This document contains 7 exercises involving calculating integrals using techniques like integration by parts, trigonometric substitutions, and partial fraction decomposition. It also contains solutions to each exercise that demonstrate the step-by-step working to evaluate the integrals. The document is from a university course on simple integration and contains examples commonly used to teach integration techniques.

1. Lebanese University - Faculty of Sciences
Section ¶
Chapter 1: Simple Integration
Solved Problems
Dr. Kamel ATTAR
attar.kamel@gmail.com
F Wednesday 24/Mars/2021 F

2. 2Ú20
Exercises
Solutions
1 Exercises
2 Solutions
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

3. 3Ú20
Exercises
Solutions
. Exercise 1. Calculate the following integrals:
a)
Z
sin2
(x) dx b)
Z
cosh2
(x) dx c)
Z
tan2
(x) dx .
Go to Solution
. Exercise 2. Evaluate by using change of variable the following integrals :
a)
Z
(x3
+ x)5
(3x2
+ 1) dx b)
Z p
2x + 1 dx c)
Z
2x dx
3
p
x2 + 1
d)
Z
arccos(x) − x
p
1 − x2
dx e)
Z
sin(2x)esin2
x
dx .
Go to Solution
. Exercise 3. Evaluate I(x) =
Z
x
2 − x2 + 2x
dx then deduce J(x) =
Z
dx
1 + x + 2
√
1 − x
Go to Solution
. Exercise 4. Calculate using the integration by parts the following integrals:
a)
Z
x sin 2xdx b)
Z
x ln x dx c)
Z
x2
ln(x)dx d)
Z
(x2
+ 7x − 5) cos(2x)dx
e)
Z
x2
e3x
dx f)
Z
x arcsin x
p
1 − x2
dx g)
Z
arcsin2 x
2
p
4 − x2
dx h)
Z
x ln
1 + x
1 − x
dx .
Go to Solution
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

4. 4Ú20
Exercises
Solutions
. Exercise 5. Using twice integration by parts, evaluate:
Z
e−x
cos(x)dx .
Go to Solution
. Exercise 6. Calculate: a)
Z
2x2
x4 − 1
dx b)
Z
dx
x2(x − 1)3
c)
Z
1
(x2 + 1)2
dx
Go to Solution
. Exercise 7. Calculate: a)
Z
dx
x(x2 + 2x + 5)
b)
Z
dx
x(x2 + 1)2
c)
Z
dx
x(x5 + 1)2
.
Go to Solution
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

5. 5Ú20
Exercises
Solutions
Solution 1.
a) We have that the power of sin is even, then we can write the integral as follows:
Z
sin2
(x) dx =
Z
1 − cos(2x)
2
dx =
x
2
−
sin(2x)
4
+ C.
b) Same for the hyperbolic function
Z
cosh2
(x) dx =
Z
1 + cosh(2x)
2
dx =
x
2
+
sinh(2x)
4
+ C.
c) We have
tan2
(x) =
sin x
cos x
2
=
1 − cos2 x
cos2 x
=
1
cos2 x
− 1 .
Then Z
tan2
(x) dx =
Z
1
cos2 x
dx −
Z
dx = tan x − x + C.
Go Back
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

6. 6Ú20
Exercises
Solutions
Solution 2.
a) We set u = x3
+ x. Then du = (3x2
+ 1)dx , so that by substitution we have
Z
(x3
+ x)5
(3x2
+ 1) dx =
Z
u5
du =
u6
6
+ C =
(x3 + x)6
6
+ C.
b) We set u = 2x + 1. Then du = 2dx , so that by substitution we have
Z p
2x + 1 dx =
Z p
2x + 1 dx =
1
2
Z
u
1
2 u0
=
1
3
u3/2
+ C =
1
3
(2x + 1)3/2
+ C .
c) Substitute u = x2
+ 1 then u0
= 2x and so
Z
2x dx
3
p
x2 + 1
=
Z
u0
u1/3
=
Z
u−1/3
u0
=
u2/3
2/3
+ C =
3
2
u2/3
+ C =
3
2
(x2
+ 1)2/3
+ C.
d) Z
arccos(x) − x
p
1 − x2
dx =
Z
arccos(x)
p
1 − x2
dx +
Z
−x
p
1 − x2
dx
We set u = arccos x and v = 1 − x2
then du = −
1
p
1 − x2
dx and dv = −2xdx. Thus
Z
arccos(x) − x
p
1 − x2
dx = −
Z
u du +
Z
dv
2
√
v
= −
arccos2 x
2
+
p
1 − x2 + C .
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

7. 7Ú20
Exercises
Solutions
e) We havesin(2x) = 2 cos(x) sin(x), we obtain
Z
sin(2x)esin2
x
dx =
Z
2 cos x sin x esin2
x
dx
Using change of variable t = sin2
x, we get dt = 2 cos x sin x dx. Thus
Z
sin(2x)esin2
x
dx =
Z
et
dt = et
+ C = esin2
x
+ C
Go Back
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

8. 8Ú20
Exercises
Solutions
Solution 3.
a)
I(x) =
Z
xdx
−x2 + 2x + 2
= −
1
2
Z
−2xdx
−x2 + 2x + 2
u = −x2
+ 2x + 2
= −
1
2
Z
−2x + 2
−x2 + 2x + 2
dx +
Z
1
−x2 + 2x + 2
dx
= −
1
2
Z
u0
u
+
Z
1
√
3
2
− (x − 1)2
dx u = x − 1 u0
= 1 , a =
√
3
= −
1
2
ln |u| +
Z
u0
√
3
2
− u2
= −
1
2
ln | − x2
+ 2x + 2| +
1
√
3
arg sh
x − 1
√
3
+ C .
b) We set t =
p
1 − x then x + 1 = 2 − t2
and dx = −2tdt. Hence
J(x) =
Z
−2t dt
2 − t2 + 2t
= −2I(t) = ln | − t2
+ 2t + 2| −
2
√
3
arg sh
t − 1
√
3
+ C
Go Back
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

9. 9Ú20
Exercises
Solutions
Solution 4.
a) We need to choose u. In this question we don’t have any of the functions suggested in the priorities
list above. We could let u = x or u = sin 2x, but usually only one of them will work. In general, we
choose the one that allows to be of a simpler form than u. So, we choose u = x and so v0
will be the
rest of the integral, v0
= sin 2x. We have
u = x
×
v0
= sin 2x
u0
= 1 ←
−
−
R v = −
1
2
cos 2x
Substituting these 4 expressions into the integration by parts formula, we get
Z
x sin 2xdx = x ×
−
1
2
cos 2x
−
Z
−
1
2
cos 2x
× 1dx
= −
1
2
x cos 2x +
1
2
Z
cos 2x dx = −
1
2
x cos 2x +
1
4
sin 2x .
b)
u = ln x
×
v0
= x
u0
=
1
x
←
−
−
R v =
x2
2
Then, applying the formula
Z
x ln x dx =
x2
2
ln x −
Z
x2
2
·
1
x
dx =
x2
2
ln x −
x2
4
+ C
Go Back
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

10. 10Ú20
Exercises
Solutions
c)
u = ln x
×
v0
= x2
u0
=
1
x
←
−
−
R v =
x3
3
Then
Z
x2
ln x dx =
x3
3
ln x −
1
3
Z
x2
dx =
x3
3
ln x −
x3
9
+ C.
d)
u x2
+ 7x − 5 cos(2x) v0
+
u0
2x + 7
1
2
sin(2x) v
−
u00
2 −
1
4
cos(2x)
Z
v
+
u000
0 −
1
8
sin(2x)
Z Z
v
Z
(x2
+ 7x − 5) cos(2x) = (x2
+ 7x − 5)
1
2
sin(2x)
− (2x + 7)
−
1
4
cos(2x)
+ (2)
=
1
2
x2
+ 7x −
11
2
sin(2x) +
1
2
x +
7
2
cos(2x) + C
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

11. 11Ú20
Exercises
Solutions
e) We have to make a choice and let one of the functions in the product equal u and one equal
to v0
. As a general rule we let u be the function which will become simpler when we
differentiate it. In this case it makes sense to let
u = x2
×
v0
= e3x
u0
= 2x ←
−
−
R v =
1
3
e3x
Then, using the formula for integration by parts,
Z
x2
e3x
dx =
1
3
e3x
· x2
−
Z
1
3
e3x
· 2x dx =
1
3
x2
e3x
−
2
3
Z
xe3x
dx .
The resulting integral is still a product. It is a product of the functions
2
3
x and e3x
. We can
use the formula again. This time we choose
u =
2
3
x
×
v0
= e3x
u0
=
2
3
←
−
−
R v =
1
3
e3x
So
Z
x2
e3x
dx =
1
3
x2
e3x
−
2
3
Z
xe3x
dx =
1
3
x2
e3x
−
2
3
x ·
1
3
e3x
−
Z
1
3
e3x
·
2
3
dx
=
1
3
x2
e3x
−
2
9
xe3x
+
2
27
e3x
+ C.
Go Back
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

12. 12Ú20
Exercises
Solutions
f)
u = arcsin(x)
×
v0
=
x
p
1 − x2
u0
=
1
p
1 − x2
←
−
−
R v = −
p
1 − x2
Then Z
x arcsin x
p
1 − x2
dx = −
p
1 − x2 arcsin x + x + C
g) Let I(x) =
Z
arcsin2 x
2
p
4 − x2
dx. Using integration by parts
u = arcsin
x
2
×
v0
= arcsin
x
2
1
2
q
1 − x
2
2
u0
=
1
2
q
1 − x
2
2
←
−
−
R v =
1
2
arcsin2 x
2
Thus
I(x) =
1
2
arcsin3 x
2
−
1
4
I(x) =
⇒ I(x) =
2
5
arcsin3 x
2
+ C .
Go Back
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

13. 13Ú20
Exercises
Solutions
h) We set
u = ln
1 + x
1 − x
×
v0
= x
u0
=
2
1 − x2
←
−
−
R v =
x2
2
We obtain
Z
x ln
1 + x
1 − x
dx =
x2
2
ln
1 + x
1 − x
+
Z
x2
1 − x2
dx
=
x2
2
ln
1 + x
1 − x
−
Z
1 − 1 − x2
1 − x2
dx
=
x2
2
ln
1 + x
1 − x
− x +
Z
1
1 − x2
dx
=
x2
2
ln
1 + x
1 − x
− x +
1
2
ln

17. 1 + x
1 − x