Solved exercises

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

21. + C
Go Back
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

22. 14Ú20
Exercises
Solutions
Solution 5. I(x) =
Z
e−x
cos(x) dx. We set
u = e−x
×
v0
= cos(x)
u0
= −e−x
←
−
−
R v = sin(x)
Then
I(x) = e−x
sin(x) +
Z
e−x
sin(x) dx .
Same for
Z
e−x
sin(x) dx we set
u = e−x
×
v0
= sin(x)
u0
= −e−x
←
−
−
R v = − cos(x)
we obtain
Z
e−x
sin(x) dx = −e−x
cos(x) −
Z
e−x
cos(x) = −e−x
cos(x) − I(x)
Thus
I(x) = e−x
sin(x) − e−x
cos(x) − I(x) + C
Finally
I(x) =
1
2
sin(x) − cos(x)
e−x
+ C
Go Back
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

23. 15Ú20
Exercises
Solutions
Solution 6.
a) The partial fraction decomposition has the form
2x2
x4 − 1
=
2x2
(x − 1)(x + 1)(x2 + 1)
=
A
x − 1
+
D
x + 1
+
Bx + C
x2 + 1
.
We clear fractions and get
2x2
= A(x + 1)(x2
+ 1) + D(x − 1)(x2
+ 1) + (Bx + C)(x − 1)(x + 1) .
For x = 1 we obtain, A =
1
2
and for x = −1 we obtain, D = −
1
2
. Then,
2x2
(x − 1)(x + 1)(x2 + 1)
=
1/2
x − 1
+
−1/2
x + 1
+
Bx + C
x2 + 1
.
For x = 0 and x = 2, we have C = 1 and B = 0. Thus
2x2
(x − 1)(x + 1)(x2 + 1)
=
1/2
x − 1
+
−1/2
x + 1
+
1
x2 + 1
.
Finally,
Z
2x2
x4 − 1
dx =
Z
1/2
x − 1
dx +
Z
−1/2
x + 1
dx +
Z
1
x2 + 1
dx
=
1
2
ln |x − 1| −
1
2
ln |x + 1| + arctan(x) + C .
Go Back
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

24. 16Ú20
Exercises
Solutions
b) Let g(x) =
1
x2(x − 1)3
. The partial fraction decomposition has the form
g(x) =
A
x2
+
B
x
+
C
(x − 1)3
+
D
(x − 1)2
+
E
x − 1
A = lim
x→0
x2
g(x) = lim
x→0
1
(x − 1)3
= −1
B = lim
x→0
h
x2
g(x)
i0
= lim
x→0
−
3
(x − 1)4
= −3
C = lim
x→1
(x − 1)3
g(x) = lim
x→1
1
x2
= 1
D = lim
x→1
h
(x − 1)3
g(x)
i0
= lim
x→1
−
2
x3
= −2
E = lim
x→1
1
2
h
(x − 1)3
g(x)
i00
=
1
2
lim
x→1
6
2x4
= 3
Thus
1
x2(x − 1)3
= −
1
x2
−
3
x
+
1
(x − 1)3
−
2
(x − 1)2
+
3
x − 1
.
Finally
Z
1
x2(x − 1)3
dx =
1
x
− 3 ln |x| −
1
2
1
(x − 1)2
+
2
x − 1
− 3 ln |x − 1| + C .
Go Back
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

25. 17Ú20
Exercises
Solutions
c) We have Z
1
(1 + x2)2
dx =
Z
1
1 + x2
dx
1 + x2
=
Z
1
1 + x2
(arctan x)0
Set t = arctan x then dt =
dx
1 + x2
and so
Z
1
(1 + x2)2
dx =
Z
dt
1 + tan2(t)
=
Z
cos2
(t) dt =
t
2
+
sin(2t)
4
+ C
=
t
2
+
tan t
2
1 + tan2(t)
+ C =
arctan x
2
+
x
2(1 + x2)
+ C
Go Back
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

26. 18Ú20
Exercises
Solutions
Solution 7.
a) The partial fraction decomposition has the form
1
x(x2 + 2x + 5)
=
A
x
+
Bx + C
x2 + 2x + 5
. We clear
fractions and get
1 = A(x2
+ 2x + 5) + (Bx + C)x .
For x = 0, we obtain A =
1
5
and for x = 1 and x = −1 we get B = −
1
5
and C = −
2
5
. Then
1
x(x2 + 2x + 5)
=
1
5x
−
x + 1
5(x2 + 2x + 5)
=
1
5x
−
2x + 2
10(x2 + 2x + 5)
−
1
5
(x + 1)2 + 4
.
Thus
Z
dx
x(x2 + 2x + 5)
=
1
5
ln |x| −
1
10
ln(x2
+ 2x + 5) −
1
10
arctan
x + 1
2
+ C .
b) Set t = 1 + x2
, then dt = 2xdx and x2
= t − 1 we obtain
Z
dx
x(x2 + 1)2
=
Z
xdx
x2(x2 + 1)2
=
Z
dt
(t − 1)t2
= −
1
2
Z
1
t2
+
1
t
+
1
1 − t
dt =
1
2t
−
l
Go Back
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

27. 19Ú20
Exercises
Solutions
c) Let t = x5
+ 1, we have
Z
1
x(x5 + 1)2
dx =
1
5
Z
dt
(t − 1)t2
=
A
t − 1
+
B
t
+
C
t2
By comparison, we find A =
1
5
, C = −
1
5
and B =
1
5
Hence
1
5
Z
dt
(t − 1)t2
=
Z
1/5
t − 1
dt +
Z
1/5
t
dt −
Z
1/5
t2
dt
and so
1
5
Z
dt
(t − 1)t2
=
1
5
ln |t − 1| +
1
5
ln |t| +
1
5t
+ C .
Go Back
Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems

28. Thank you! Questions?