 2π. Standard computations
now give: ∫ 2
−2
(x + 3)√
4− x2dx = 6π.
8
33. If f is continuous and∫ 4
0f(x)dx = 10, find
∫ 2
0f(2x)dx.
Solution: Do the substitution 2x = t, and observe that the limits of the
integral must be changed, in the wanted integral to get∫ 2
0
f(2x)dx =1
2
∫ 4
0
f(t)dt =1
2× 10 = 5.
34. If f is continuous and∫ 9
0f(x)dx = 4, find
∫ 3
0xf(x2)dx.
Solution: Let x2 = t in the wanted integral to get:∫ 3
0
xf(x2)dx =1
2
∫ 9
0
f(t)dt =1
2× 4 = 2.
35. If a and b are positive numbers, show that∫ 1
0
xa(1− x)bdx =
∫ 1
0
xb(1− x)adx.
Solution: Let 1−x = t ⇒ dx = −dt. Observe that the limits should be
changed as follows: x = 0 ⇒ t = 1− 0 = 1 and x = 1 ⇒ t = 1− 1 = 0.
Hence∫ 1
0
xa(1− x)bdx =
∫ 0
1
(1− t)atb(−dt) = −∫ 0
1
(1− t)atbdt
=
∫ 1
0
tb(1− t)adt
=
∫ 1
0
xb(1− x)adx.
Note: When having definite integrals, the letter used for the variable is
not a big matter. In other words,∫ 1
0tb(1− t)adt =
∫ 1
0xb(1− x)adx.
This observation is not necessarily true when dealing with indefinite in-
tegrals!
36. Show that ∫ π
0
xf(sin x)dx =π
2
∫ π
0
f(sin x)dx.
9
Solution: Let π − x = t ⇒ dx = −dt. Hence∫ π
0
xf(sin x)dx =
∫ 0
π
(π − t)f(sin(π − t))(−dt)
= π
∫ π
0
f(sin t)dt−∫ π
0
tf(sin t)dt,
therefore, ∫ π
0
xf(sin x)dx +
∫ π
0
tf(sin t)dt = π
∫ π
0
f(sin t)dt
⇒ 2
∫ π
0
xf(sin x)dx = π
∫ π
0
f(sin x)dx
⇒∫ π
0
xf(sin x)dt =π
2
∫ π
0
f(sin x)dx.
1.2 Integration by parts
The rule for integration by parts is∫udv = uv −
∫vdu.
This suggests that we divide the integrand into two parts. One (u) to be
differentiated and the other one (dv) to be integrated.
The following are problems taken from section 8.2, Calculus 8th edition,
Howard Anton.
1.∫
xe−2xdx. Whenever we have a product of a polynomial with an expo-
nential function we do parts if the exponent of the exponential is linear.
If the exponent is not linear we do a substitution.
We let the polynomial = u and the exponential = dv.
Thus,
x e−2x
↓(differentiate) ↘+ ↓(integrate)
1 − −→∫ −1
2e−2x
.
10
In this table: The first column represents the differentiation process and
the second represents the integral part. Therefore,∫xe−2xdx = −1
2xe−2x −
∫−1
2e−2xdx
= −1
2xe−2x − 1
4e−2x + C.
2.∫
xe3xdx. Do similar steps:
x e3x
↓ ↘+ ↓
1 − −→∫
13e3x
to get ∫xe3xdx =
1
3xe3x − 1
3
∫e3xdx
= xe3x − 1
9e3x + C.
3.∫
x2exdx. We apply the above steps twice as the table indicates:
x2 ex
↓ ↘+ ↓
2x ex
↓ ↘− ↓
2 −→+∫
ex
.
∫x2exdx = x2ex − 2xex + 2
∫exdx
= x2ex − 2xex + 2ex + C.
4.∫
x2e−2xdx.
x2 e−2x
↓ ↘+ ↓
2x −12e−2x
↓ ↘− ↓
2 −→+∫
14e−2x
.
∫x2e−2xdx = −1
2x2e−2x − 1
2xe−2x +
1
2
∫e−2xdx
= −1
2x2e−2x − 1
2xe−2x − 1
4e−2x + C.
5.∫
x sin 3xdx
11
When the integrand is a product of a polynomial and a trigonometric
function we let the polynomial = u and the trig. function = dv,
provided that the argument of the trig. function is linear.
Thus:
x sin 3x
↓ ↘+ ↓
1 −→∫ − −1
3cos 3x
∫x sin 3xdx = −1
3x cos 3x +
1
3
∫cos 3xdx
= −1
3x cos 3x +
1
9sin 3x + C.
6.∫
x cos 2xdx. Similar steps:
x cos 3x
↓ ↘+ ↓
1 −→∫ − 1
2sin 2x
∫x cos 2xdx =
1
2x sin 2x− 1
2
∫sin 2xdx
=1
2x sin 2x +
1
4cos 2x + C.
7.∫
x2 cos xdx. We apply integration by parts twice to get:
x2 cos x
↓ ↘+ ↓
2x sin x
↓ ↘− ↓
2 −→∫ + − cos x
∫x2 cos xdx = x2 sin x + 2x cos x− 2
∫cos xdx
= x2 sin x + 2x cos x− 2 sin x + C.
8.∫
x2 sin xdx. Similar steps:
x2 sin x
↓ ↘+ ↓
2x − cos x
↓ ↘− ↓
2 −→∫ + − sin x
∫x2 sin xdx = −x2 cos x + 2x sin x− 2
∫sin xdx
= −x2 cos x + 2x sin x + 2 cos x + C.
9.∫
x ln xdx.
12
When the integrand is a product of a polynomial and a logarithmic function we let
the polynomial = dv and the logarithm = u as follows:
ln x x
↓ ↘+ ↓1x
−→∫ − 1
2x2
∫x ln xdx =
1
2x2 ln x− 1
2
∫xdx
=1
2x2 ln x− 1
4x2 + C.
10.∫ √
x ln xdx.
ln x√
x
↓ ↘+ ↓1x
−→∫ − 2
3x3/2
∫ √x ln xdx =
2
3x3/2 ln x− 2
3
∫ √xdx
=2
3x3/2 ln x− 4
9x3/2 + C.
11.∫
(ln x)2dx
(ln x)2 1
↓ ↘+ ↓2 ln x
x−→∫ −
x∫
(ln x)2dx = x(ln x)2 − 2
∫ln xdx
= x(ln x)2 − 2(x ln x− x) + C.
We used the fact ln xdx = x ln x − x + C. This can be done using
integration by parts.
12.∫
ln x√xdx
ln x x−1/2
↓ ↘+ ↓1x
−→∫ −
2x1/2
∫ln x√
xdx = 2
√x ln x− 2
∫x−1/2dx
= 2√
x ln x− 4√
x + C.
13
13.∫
ln(3x−2)dx
ln(3x− 2) 1
↓ ↘+ ↓3
3x−2−→∫ −
x∫ln(3x− 2)dx = x ln(3x− 2)−
∫3x
3x− 2dx
= x ln(3x− 2)−∫
3x− 2 + 2
3x− 2dx
= x ln(3x− 2)−∫
dx−∫
2
3x− 2dx
= x ln(3x− 2)− x− 2
3ln(3x− 2) + C.
14.∫
ln(x2+4)dx
ln(x2 + 4) 1
↓ ↘+ ↓2x
x2+4−→∫ −
x
∫ln(x2 + 4)dx = x ln(x2 + 4)−
∫2x2
x2 + 4dx
= x ln(x2 + 4)− 2
∫ (1− 4
x2 + 4
)dx
= x ln(x2 + 4)− 2
(x− arctan(
1
2x)
)+ C.
15.∫
arcsin xdx
arcsin x 1
↓ ↘+ ↓1√
1−x2 −→∫
x
∫arcsin xdx = x arcsin x−
∫x√
1− x2dx
= x arcsin x +√
1− x2 + C,
where we have used a substitution in the last integral.
16.∫
arccos(2x)dx.
arccos(2x) 1
↓ ↘+ ↓−2√
1−4x2 −→∫ −
x
∫arccos(2x)dx = x arccos(2x) +
∫2x√
1− 4x2dx
= x arccos(2x)− 1
2
√1− 4x2 + C
17.∫
arctan(3x)dx
arctan(3x) 1
↓ ↘+ ↓3
9x2+1−→∫ −
x
14
∫arctan(3x)dx = x arctan(3x)−
∫3x
9x2 + 1dx
= x arctan(3x)− 1
6ln(9x2 + 1) + C.
18.∫
x arctan xdx
arctan x x
↓ ↘+ ↓1
x2+1−→∫ − x2
2
∫x arctan xdx =
1
2x2 arctan x− 1
2
∫x2
x2 + 1dx
=1
2x2 arctan x− 1
2
∫ (1− 1
x2 + 1
)dx
=1
2x2 arctan x− 1
2(x− arctan x) + C.
19.∫
ex sin xdx
sin x ex
↓ ↘+ ↓
cos x ex
↓ ↘− ↓
− sin x −→∫ +
ex
∫ex sin xdx = ex sin x− ex cos x−
∫ex sin xdx.
Thus, we have∫
ex sin xdx on both sides. Collect it in one side:
2
∫ex sin xdx = ex(sin x− cos x) + C
⇒∫
ex sin xdx =1
2ex (sin x− cos x) + C.
20.∫
e3x cos(2x)dx
cos(2x) e3x
↓ ↘+ ↓
−2 sin x 13e3x
↓ ↘− ↓
−4 cos(2x) −→∫ + 1
9e3x
∫e3x cos(2x)dx =
1
3e3x cos(2x) +
2
9e3x sin(2x)− 4
9
∫e3x cos(2x)dx.
Similar idea to the above:
13
9
∫e3x cos(2x)dx = e3x
(1
3cos(2x) +
2
9sin(2x)
)+ C
⇒∫
e3x cos(2x)dx =9
13× e3x
(1
3cos(2x) +
2
9sin(2x)
)+ C
15
21.∫
eax sin(bx)dx
sin(bx) eax
↓ ↘+ ↓
b cos(bx) 1aeax
↓ ↘− ↓
−b2 sin x −→∫ + 1
a2 eax
∫eax sin(bx)dx =
1
aeax sin(bx)− b
a2eax cos(bx)− b2
a2
∫eax sin(bx)dx;
hence, (1 +
b2
a2
)∫eax sin(bx)dx = eax
(1
asin(bx)− b
a2cos(bx)
)+ C
⇒∫
eax sin(bx)dx =a2
a2 + b2eax
(1
asin(bx)− b
a2cos(bx)
)+ C.
22.∫
e−3θ sin(5θ)dθ. Apply the above ideas!
23.∫
sin(ln x)dx.
Whenever the argument of a trig. function is not linear
we start by substituting this argument.
Thus
ln x = t ⇒ dx = xdt ⇒ dx = etdt
∫sin(ln x) =
∫sin(t)× etdt =
∫et sin tdt
=1
2et (sin t− cos t) + C
=1
2eln x (sin(ln x)− cos(ln x)) + C
=1
2x (sin(ln x)− cos(ln x)) + C.
24.∫
cos(ln x)dx. Similar to the above!
25.∫
x sec2 xdx
x sec2 x
↓ ↘+ ↓
1 −→∫ −
tan x
16
∫x sec2 xdx = x tan x−
∫tan xdx
= x tan x− ln |secx|+ C.
Reminder: Recall that∫tan xdx =
∫sin x
cos xdx
may be done by doing the substitution cos x = t.
26.∫
x tan2 xdx
x tan2 x
↓ ↘+ ↓
1 −→∫ −
tan x− x
∫x tan2 xdx = x(tan x− x)−
∫(tan x− x) dx
= x(tan x− x)−(
ln | sec x| − x2
2
)+ C.
27.∫
x3ex2dx
Whenever the exponent of e is not linear, we substitute it!
Thus, x2 = t ⇒ dx = dt2x
makes∫x3ex2
dx =
∫x3et dt
2x=
1
2
∫tetdt =
1
2et(t− 1) + C
=1
2ex2
(x2 − 1) + C.
Note: We used integration by parts to do∫
tetdt. The details are left to
the student.
28.∫
xex
(x+1)2dx.
xex (x + 1)−2
↓ ↘+ ↓
(x + 1)ex −→∫ − −(x + 1)−1
∫xex
(x + 1)2dx = − xex
x + 1+
∫exdx
= − xex
x + 1+ ex + C =
ex
x + 1+ C.
29.∫ 2
0xe2xdx. We do
∫xe2xdx by parts and then we plug the limits:
17
x e2x
↓ ↘+ ↓
1 −→∫ − 1
2e2x
∫xe2xdx =
1
2xe2x − 1
2
∫e2xdx
=1
2xe2x − 1
4e2x + C.
Therefore, ∫ 2
0
xe2xdx = e2x
(1
2x− 1
4
)]2
0
=3
4e4 − −1
4=
1
4(3e4 + 1).
30.∫ 1
0xe−5xdx
x e−5x
↓ ↘+ ↓
1 −→∫ − −1
5e−5x
∫xe−5xdx = −1
5xe−5x − 1
25e−5x + C,
therefore, ∫ 1
0
xe−5xdx = e−5x
(−1
5x− 1
25
)]1
0
= − 6
25e−5 − −1
25=
1
25(1− 6e−5).
31.∫ e
1x2 ln xdx
ln x x2
↓ ↘+ ↓1x
−→∫ − x3
3
∫x2 ln xdx =
x3
3ln x− 1
9x3 + C,
therefore, ∫ e
1
x2 ln xdx =
(1
3ln x− 1
9
)x3
]e
1
=2
9e3 − −1
9=
1
9(1 + 2e3).
32.∫ e√
eln xx2 dx
ln x 1x2
↓ ↘+ ↓1x
−→∫ − − 1
x
∫ln x
x2dx = −1
xln x +
∫1
x2dx
= −1
xln x− 1
x+ C,
18
therefore, ∫ e
√e
ln x
x2dx = −1
x(ln x + 1)
]e
√e
= −2
e+
3
2√
e.
33.∫ 1
−1ln(x+2)dx
ln(x + 2) 1
↓ ↘+ ↓1
x+2−→∫ −
x + 2
∫ln(x + 2)dx = (x + 2) ln(x + 2)− x + C,
therefore, ∫ 1
−1
ln(x + 2)dx = (x + 2) ln(x + 2)− x]1−1
= 3 ln 3− 1− 1 = 3 ln 3− 2.
34.∫ √3/2
0arcsin xdx. We have seen that∫
arcsin xdx = x arcsin x +√
1− x2 + C.
For the details see problem 15. Thus,∫ √3/2
0
arcsin xdx = x arcsin x +√
1− x2]√3/2
0
=
√3
2arcsin(
√3/2) +
√1/4− (0 + 1)
=
√3π
6− 1
2.
35.∫ 4
2sec−1(
√θ)dθ
sec−1√
θ 1
↓ ↘+ ↓1
2θ√
θ−1−→∫ −
θ
∫sec−1
√θdθ = θ sec−1
√θ − 1
2
∫1√
θ − 1dθ
= θ sec−1√
θ −√
θ − 1 + C.
Therefore,∫ 4
2
sec−1(√
θ)dθ = θ sec−1√
θ −√
θ − 1]4
2
= 2 sec−1 2−√
3− 2 sec−1√
2 + 1
=2π
3−√
3− 2π
4+ 1
=π
6+ 1−
√3.
19
36.∫ 2
1x sec−1 xdx
37.∫ π
0x sin 2xdx
38.∫ π
0(x + x cos x)dx
39.∫ 3
1
√x tan−1
√xdx
tan−1√
x√
x
↓ ↘+ ↓1
2√
x(x2+1)−→∫ − 1
2√
x
∫ √x tan−1
√xdx =
tan−1√
x
2√
x− 1
4
∫1
x(x2 + 1)dx
=tan−1
√x
2√
x− 1
4
∫1 + x2 − x2
x(x2 + 1)dx
=tan−1
√x
2√
x− 1
4
∫ (1
x− x
x2 + 1
)dx
=tan−1
√x
2√
x− 1
4
(ln |x| − 1
2ln(x2 + 1)
)+ C.
Hence,∫ 3
1
√x tan−1
√xdx =
tan−1√
x
2√
x− 1
4
(ln |x| − 1
2ln(x2 + 1)
)]3
1
=
(tan−1
√3
2√
3− 1
4
(ln 3− 1
2ln 10
))
−(
tan−1 1
2− 1
4× −1
2ln 2
)= ......
40.∫ 2
0ln(x2 + 1)dx.
41. (a)∫
e√
xdx
Let√
x = t ⇒ x = t2 ⇒ dx = 2tdt.
Hence, ∫e√
xdx =
∫et2tdt = 2
∫tetdt
= 2(tet − et) + C
= 2e√
x(√
x− 1) + C.
(b)∫
cos√
xdx
20
√x = t ⇒ x = t2 ⇒ dx = 2tdt.
∫cos√
xdx = 2
∫t cos tdt
= 2(t sin t + cos t) + C
= 2(√
x sin√
x + cos√
x) + C
58. In each part, use integration by parts or other methods to derive the
reduction formula:
(a) ∫secn xdx =
secn−2 x tan x
n− 1+
n− 2
n− 1
∫secn−1 xdx.
secn−2 x sec2 x
↓ ↘+ ↓
(n− 2) secn−2 x tan x −→∫ −
tan x
∫secn xdx = secn−2 x tan x− (n− 2)
∫secn−2 x tan2 xdx
= secn−2 x tan x− (n− 2)
∫secn−2 x(sec2 x− 1)dx
= secn−2 x tan x− (n− 2)
∫secn xdx + (n− 2)
∫secn−2 xdx,
hence ∫secn xdx + (n− 2)
∫secn xdx = secn−2 x tan x + (n− 2)
∫secn−2 xdx
⇒ (n− 1)
∫secn xdx = secn−2 x tan x + (n− 2)
∫secn−2 xdx
⇒∫
secn xdx =secn−2 x tan x
n− 1+
n− 2
n− 1
∫secn−2 xdx.
(b) ∫tann xdx =
tann−1 x
n− 1−∫
tann−2 xdx.
21
∫tann xdx =
∫tann−2 x tan2 xdx =
∫tann−2 x(sec2 x− 1)dx
=
∫tann−2 x sec2 xdx−
∫tann−2 xdx
=tann−1 x
n− 1−∫
tann−2 xdx,
where we have used the substitution tan x = t to do the integral∫tann−2 x sec2 xdx.
(c) ∫xnexdx = xnex − n
∫xn−1exdx.
Integrate by parts:
xn ex
↓ ↘+ ↓
nxn−1 −→∫ −
ex
∫xnexdx = xnex − n
∫xn−1exdx.
1.3 Trigonometric Integrals
1.∫
cos3 x sin xdx. Since the derivative of cos x is sin x which is a part of
the integrand we do the substitution:
cos x = t ⇒ − sin xdx = dt ⇒ dx = − dtsin x
.
Thus, ∫cos3 x sin xdx =
∫t3 sin x
−dt
sin x
= −∫
t3dt = −t4
4+ C
= −cos4 x
4+ C.
2.∫
sin5 3x cos 3xdx.
22
sin 3x = t ⇒ 3 cos 3xdx = dt ⇒ dx = dt3 cos 3x
.
Hence ∫sin5 3x cos 3xdx =
∫t5 cos 3x
dt
3 cos 3xdx =
1
3
∫t5dt
=1
18t18 + C =
1
18sin6 3x + C.
3.∫
sin2 5xdx.
To do the integral∫
sineven xdx we use the identity sin2 x = 12(1− cos 2x).
Thus, ∫sin2 5xdx =
1
2
∫(1− cos 10x)dx
=1
2
(x− sin 10x
10
)+ C.
4.∫
cos2 3xdx.
To do∫
coseven xdx we use the identity cos2 x = 12(1 + cos 2x).
Thus, ∫cos2 3xdx =
1
2
∫(1 + cos 6x)dx
=1
2
(x +
sin 6x
6
)+ C.
5.∫
sin3 axdx.
To do∫
sinodd xdx we let cos x = t ⇒ dx = − dtsin x
and then we use the identity
sin2 x = 1− cos2 x = 1− t2.
23
Let cos ax = t ⇒ dx = − dta sin ax
(a 6= 0),∫sin3 axdx =
∫sin3 ax
−dt
a sin ax
= −1
a
∫sin2 axdt = −1
a
∫(1− cos2 ax)dt
= −1
a
∫(1− t2)dt = −1
a(t− t3/3) + C
= −1
a
(cos ax− cos3 ax
3
)+ C.
6.∫
cos3 atdt. Let sin at = x ⇒ dt = dxa cos at
, hence∫cos3 atdt =
∫cos3 at
dx
a cos at=
1
a
∫cos2 atdx
=1
a
∫(1− sin2 at)dx =
1
a
∫(1− x2)dx
=1
a(x− x3/3) + C =
1
a
(sin at− sin3 at
3
)+ C.
7.∫
sin ax cos axdx. Let sin ax = t ⇒ a cos axdx = dt ⇒ dx = 1a cos ax
dt,
hence ∫sin ax cos axdx =
∫t cos ax
1
a cos axdt
=1
a
∫tdt =
1
2at2 + C
=1
2asin2 ax + C.
8.∫
sin3 x cos3 xdx.
To do∫
sinodd x cosodd xdx we let the trig. function whose power is bigger equal t.
Thus, let sin x = t ⇒ cos xdx = dt ⇒ dx = dtcos x
, hence∫sin3 x cos3 xdx =
∫t3 cos3 x
dt
cos x=
∫t3 cos2 xdt
=
∫t3(1− t2)dt =
∫(t3 − t5)dt
=
(t4
4− t6
6
)+ C
=
(sin4 x
4− sin6 x
6
)+ C.
24
9.∫
sin2 t cos3 tdt.
To do∫
sineven x cosodd xdx we let sin x = t,
then we use cos2 x = 1− sin2 x = 1− t2.
Thus,∫sin2 x cos3 xdx =
∫t2 cos3 x
dt
cos x=
∫t2 cos2 xdt
=
∫t2(1− sin2 x)dt =
∫t2(1− t2)dt
= t3/3− t5/5 + C =sin3 x
3− sin5 x
5+ C.
10.∫
sin3 x cos2 xdx
To do∫
sinodd x coseven xdx we let cos x = t ⇒ dx = − 1sin x
dt
then we use sin2 x = 1− cos2 x = 1− t2.
Thus ∫sin3 x cos2 xdx =
∫sin3 xt2
−dt
sin x= −
∫sin2 xt2dt
= −∫
(1− t2)t2dt = −(
t3
3− t5
5
)+ C
= −(
cos3 x
3− cos5 x
5
)+ C.
11.∫
sin2 x cos2 xdx.
To do∫
sineven x coseven xdx, we use the identities
sin2 x = 12(1− cos 2x), cos2 x = 1
2(1 + cos 2x).
Thus, ∫sin2 x cos2 xdx =
∫1
2(1− cos 2x)
1
2(1 + cos 2x)dx
=1
4
∫(1− cos2 2x)dx =
1
4
∫sin2 2xdx
=1
4
∫1
2(1− cos 4x)dx
=1
8
(x− sin 4x
4
)+ C.
25
12.∫
sin2 x cos4 xdx. Apply the above remark,∫sin2 x cos4 xdx =
∫1
2(1− cos 2x)
[1
2(1 + cos 2x)
]2
dx
=1
8
∫(1− cos 2x)(1 + cos 2x)(1 + cos 2x)dx
=1
8
∫(1− cos2 2x)(1 + cos 2x)dx
=1
8
∫sin2 2x(1 + cos 2x)dx
=1
8
∫(sin2 2x + sin2 2x cos 2x)dx.
Now, for∫sin2 2xdx =
1
2
∫(1− cos 4x)dx =
1
2
(x− sin 4x
4
)+ C.
For∫
sin2 2x cos 2xdx we let sin 2x = t to get∫sin2 2x cos 2xdx =
1
6sin3 2x + C.
Thus ∫sin2 x cos4 xdx =
1
8
[1
2
(x− sin 4x
4
)+
1
6sin3 2x
]+ C.
13.∫
sin 2x cos 3xdx.
To do integrals of the form∫
sin ax cos bxdx we use the identity
sin ax cos bx = 12[sin(a + b)x + sin(a− b)x] .
Thus ∫sin 2x cos 3xdx =
1
2
∫[sin 5x + sin(−x)] dx
=1
2
[cos 5x
5+ cos x
]+ C.
14.∫
sin 3x cos 2xdx. We do similar steps:∫sin 3x cos 2xdx =
1
2[sin 5x + sin x] dx
=1
2
[− cos 5x
5− cos x
]+ C.
26
15.∫
sin x cos(x/2)dx. Apply the above identity:∫sin x cos(x/2)dx =
1
2
∫[sin(3x/2) + sin(x/2)] dx
=1
2
[−cos(3x/2)
3/2− cos(x/2)
1/2
]+ C.
16.∫
cos1/3 x sin xdx. Do the substitution:
cos x = t ⇒ dx = − dtsin x
Thus ∫cos1/3 x sin xdx =
∫t1/3 sin x
−dt
sin x
= −3
4t4/3 + C = −3
4cos4/3 x + C.
17.∫ π/2
0cos3 xdx.
To do∫
cosodd xdx we let sin x = t ⇒ dx = dtcos x
.
Therefore,∫cos3 xdx =
∫cos3 x
dt
cos x=
∫cos2 xdt =
∫(1− t2)dt
= t− t3
3+ C = sin x− sin3 x
3+ C.
Hence, ∫ π/2
0
cos3 xdx = sin x− sin3 x
3
]π/2
0
=2
3− 0 =
2
3.
27
18. ∫ π/2
0
sin2 x
2cos2 x
2dx =
∫ π/2
0
1
2(1− cos x)
1
2(1 + cos x)dx
=1
4
∫ π/2
0
(1− cos2 x)dx
=1
4
∫ π/2
0
sin2 xdx
=1
4
1
2
∫(1− cos 2x)dx
=1
8
(x− sin 2x
2
)]π/2
0
=π
16.
19.∫ π/3
0sin4 3x cos3 3xdx. We do the substitution
sin 3x = t ⇒ dx = dt3 cos x
.
∫sin4 3x cos3 3xdx =
∫t4 cos3 3x
dt
3 cos 3x=
1
3
∫t4 cos2 3xdx
=1
3
∫t4(1− t2)dt =
1
3
(t5
5− t7
7
)+ C
=1
3
(sin5 3x
5− sin7 3x
7
)+ C.
Therefore,∫ π/3
0
sin4 3x cos3 3xdx =1
3
(sin5 3x
5− sin7 3x
7
)]π/3
0
= 0.
20.∫ π
−πcos2 5xdx. ∫
cos2 5xdx =1
2
∫(1 + cos 10x)dx
=1
2
(x +
sin 10x
10
)+ C.
Hence ∫ π
−π
cos2 5xdx =1
2
(x +
sin 10x
10
)]π
−π
= π.
28
21.∫ π/6
0sin 4x cos 2xdx.∫
sin 4x cos 2xdx =1
2
∫[sin 6x + sin 2x] dx
=1
2
[− cos 6x
6− sin 2x
2
]+ C.
Thus ∫ π/6
0
sin 4x cos 2xdx =1
2
[− cos 6x
6− sin 2x
2
]π/6
0
=1
2
[1
6−√
3
4
]− 1
2[−1
6− 0]
=1
6−√
3
4.
22.∫ 2π
0sin2 kxdx. ∫
sin2 kxdx =1
2
∫(1− cos 2kx)dx
=1
2
(x− sin 2kx
2k
)+ C.
Hence, ∫ 2π
0
sin2 kxdx =1
2
(x− sin 2kx
2k
)]2π
0
= π.
23.∫
sec2(2x − 1)dx. Doing the substitution 2x − 1 = t yields∫
sec2(2x −
1)dx = 12tan(2x− 1) + C.
24.∫
tan 5xdx =∫
sin 5xcos 5x
dx. doing the substitution cos 5x = t yields
1
5
∫tan 5xdx = ln | sec 5x|+ C.
25.∫
e−x tan(e−x)dx. Do the substitution e−x = t to get∫e−x tan(e−x)dx = −
∫tan tdt = − ln | sec t|+ C = − ln | sec(e−x)|+ C.
29
26.∫
cot 3xdx =∫
cos 3xsin 3x
dx. Do the substitution cos 3x = t to get∫cot 3xdx =
∫cos 3x
t
dt
3 cos 3x
=1
3ln |t|+ C =
1
3ln | sin 3x|+ C.
27.∫
sec 4xdx. We multiply with and divide by sec 4x + tan 4x to get∫sec 4xdx =
∫sec 4x(sec 4x + tan 4x)
sec 4x + tan 4xdx
=
∫sec2 4x + sec 4x tan 4x
tan 4x + sec 4xdx
=1
4ln | sec 4x + tan 4x|+ C,
where we have used the substitution sec 4x + tan 4x = t in the last
integral.
28.∫
sec√
x√x
dx. Do the substitution
√x = t ⇒ x = t2 ⇒ dx = 2tdt.
Therefore∫sec√
x√x
dx = 2
∫sec tdt = 2
∫sect(sec t + tan t)
sec t + tan tdt
= 2 ln | sec t + tan t|+ C = 2 ln | sec√
x + tan√
x|+ C.
29.∫
tan2 x sec2 xdx. Do the substitution:
tan x = t ⇒ sec2 xdx = dt ⇒ dx = dtsec2 x
.
Hence ∫tan2 x sec2 xdx =
∫t2 sec2 x
dt
sec2 x=
∫t2dt
=t3
3+ C =
tan3 x
3+ C.
30.∫
tan5 x sec4 xdx.
30
To do integrals of the form∫
tanodd x seceven xdx, do one of the substitutions
tan x = t or sec x = t.
Let us do tan x = t ⇒ sec2 xdx = dt ⇒ dx = dtsec2 x
. Hence∫tan5 x sec4 xdx =
∫t5 sec4 x
dt
sec2 x=
∫t5 sec2 xdt
=
∫t5(tan2 x + 1)dt =
∫t5(t2 + 1)dt
=t8
8+
t6
6+ C
=tan8 x
8+
tan6 x
6+ C.
31.∫
tan 4x sec4 4xdx. Similar to the above. Let’s do
sec 4x = t ⇒ dx =dt
4 sec 4x tan 4x=
dt
4t tan 4x.
Hence, ∫tan 4x sec4 4xdx =
1
4
∫tan 4xt4
dt
t tan 4x=
1
4
∫t3dt
=1
4
t4
4+ C =
sec4 4x
16+ C.
32.∫
tan4 x sec4 xdx.
To do∫
taneven x seceven xdx we let tan x = t, then we
use the identity sec2 x = tan2 x + 1.
Thus, ∫tan4 x sec4 xdx =
∫t4 sec4 x
dt
sec2 x=
∫t4 sec2 xdt
=
∫t4(tan2 x + 1)dt =
∫t4(t2 + 1)dt
=t7
7+
t5
5+ C =
tan7 x
7+
tan5 x
5+ C.
33.∫
sec5 x tan3 xdx.
31
To do integrals of the form∫
tanodd x secodd xdx we use the substitution
sec x = t then we use the identity tan2 x = sec2 x− 1.
Whence∫sec5 x tan3 xdx =
∫t5 tan3 x
dt
sec x tan x=
∫t5 tan3 x
dt
t× tan x
=
∫t4 tan2 xdt =
∫t4(sec2 x− 1)dt
=
∫t4(t2 − 1)dt =
t7
7− t5
5+ C
=sec7 x
7− sec5 x
5+ C.
34. tan5 x sec xdx. Same as above: sec x = t ⇒ secx tan xdx = dt ⇒ dx =
dtt×tan x
. Hence∫tan5 x sec xdx =
∫tan5 xt
dt
t× tan x=
∫tan4 xdt
=
∫(tan2 x)2dt =
∫(sec2 x− 1)2dt =
∫(t2 − 1)2dt
=
∫(t4 − 2t2 + 1)dt =
t5
5− 2t3
3+ t + C
=sec5 x
5− 2 sec3 x
3+ sec x + C.
35.∫
tan4 x sec xdx.
For integrals of the form∫
taneven x secodd xdx we replace each tan2 x with sec2 x− 1
in order to get an integral involving only powers of sec x.
Thus∫tan4 x sec xdx =
∫(tan2 x)2 sec xdx =
∫(sec2 x− 1)2 sec xdx
=
∫(sec4 x− 2 sec2 x + 1) sec xdx
=
∫(sec5 x− 2 sec3 x + sec x)dx. (1.3.1)
32
Each of these three integrals has the form∫
secodd xdx. For∫
sec xdx see
example 27 above. Thus∫sec xdx = ln | sec x + tan x|+ C.
Now,
To do integrals of the form∫
secodd>1 xdx we use integration by parts.
sec2 x = dv and the rest is u.
Thus, for∫
sec3 xdx we use sec2 x = dv, sec x = u:
sec x sec2 x
↓ ↘+ ↓
sec x tan x −→∫ −
tan x
Therefore∫sec3 xdx = sec x tan x−
∫sec x tan2 xdx
= sec x tan x−∫
sec x(sec2 x− 1)dx
= sec x tan x−∫
sec3 xdx +
∫sec xdx
thus, the term∫
sec3 x appears on both sides. Collect it to get
2
∫sec3 xdx = sec x tan x +
∫sec xdx
⇒∫
sec3 xdx =1
2(sec x tan x + ln | sec x + tan x|) + C.
It remains to do∫
sec5 xdx. This may be done in a similar way:
sec3 x sec2 x
↓ ↘+ ↓
3 sec3 x tanx −→∫ − tanx
∫sec5 xdx = sec3 x tanx− 3
∫sec3 x tan2 xdx
= sec3 x tanx− 3∫
sec3 x(sec2 x− 1)dx
where we have used the idea for doing integrals of the form∫
secodd x taneven xdx
in the last line. Thus∫sec5 xdx = sec3 x tan x− 3
∫sec5 xdx + 3
∫sec3 xdx.
33
We have∫
sec5 xdx on both sides, collect it. Also we have∫
sec3 xdx
which has been done above:
4
∫sec5 xdx = sec3 x tan x +
3
2(sec x tan x + ln | sec x + tan x|) + C
⇒∫
sec5 xdx =1
4
[sec3 x tan x +
3
2(sec x tan x + ln | sec x + tan x|)
]+ C.
Thus, collecting the terms needed to do 1.3.1 we see that:∫tan4 x sec xdx =
1
4
[sec3 x tan x +
3
2(sec x tan x + ln | sec x + tan x|)
]− (sec x tan x + ln | sec x + tan x|) + ln | sec x + tan x|+ C.
36.∫
tan2 x sec3 xdx. Since the power of tan is even and that of sec is odd
we may apply the remark above to get∫tan2 x sec3 xdx =
∫(sec2 x− 1) sec3 xdx
=
∫sec5 xdx−
∫sec3 xdx
=1
4
[sec3 x tan x +
3
2(sec x tan x + ln | sec x + tan x|)
]− 1
2(sec x tan x + ln | sec x + tan x|) + C,
where we have used the formulae of∫
sec5 xdx and∫
sec3 xdx made in
the above example.
37.∫
tan t sec3 tdt. Since both powers are odd we let sec t = x ⇒ sec t tan tdt =
dx ⇒ dt = dxx tan t
. Thus∫tan t sec3 tdt =
∫tan tx3 dx
x× tan t
=
∫x2dx =
x3
3+ C =
sec3 t
3+ C.
38. tan x sec5 xdx. Same as above:∫tan x sec5 xdx =
∫tan xt5
dt
t× tan x
=
∫t4dt =
t5
5+ C =
sec5 x
5+ C.
34
39.∫
sec4 xdx.
To do integrals of the form∫
seceven xdx we let tan x = t.
tan x = t ⇒ dx = dtsec2 x
and sec2 x = t2 + 1. Therefore∫sec4 xdx =
∫sec4 x
dt
sec2 x=
∫sec2 xdt
=
∫(t2 + 1)dt =
t3
3+ t + C
=tan3 x
3+ tan x + C.
40.∫
sec5 xdx. This has been done in example 35.
41.∫
tan3 4xdx.
For integrals of the form∫
tanany power>1 xdx we separate a tan2 xdx
and replace it by sec2 x− 1. We repeat this until the problem is solved!
∫tan3 4xdx =
∫tan 4x(tan2 4x)dx =
∫tan 4x(sec2 4x− 1)dx
=
∫tan 4x sec2 4xdx−
∫tan 4xdx
=1
4tan2 4x− 1
4ln | sec 4x|+ C
where we have used the substitutions tan 4x = t and cos 4x = t, respec-
tively, to do the last two integrals.
42.∫
tan4 xdx. Same idea as above:∫tan4 xdx =
∫tan2 x tan2 xdx =
∫tan2 x(sec2 x− 1)dx
=
∫tan2 x sec2 xdx−
∫tan2 xdx
=
∫tan2 x sec2 xdx−
∫(sec2 x− 1)dx
=1
3tan3 x− (tan x− x) + C.
35
43.∫ √
tan x sec4 xdx. Do the substitution
tan x = t ⇒ sec2 xdx = dt ⇒ dx =dt
sec2 x.
Hence∫ √tan x sec4 xdx =
∫ √t sec4 x
dt
sec2 x
=
∫ √t sec2 xdt =
∫ √t(tan2 x + 1)dt
=
∫ √t(t2 + 1)dt =
2
7t7/2 +
2
3t3/2 + C
=2
7tan7/2 x +
2
3tan3/2 x + C.
44.∫
tan x sec3/2 xdx. Do the substitution
sec x = t ⇒ sec x tan xdx = dt ⇒ dt
t tan x
hence∫tan x sec3/2 xdx =
∫tan xt3/2 dt
t tan x
=
∫ √tdt =
2
3t3/2 + C =
2
3sec3/2 x + C.
45. ∫ π/8
0
tan2 2xdx =
∫ π/8
0
(sec2 2x− 1)dx
=1
2tan 2x− x
]π/8
0
=1
2− π
8.
46. ∫ π/6
0
sec3 2x tan 2xdx =1
6sec3 2x
]π/6
0
=7
6.
36
47.∫ π/2
0tan5 x
2dx.∫
tan5 xdx =
∫tan3 x(sec2 x− 1)dx
=
∫tan3 x sec2 xdx−
∫tan3 xdx
=1
4tan4 x−
∫tan x(sec2 x− 1)dx
=1
4tan4 x− 1
2tan2 x + ln | sec x|+ C.
Hence∫tan5 x
2dx =
1
2
[1
4tan4 x
2− 1
2tan2 x
2+ ln | sec x
2
]+ C,
which gives∫ π/2
0
tan5 x
2dx =
1
2
[1
4tan4 x
2− 1
2tan2 x
2+ ln | sec x
2|]π/2
0
=1
2(−1
4+ ln
√2).
48. ∫ 1/4
0
sec πx tan πxdx =1
πsec πx
]1/4
0
=1
π(1−
√2).
49.∫
cot3 x csc3 xdx. Do the substitution
csc x = t ⇒ − csc x cot xdx = dt ⇒ dx =dt
−t cot x.
Hence∫csc3 x cot3 xdx =
∫t3 cot3 x
dt
−t cot x= −
∫t2 cot2 xdt
= −∫
t2(t2 − 1)dt = −(
t5
5− t3
3
)+ C
= −(
csc5 x
5− csc3 x
3
)+ C.
37
50. ∫cot2 3x sec 3xdx =
∫cos2 3x
sin2 3x
1
cos 3xdx
=
∫cos 3x
sin2 3xdx =
∫cot 3x csc 3xdx
= −1
3csc 3x + C.
51. ∫cot3 xdx =
∫cot x cot2 xdx =
∫cot x(csc2 x− 1)dx
=
∫cot x csc2 xdx−
∫cot xdx
= −cot2 x
2− ln | sin x|+ C.
52.∫
csc4 xdx. Lett
cot x = t ⇒ − csc2 xdx = dt ⇒ dx = − dt
csc2 x.
Hence ∫csc4 xdx =
∫csc4 x
−dt
csc2 x= −
∫csc2 xdt
= −∫
(t2 + 1)dt = −(
t3
3+ t
)+ C
= −(
cot3 x
3+ cot x
)+ C.
53. Let m and n be distinct integers, show that
(a)∫ 2π
0sin mx cos nxdx = 0. Use the identity
sin a cos b = 12[sin(a− b) + sin(a + b)].
38
∫ 2π
0
sin mx cos nxdx =1
2
∫ 2π
0
[sin(m− n)x + sin(m + n)x] dx
=1
2
[−cos(m− n)x
m− n− cos(m + n)x
m + n
]2π
0
=1
2
[− 1
m− n− 1
m + n
]− 1
2
[−cos(m− n)2π
m− n− cos(m + n)2π
m + n
]=
1
2
[− 1
m− n− 1
m + n
]− 1
2
[− 1
m− n− 1
m + n
]= 0.
(b)∫
cos mx cos nxdx = 0. Use the identity
cos a cos b = 12[cos(a− b) + cos(a + b)]
and apply similar ideas as above.
(c)∫
sin mx sin nxdx = 0. Use the identity
sin a sin b = 12[cos(a− b)− cos(a + b)]
and apply similar ideas as above.
54. Solve∫
sin nx cos mxdx when n = m 6= 0. Let’s denote m and n both by
a so that we have∫sin nx cos mxdx =
∫sin ax cos axdx
=1
2asin2 ax + C,
where we have used the substitution sin ax = t to do the last integral.
39
1.4 Trigonometric substitution
The idea of this section is to compare some algebraic quantities with one of
the following trigonometric identities:
1− sin2 t = cos2 t,
1 + tan2 t = sec2 t,
sec2 t− 1 = tan2 t.
We shall use the symbol ∼ to say that two quantities look like each other. For
instance, 1 − x2 ∼ 1 − sin2 t, 2 − 3x2 ∼ 1 − sin2 t, x2 + 3 ∼ tan2 t + 1 and
x2 − 3 ∼ sec2 t− 1.
The examples illustrate how to choose the suitable substitution once a similar
term has been found.
The following remark should be emphasized before proceeding:
We know that√
cos2 t = | cos t|. In the sequel, we shall always write√
cos2 t = cos t.
The reason for this is that when we do a substitution like x = sin t we intend to have
a well defined inverse sin−1 t of sin t. In order to have this, t should be in [−π2, π
2] where
cos t is always positive!
1.∫ √
4− x2dx. We observe that
4− x2 ∼ 1− sin2 t ⇒ 4− x2 ∼ 4− 4 sin2 t ⇒ −x2 ∼ −4 sin2 t ⇒ x ∼ 2 sin t.
Thus, do the substitution x = 2 sin t ⇒ dx = 2 cos tdt.
Therefore,∫ √4− x2dx =
∫ √4− 4 sin2 t× 2 cos tdt
= 4
∫ √cos2 t cos tdt = 4
∫cos2 tdt
= 4
∫1
2(1 + cos 2t)dt
= 2
(t +
1
2sin 2t
)+ C.
40
In order to find this final answer in terms of x we observe that
x = 2 sin t ⇒ x
2= sin t ⇒ t = sin−1 x
2
and that
sin 2t = 2 sin t cos t ⇒ sin 2t = 2x
2
√1− sin2 t = x
√1− x2
4.
Thus, ∫ √4− x2dx = 2
(sin−1 x
2+
1
2x
√1− x2
4
)+ C.
2.∫ √
1− 4x2dx. We observe that
1− 4x2 ∼ 1− sin2 t ⇒ −4x2 ∼ sin2 t ⇒ x2 ∼ 14sin2 t ⇒ x ∼ 1
2sin t.
Thus, do the substitution x = 12sin t ⇒ dx = 1
2cos tdt.
Therefore,∫ √1− 4x2dx =
∫ √1− 4
1
4sin2 t× 1
2cos tdt
=1
2
∫cos2 tdt =
1
4
∫(1 + cos 2t)dt
=1
4
(t +
1
2sin 2t
)+ C.
In order to write this final answer in terms of x, we observe that
x =1
2sin t ⇒ t = sin−1 2x
and
sin 2t = 2 sin t cos t = 2× 2x√
1− sin2 t = 4x√
1− 4x2.
Thus ∫ √1− 4x2dx =
1
4
(sin−1 2x + 2x
√1− 4x2
)+ C.
3.∫
x2√
16−x2 dx.
41
16− x2 ∼ 1− sin2 t ⇒ 16− x2 ∼ 16− 16 sin2 t ⇒ x ∼ 4 sin t.
Do the substitution x = 4 sin t ⇒ dx = 4 cos tdt to get
∫x2
√16− x2
dx =
∫16 sin2 t√
16− 16 sin2 t4 cos tdt
= 16
∫sin2 tdt = 8
∫(1− cos 2t)dt
= 8
(t− 1
2sin 2t
)+ C
= 8(sin−1 x
4− sin t cos t
)+ C
= 8
(sin−1 x
4− x
4
√1− x2
16
)+ C.
4.∫
dxx2√
9−x2 .
9− x2 ∼ 1− sin2 t ⇒ 9− x2 ∼ 9− 9 sin2 t ⇒ x2 ∼ 9 sin2 t.
Thus do the substitution x = 3 sin t ⇒ dx = 3 cos tdt.
Hence ∫dx
x2√
9− x2=
∫3 cos tdt
9 sin2 t√
9− 9 sin2 t
=1
9
∫1
sin2 tdt =
1
9
∫csc2 tdt
= −1
9cot t + C = −1
9
cos t
sin t+ C
= −1
9
√1− x2/9
x/3+ C.
5.∫
dx(x2+4)2
.
x2 + 4 ∼ tan2 t + 1 ⇒ x2 + 4 ∼ 4 tan2 t + 4 ⇒ x2 ∼ 4 tan2 t.
Thus, do the substitution x = 2 tan t ⇒ dx = 2 sec2 tdt.
42
Whence ∫dx
(x2 + 4)2=
∫2 sec2 tdt
(4 tan2 t + 4)2
=1
8
∫sec2 t
(tan2 t + 1)2=
1
8
∫1
sec2 tdt
=1
8
∫cos2 tdt =
1
16(t +
1
2sin 2t) + C
=1
16
(tan−1 x
2+
2x
x2 + 4
)+ C.
6.∫
x2√
5+x2 dx.
5 + x2 ∼ 1 + tan2 t ⇒ 5 + x2 ∼ 5 + 5 tan2 t ⇒ x ∼ 5 tan2 t.
Do the substitution x =√
5 tan t ⇒ dx =√
5 sec2 tdt to get
∫x2
√5 + x2
dx =
∫5 tan2 t√
5 + 5 tan2 t
√5 sec2 tdt
= 5
∫tan2 t sec2 t√
1 + tan2 tdt
= 5
∫tan2 t sec tdt
= 5
∫(sec2 t− 1) sec tdt
= 5
∫(sec3− sec t)dt
= 5
(1
2(sec t tan t− ln | sec t + tan t|)− ln | sec t + tan t|
)+ C
=5
2sec t tan t− 7
2ln | sec t + tan t|+ C
=5
2
√1 +
x2
5
x√5− 7
2ln
∣∣∣∣∣√
1 +x2
5+
x√5
∣∣∣∣∣+ C.
7.∫ √
x2−9x
dx. Do the substitution x = 3 sec t ⇒ dx = 3 sec t tan tdt. Hence∫ √x2 − 9
xdx =
∫3 tan t
3 sec t3 sec t tan tdt
= 3
∫tan2 tdt = 3
∫(sec2 t− 1)dt
= 3(tan t− t) + C
= 3
(√x2
9− 1− sec−1 x
3
)+ C.
43
8.∫
dxx2√
x2−16. let x = 4 sec t ⇒ dx = 4 sec t tan tdt. Hence∫
dx
x2√
x2 − 16=
∫4 sec t tan t
16 sec2 t√
16 sec2 t− 16
=1
16
∫1
sec tdt =
1
16
∫cos t + C
=1
16sin t + C =
1
16
√1− 16
x2+ C.
9.∫
3x3√
1−x2 dx. We may use the substitution 1− x2 = t ⇒ dx = − dt2x
to get∫3x3
√1− x2
dx =
∫3x3
√t
−dt
2x= −3
2
∫x2
√tdt
= −3
2
∫1− t√
tdt = −3
2
(1√t−√
t
)dt
= −3
2
(2√
t− 2
3t3/2
)+ C
= −3
2
(2√
1− x2 − 2
3
√(1− x2)3
)+ C.
10.∫
x3√
5− x2dx. Do 5− x2 = t ⇒ dx = − dt2x
, hence∫x3√
5− x2dx = −∫
x3√
tdt
2x= −1
2
∫x2√
tdt
= −1
2
∫x2√
tdt = −1
2
∫(5− t)
√tdt
= −1
2
∫(5t1/2 − t3/2)dt
= −1
2
(10
3t3/2 − 2
5t5/2
)+ C
= −1
2
(10
3
√(5− x2)3 − 2
5
√(5− x2)5
)+ C.
11.∫
dxx2√
9x2−4. Let
x =2
3sec t ⇒ dx =
2
3sec t tan tdt.
44
Hence ∫dx
x2√
9x2 − 4=
∫2/3 sec t tan tdt
4/9 sec2 t√
4 sec2 t− 4
=3
4
∫1
sec tdt =
3
4
∫sec tdt
=3
4ln | sec t + tan t|+ C
=3
4ln
∣∣∣∣∣3x2 +
√9x2
4− 1
∣∣∣∣∣+ C.
12.∫ √
1+t2
tdt. Let
t = tan θ ⇒ dt = sec2 θdθ.
Thus∫ √1 + t2
tdt =
∫sec θ
tan θsec2 θdθ
=
∫1/ cos θ
sin θ/ cos θ
1
cos2 θdθ =
∫1
sin θ cos2 θdθ
=
∫sin2 θ + cos2 θ
sin θ cos2 θdθ =
∫ (sin θ
cos2 θ+
1
sin θ
)+ C
=1
cos θ+ ln | csc θ − cot θ|+ C
=√
1 + t2 + ln∣∣∣√1 + 1/t2 − 1/t
∣∣∣+ C.
13.∫
1(1−x2)3/2 dx. Let
x = sin t ⇒ dx = cos tdt.
Whence ∫1
(1− x2)3/2dx =
∫1
cos3 tcos tdt =
∫sec2 tdt
= tan t + C
=x√
1− x2+ C.
14.∫
dxx2√
x2+25. Let
x = 5 tan t ⇒ dx = 5 sec2 tdt.
45
Hence ∫dx
x2√
x2 + 25=
∫5 sec2 tdt
25 tan2 t√
25 + 25 tan2 t
=1
25
∫sec t
tan2 tdt =
1
25
∫cos t
sin2 tdt
=1
25
∫csc t cot tdt = − 1
25csc t + C
= − 1
25
√1 +
x2
25+ C.
15.∫
dx√x2−9
. Let
x = 3 sec t ⇒ dx = 3 sec t tan tdt.
Hence ∫dx√
x2 − 9=
∫3 sec t tan t
3 tan tdt =
∫sec tdt
= ln | sec t + tan t|+ C
= ln
∣∣∣∣∣x3 +
√x2
9− 1
∣∣∣∣∣+ C.
16.∫
dx1+2x2+x4 . By letting x = tan t we see that∫
dx
1 + 2x2 + x4=
∫dx
(x2 + 1)2=
∫sec2 t
sec4 tdt
=
∫cos2 tdt =
1
2
∫(1 + cos 2t)dt
=1
2
(t +
sin 2t
2
)+ C
=1
2
(tan−1 x +
x
x2 + 1
)+ C.
17.∫
dx(4x2−9)3/2 . Let
x =3 sec t
2⇒ dx =
3
2sec t tan tdt.
∫dx
(4x2 − 9)3/2=
∫3/2 sec t tan tdt
27 tan3 t
=1
18
∫sec t
tan2 tdt =
1
18
∫cot t csc tdt
= − 1
18csc t + C = − 1
18
2x√4x2 − 9
+ C.
46
18.∫
3x3√
25−x2 dx. Do the substitution
25− x2 = t ⇒ dx = − dt
2xand x2 = 25− t.
We leave the details to the student.
19.∫
ex√
1− e2xdx. Use the substitution
ex = sin t ⇒ dx =cos tdt
ex
to get ∫ex√
1− e2xdx =
∫ex√
1− sin2 tcos tdt
ex
=
∫cos2 tdt =
1
2
∫(1 + cos 2t)dt
=1
2
(t +
sin 2t
2
)+ C
=1
2
(sin−1 ex + ex
√1− e2x
)+ C.
20.∫
cos θ√2−sin2 θ
dθ. Do the substitution
sin θ =√
2 sin t ⇒ dθ =
√2 cos tdt
cos θ
to get ∫cos θ√
2− sin2 θdθ =
∫cos θ√
2− 2 sin2 t
√2 cos t
cos θdt
=
∫dt = t + C
= sin−1 sin θ√2
+ C.
21.∫ 1
05x3√
1− x2dx. Do the substitution
1− x2 = t ⇒ dx = − dt
2x
47
to get∫5x3√
1− x2dx = −∫
5x3√
tdt
2x
= −5
2
∫(1− t)
√tdt
= −5
2
(2
3t3/2 − 2
5t5/2
)+ C
= −5
2
(2
3
√(1− x2)3 − 2
5
√(1− x2)5
)+ C.
Hence∫ 1
0
5x3√
1− x2dx = −5
2
(2
3
√(1− x2)3 − 2
5
√(1− x2)5
)]1
0
= −2
3.
22.∫ 1/2
0dx
(1−x2)2. Let
x = sin t ⇒ dx = cos tdt,
hence ∫dx
(1− x2)2=
∫cos tdt
cos4 t=
∫sec3 tdt
=1
2[sec t tan t− ln | sec t + tan t|] + C.
Therefore ∫ 1/2
0
dx
(1− x2)2=
1
2
[x
1− x2− ln
1 + x√1− x2
]1/2
0
=1
3+
1
4ln 3.
23.∫ 2√
2dx
x2√
x2−1. Let
x = sec t ⇒ dx = sec t tan tdt.
Thus ∫dx
x2√
x2 − 1=
∫sec t tan tdt
sec2 t tan t
=
∫cos tdt = sin t + C
=√
1− 1/x2 + C.
48
By plugging the limits√
2 and 2 we get∫ 2
√2
dx
x2√
x2 − 1=
1
2(√
3−√
2).
24.∫ 2√
2
√2x2−4
xdx. Let x =
√2 sec t ⇒ dx =
√2 sec t tan tdt. Therefore∫ √
2x2 − 4
xdx =
∫2 tan t√2 sec t
√2 sec t tan tdt
= 2
∫tan2 tdt = 2
∫(sec2 t− 1)dt
= 2(tan t− t) + C
= 2(√
x2/2− 1− sec−1(x/√
2))
+ C.
Thus ∫ 2
√2
√2x2 − 4
xdx = 2
[√x2/2− 1− sec−1(x/
√2)]2√
2
= 2(1− π/4).
33.∫
dxx2−4x+5
dx. We complete the square for x2 − 4x + 5 in order to use
trigonometric substitution.
x2 − 4x + 5 = x2 − 4x + 4− 4 + 5 = (x− 2)2 + 1.
Therefore∫dx
x2 − 4x + 5=
∫dx
(x− 2)2 + 1(let x− 2 = tan t)
=
∫sec2 tdt
sec2 t= t + C
= arctan(x− 2) + C.
34.∫
dx√2x−x2 . We complete the square for 2x− x2 to get∫
dx√2x− x2
=
∫dx√
1− (x− 1)2(let x− 1 = sin t)
=
∫cos tdt
cos tdt = t + C
= arcsin(x− 1) + C.
49
35.∫
dx√3+2x−x2 .∫
dx√3 + 2x− x2
=
∫dx√
4− (x− 1)2(let x− 1 = 2 sin t)
=
∫2 cos tdt√4 cos2 t
= t + C
= arcsin
(x− 1
2
)+ C.
36.∫
dx16x2+16x+5
.∫dx
16x2 + 16x + 5=
∫dx
16(x + 1/2)2 + 1(let x + 1/2 = tan t/4)
=1
4
∫sec2 tdt
sec2 t=
1
4t + C
=1
4arctan [4(x + 1/2)] + C.
1.5 Partial fractions
For questions 1-8, write out the form of the partial fraction decomposition.
1.
3x− 1
(x− 3)(x + 4)=
A
x− 3+
B
x + 4.
2.
5
x(x2 − 4)=
5
x(x− 2)(x + 2)
=A
x+
B
x− 2+
C
x + 2.
3.
2x− 3
x3 − x2=
2x− 3
x2(x− 1)
=A
x+
B
x2+
C
x− 1.
50
4.
x2
(x + 2)3=
A
x + 2+
B
(x + 2)2+
C
(x + 2)2.
5.
2x− 3
x3(x2 + 2)=
A
x+
B
x2+
C
x3+
Dx + E
x2 + 2.
6.
3x
(x− 1)(x2 + 6)=
A
x− 1+
Bx + C
x2 + 6.
7.
4x3 − x
(x2 + 5)2=
Ax + B
(x2 + 5)+
Cx + D
(x2 + 5)2.
8.
1− 3x4
(x− 2)(x2 + 1)2=
A
x− 2+
Bx + C
x2 + 1+
Dx + E
(x2 + 1)2.
51
Evaluate the following integrals:
9.∫
dxx2−3x−4
. We begin by writing the partial fraction decomposition for
1/(x2 − 3x− 4).
1
x2 − 3x− 4=
1
(x− 4)(x + 1)=
A
x− 4+
B
x + 1
⇒ 1 = A(x + 1) + B(x− 4).
In order to find A and B we plug two values for x:
x = 4 ⇒ 1 = 5A ⇒ A =1
5, x = −1 ⇒ 1 = −5B ⇒ B = −1
5.
Therefore ∫dx
x2 − 3x− 4=
∫ (1/5
x− 4− 1/5
x + 1
)dx
=1
5ln |x− 4| − 1
5ln |x + 1|+ C
=1
5ln
∣∣∣∣x− 4
x + 1
∣∣∣∣+ C.
10.∫
dxx2−6x−7
. We do as above:
1
x2 − 6x− 7=
1
(x− 7)(x + 1)=
A
x− 7+
B
x + 1
⇒ 1 = A(x + 1) + B(x− 7).
x = −1 ⇒ 1 = −8B ⇒ B = −1
8; x = 7 ⇒ 1 = 8A ⇒ A =
1
8.
Hence ∫dx
x2 − 6x− 7=
∫ (1/8
x− 7− 1/8
x + 1
)dx
=1
8ln |x− 7| − 1
8ln |x + 1|+ C
=1
8ln
∣∣∣∣x− 7
x + 1
∣∣∣∣+ C.
11.∫
11x+172x2+7x−4
dx. Applying same ideas as above, we get∫11x + 17
2x2 + 7x− 4dx =
∫ (3
x + 4+
5
2x− 1
)dx
= 3 ln |x + 4|+ 5
2ln |2x− 1|+ C.
52
12.∫
5x−53x2−8x−3
dx.∫5x− 5
3x2 − 8x− 3dx =
∫ (1
x− 3+
2
3x + 1
)dx
= ln |x− 3|+ 2
3ln |3x + 1|+ C.
13.∫
2x2−9x−9x3−9x
dx. We observe that
2x2 − 9x− 9
x3 − 9x=
2x2 − 9x− 9
x(x− 3)(x + 3)
=A
x+
B
x− 3+
C
x + 3
which means
2x2 − 9x− 9 = A(x− 3)(x + 3) + Bx(x + 3) + Cx(x− 3).
To find the values of A, B and C we plug
x = 0 ⇒ −9 = −9A ⇒ A = 1.
x = 3 ⇒ −18 = 18B ⇒ B = −1.
x = −3 ⇒ 36 = 18C ⇒ C = 2.
Therefore∫2x2 − 9x− 9
x3 − 9xdx =
∫ (1
x− 1
x− 3+
2
x + 3
)dx
= ln |x| − ln |x− 3|+ 2 ln |x + 3|+ C.
14.∫
dxx(x2−1)
.
1
x(x2 − 1)=
A
x+
B
x− 1+
C
x + 1
⇒ 1 = A(x2 − 1) + Bx(x + 1) + Cx(x− 1)
= A = −1(x = 0); B =1
2(x = 1); C =
1
2(x = −1).
∫dx
x(x2 − 1)=
∫ (−1
x+
1
2(x− 1)+
1
2(x + 1)
)dx
= − ln |x|+ 1
2ln |x− 1|+ 1
2ln |x + 1|+ C.
53
15.∫
x2−8x+3
dx. Since the degree of the numerator is larger than the degree of
the denominator, we use long division to get∫x2 − 8
x + 3=
∫ (x− 3 +
1
x + 3
)dx
=x2
2− 3x + ln |x + 3|+ C.
16.∫
x2+1x−1
dx. Use long division as above to get∫x2 + 1
x− 1dx =
∫ (x + 1 +
2
x− 1
)dx
=x2
2+ x + 2 ln |x− 1|+ C.
17.∫
3x2−10x2−4x+4
dx. Use long division:∫3x2 − 10
x2 − 4x + 4dx =
∫ (3 +
12x− 22
x2 − 4x + 4
)dx.
Now use partial fractions to simplify:
12x− 22
x2 − 4x + 4=
12x− 22
(x− 2)2=
A
x− 2+
B
(x− 2)2
⇒ 12x− 22 = A(x− 2) + B
⇒ B = 2(x = 2) ⇒ −10 = −A + B(x = 1) ⇒ A = 12.
Hence ∫3x2 − 10
x2 − 4x + 4dx =
∫ (3 +
12
x− 2+
2
(x− 2)2
)dx
= 3x + 12 ln |x− 2| − 2
x− 2+ C.
18.∫
x2
x2−3x+2dx. Since the degree of the top is the same as the degree of the
bottom we use long division:
x2
x2 − 3x + 2= 1 +
3x− 2
x2 − 3x + 2
= 1 +A
x− 2+
B
x− 1.
54
We find A and B:
3x− 2
x2 − 3x + 2=
A
x− 2+
B
x− 1
⇒ 3x− 2 = A(x− 1) + B(x− 2)
⇒ B = −1(x = 1), A = 4(x = 2).
Therefore∫x2
x2 − 3x + 2dx =
∫ (1 +
4
x− 2− 1
x− 1
)dx
= x + 4 ln |x− 2| − ln |x− 1|+ C.
19.∫
x5+x2+2x3−x
dx. Long division gives:
x5 + x2 + 2
x3 − x= x2 + 1 +
x2 + x + 2
x3 − x
and
x2 + x + 2
x3 − x=
A
x+
B
x− 1+
C
x + 1
⇒ x2 + x + 2 = A(x2 − 1) + Bx(x + 1) + Cx(x− 1)
⇒ (x = 0 ⇒ A = −2), (x = 1 ⇒ B = 2), (x = −1 ⇒ C = 1).
Hence∫x5 + x2 + 2
x3 − xdx =
∫ (x2 + 1 +
−2
x+
2
x− 1+
1
x + 1
)dx
=x3
3+ x− 2 ln |x|+ 2 ln |x− 1|+ ln |x + 1|+ C.
20.∫
x5−4x3+1x3−4x
dx. Imitate the above solution to get∫x5 − 4x3 + 1
x3 − 4xdx =
∫ (x2 +
1
x3 − 4x
)dx
=
∫ (x2 +
A
x+
B
x− 2+
C
x + 2
)dx
=x3
3+ A ln |x|+ B ln |x− 2|+ C ln |x + 2|+ D,
55
where A, B and C may be found by:
1
x3 − 4x=
A
x+
B
x− 2+
C
x + 2
⇒ 1 = A(x2 − 4) + Bx(x + 2) + Cx(x− 2)
⇒ (x = 0 ⇒ A = −1
4), (x = 2 ⇒ B =
1
8), (x = −2 ⇒ C = −1
8).
21.∫
2x2+3x(x−1)2
dx.∫2x2 + 3
x(x− 1)2dx =
∫ (A
x+
B
x− 1+
C
(x− 1)2
)dx
= A ln |x|+ B ln |x− 1| − C
x− 1+ D,
where A, B and C may be found by
2x2 + 3 = A(x− 1)2 + Bx(x− 1) + Cx
⇒ (x = 0 ⇒ A = −3), (x = 1 ⇒ C = 5)
(x = −1 ⇒ 5 = −2A + 2B − C ⇒ B = 2).
22.∫
3x2−x+1x3−x2 dx.∫
3x2 − x + 1
x3 − x2dx =
∫ (A
x+
B
x2+
C
x− 1
)dx
= A ln |x| − B
x+ C ln |x− 1|+ D,
and we may find the constants A, B and C as following:
3x2 − x + 1 = Ax(x− 1) + B(x− 1) + Cx2
⇒ (x = 0 ⇒ B = −1), (x = 1 ⇒ C = 3),
(x = −1 ⇒ 5 = 2A− 2B + C ⇒ A = 0).
23.∫
2x2−10x+4(x+1)(x−3)2
dx.∫2x2 − 10x + 4
(x + 1)(x− 3)2dx =
∫ (A
x + 1+
B
x− 3+
C
(x− 3)2
)dx
= A ln |x + 1|+ B ln |x− 3| − C
x− 3+ D,
56
where A, B and C may be found by:
2x2 − 10x + 4 = A(x− 3)2 + B(x + 1)(x− 3) + C(x + 1)
⇒ (x = 3 ⇒ C = −2); (x = −1 ⇒ A = 1); (x = 0 ⇒ B = 1).
24.∫
2x2−2x−1x3−x2 dx.∫
2x2 − 2x− 1
x3 − x2dx =
∫ (A
x+
B
x2+
C
x− 1
)dx
= A ln |x| − B
x+ C ln |x− 1|+ D,
where A, B and C may be found by:
2x2 − 2x− 1 = Ax(x− 1) + B(x− 1) + Cx2
⇒ (x = 0 ⇒ B = 1); (x = 1 ⇒ C = −1); (x = 2 ⇒ A = 3).
25.∫
x2
(x+1)3dx.∫
x2
(x + 1)3dx =
∫ (A
x + 1+
B
(x + 1)2+
C
(x + 1)3
)dx
= A ln |x + 1| − B
x + 1− C
2(x + 1)2+ D.
A, B and C may be found as usual:
x2 = A(x + 1)2 + B(x + 1) + C
⇒ (x = −1 ⇒ C = 1); (x = 0 ⇒ A + B + C = 0); (x = 1 ⇒ 1 = 4A + 2B + C).
Solving the above equations gives A = 1, B = −2, C = 1.
57
1.6 Improper integrals
1. In each part, determine whether the integral is improper, and if so, ex-
plain why.
(a)∫ 5
1dx
x−3. The integral is improper since 1/(x− 3), the integrand, has
an infinite discontinuity at x = 3 which is a point of [1, 5].
(b)∫ 5
1dx
x+3is not an improper integral. The integrand is continuous on
the interval of integration.
(c)∫ 1
0ln xdx is an improper integral because ln x has an infinite dis-
continuity at 0, namely, limx→0+ ln x = −∞.
(d)∫∞
1e−xdx is an improper integral because ∞ is a limit of the inte-
gral.
(e)∫∞−∞
dx3√x−1
is an improper integral because ∞ and −∞ are limits of
the integral.
(f)∫ π/4
0tan xdx is not an improper integral because tan x is continuous
on [0, π/4].
2. In each part, determine the value of p for which the integral is improper:
(a)∫ 1
01xp dx. We do the integral: For p < 1,∫ 1
0
1
xpdx = lim
b→0+
∫ 1
b
x−pdx
= limb→0+
x1−p
1− p
]1
b
= limb→0+
(1
1− p− b1−p
1− p
)=
1
1− p.
Thus, if p < 1 the integral is not improper.
58
For p > 1 the above limit is infinite, thus the integral is improper.
Also, if p = 1 the integral will be improper because limb→0+ ln b =
−∞.
Consequently, the values of p which make the above integral im-
proper are {p ≥ 1}.
(b)∫ 2
1dx
x−p. This integral will be improper if x−p = 0 for some p ∈ [1, 2].
Hence, p ∈ [1, 2].
(c)∫ 1
0e−pxdx. Since e−px is continuous on [0, 1] for all p, this integral
is always proper (not improper.)
Evaluate the integrals that converge:
3.∫∞
0e−2xdx, we evaluate:∫
e−2xdx = −1
2e−2x + C,
hence
limb→∞
∫ b
0
e−2xdx = limb→∞
−1
2e−2x
]b
0
= limb→∞
(1
2− 1
2e−2b
)=
1
2.
Therefore, the integral converges and is equal to 1/2.
4.∫∞−1
dx1+x2 . We have
limb→∞
∫ b
−1
dx
x2 + 1= lim
b→∞tan−1 x
]b−1
= limb→∞
(tan−1 b− tan−1(−1)
)=
π
2− −π
4=
3π
4.
5.∫∞
32
x2−1dx. Do the integral by partial fractions to get∫
2
x2 − 1dx = ln
∣∣∣∣x− 1
x + 1
∣∣∣∣+ C.
59
Hence,
limb→∞
∫ b
3
2
x2 − 1dx = lim
b→∞ln
∣∣∣∣x− 1
x + 1
∣∣∣∣]b
3
= limb→∞
ln
∣∣∣∣b− 1
b + 1
∣∣∣∣− ln1
2
= ln 1 + ln 2 = ln 2.
Observe that L’Hopital’s rule has been used to evaluate the above limit.
6.∫∞
0xe−x2
dx. Do the indefinite integral by substitution to get:∫xe−x2
dx = −1
2e−x2
+ C.
Hence
limb→∞
∫ b
0
xe−x2
dx = limb→∞
−1
2e−x2
]b
0
= limb→∞
(1
2− 1
2e−b2
)=
1
2.
7.∫∞
e1
x ln3 xdx. Do the indefinite integral by substitution: ln x = t to get∫
1
x ln3 xdx =
−1
2 ln2 x+ C.
Hence,
limb→∞
∫ b
e
1
x ln3 xdx = lim
b→∞
−1
2 ln2 x
]b
e
= limb→∞
(1
2− 1
2 ln2 b
)=
1
2.
8.∫∞
2dx
x√
ln x. Let ln x = t to get∫
dx
x√
ln x= 2
√ln x + C.
Hence
limb→∞
∫ b
2
1
x√
ln xdx = lim
b→∞2√
ln x]b
2(1.6.1)
= limb→∞
(2√
ln b− 2√
ln 2)
= ∞. (1.6.2)
Thus the integral diverges to ∞.
60
9.∫ 0
−∞dx
(2x−1)3.
limb→−∞
∫ 0
b
dx
(2x− 1)3= lim
b→−∞− 1
4(2x− 1)2
]0
b
= limb→−∞
(1
4(2b− 1)2− 1
4
)= −1
4.
10.∫ 3
−∞dx
x2+9.
limb→−∞
∫ 3
b
dx
x2 + 9= lim
b→−∞
1
3tan−1
(1
3x
)]3
b
=1
3lim
b→−∞
(tan−1 1− tan−1 b
3
)=
1
3
(π
4− −π
2
)=
π
4.
11.∫ 0
−∞ e3xdx.
limb→−∞
∫ 0
b
e3xdx = limb→−∞
e3x
3
]0
b
= limb→−∞
(1
3− e3b
3
)=
1
3.
12.∫ 0
−∞ex
3−2ex dx. Observe first that the only problem is −∞. That is, 3 −
2ex = 0 gives a solution that does not lie in (−∞, 0].
limb→−∞
∫ 0
b
ex
3− 2exdx
61
