Integration
SUBSTITUTION II ..
1
f (x)
· f 0 (x)
Graham S McDonald
and Silvia C Dalla
A Tutorial Module for practising the integra1
· f 0 (x)
tion of expressions of the form f (x)
● Table of contents
● Begin Tutorial
c 2004 g.s.mcdonald@salford.ac.uk
Table of contents
1.
2.
3.
4.
5.
Theory
Exercises
Answers
Standard integrals
Tips
Full worked solutions
Section 1: Theory
3
1. Theory
Consider an integral of the form
Z 0
f (x)
dx
f (x)
Letting u = f (x) then
du
= f 0 (x) , and this gives du = f 0 (x)dx
dx
So
Z
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f 0 (x)
dx =
f (x)
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Z
1
· f 0 (x) dx
f (x)
Z
=
du
u
=
ln |u| + C
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Section 1: Theory
4
The general result is
R
f 0 (x)
f (x)
dx = ln |f (x)| + C
This is a useful generalisation of the standard integral
Z
1
dx = ln |x| + C ,
x
that applies straightforwardly when the top line of what we are integrating is a multiple of the derivative of the bottom line
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Section 2: Exercises
5
2. Exercises
Click on EXERCISE links for full worked solutions (there are 11 exercises in total).
Perform the following integrations:
Exercise 1.
Z
10x + 3
dx
2
5x + 3x − 1
Exercise 2.
Z
3x2
dx
x3 − 6
Exercise 3.
Z
4x2
dx
x3 − 1
● Theory ● Standard integrals ● Answers ● Tips
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Section 2: Exercises
Exercise 4.
Z
2x − 1
dx
2
x −x−7
Exercise 5.
Z
x−1
dx
x2 − 2x + 7
Exercise 6.
Z
8x
dx
x2 − 4
Exercise 7.
Z
x2
dx
3
x +2
● Theory ● Standard integrals ● Answers ● Tips
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Section 2: Exercises
7
Exercise 8.
Z
sin x
dx
cos x
Exercise 9.
Z
cos x
dx
sin x
Exercise 10.
Z
cos x
dx
5 + sin x
Exercise 11.
Z
sinh x
dx ,
1 − cosh x
● Theory ● Standard integrals ● Answers ● Tips
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Section 3: Answers
8
3. Answers
1. ln |5x2 + 3x − 1| + C ,
2. ln |x3 − 6| + C ,
3.
4
3
ln |x3 − 1| + C ,
4. ln |x2 − x − 7| + C ,
5.
1
2
ln |x2 − 2x + 7| + C ,
6. 4 ln |x2 − 4| + C ,
7.
1
3
ln |x3 + 2| + C
8. − ln | cos x| + C ,
9. ln | sin x| + C,
10. ln |5 + sin x| + C ,
11. − ln |1 − cosh x| + C .
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Section 4: Standard integrals
9
4. Standard integrals
f (x)
n
x
1
x
x
e
sin x
cos x
tan x
cosec x
sec x
sec2 x
cot x
sin2 x
cos2 x
R
f (x)dx
xn+1
n+1
(n 6= −1)
ln |x|
ex
− cos x
sin x
− ln
|cos x|
ln tan x2 ln |sec x + tan x|
tan x
ln |sin x|
x
sin 2x
2 −
4
x
sin 2x
2 +
4
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R
f (x)
n
0
[g (x)] g (x)
g 0 (x)
g(x)
x
a
sinh x
cosh x
tanh x
cosech x
sech x
sech2 x
coth x
sinh2 x
cosh2 x
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f (x)dx
[g(x)]n+1
n+1
(n 6= −1)
ln |g (x)|
ax
(a > 0)
ln a
cosh x
sinh x
ln cosh x ln tanh x2 2 tan−1 ex
tanh x
ln |sinh x|
sinh 2x
− x2
4
sinh 2x
+ x2
4
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Section 4: Standard integrals
f (x)
1
a2 +x2
√ 1
a2 −x2
√
a2 − x2
10
R
f (x) dx
f (x)
1
a
tan−1
1
a2 −x2
R
(a > 0)
1
x2 −a2
f (x) dx
a+x 1
2a ln a−x (0 < |x| < a)
x−a 1
ln
2a
x+a (|x| > a > 0)
sin−1
x
a
√ 1
a2 +x2
√
2
2
ln x+ aa +x (a > 0)
(−a < x < a)
√ 1
x2 −a2
√
2
2
ln x+ xa −a (x > a > 0)
a2
2
−1
sin
√
+x
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x
a
x
a
a2 −x2
a2
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√
i √
a2 +x2
a2
2
x2 −a2
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a2
2
J
h
h
sinh−1
− cosh−1
I
x
a
i
√
x a2 +x2
2
a
i
√
2
2
+ x xa2−a
+
x
a
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Section 5: Tips
11
5. Tips
● STANDARD INTEGRALS are provided. Do not forget to use these
tables when you need to
● When looking at the THEORY, STANDARD INTEGRALS, ANSWERS or TIPS pages, use the Back button (at the bottom of the
page) to return to the exercises
● Use the solutions intelligently. For example, they can help you get
started on an exercise, or they can allow you to check whether your
intermediate results are correct
● Try to make less use of the full solutions as you work your way
through the Tutorial
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Solutions to exercises
12
Full worked solutions
Exercise 1.
Z
Z 0
10x + 3
f (x)
dx
is
of
the
form
dx = ln |f (x)| + C.
5x2 + 3x − 1
f (x)
Let u = 5x2 + 3x − 1 then
Z
∴
du
= 10x + 3, and du = (10x + 3)dx
dx
10x + 3
dx =
2
5x + 3x − 1
Z
du
u
=
ln |u| + C
=
ln |5x2 + 3x − 1| + C .
Return to Exercise 1
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Solutions to exercises
13
Exercise 2.
Z
Z 0
3x2
f (x)
dx
is
of
the
form
dx = ln |f (x)| + C.
x3 − 6
f (x)
Let u = x3 − 6 then
Z
∴
du
= 3x2 and du = 3x2 dx
dx
3x2
dx =
x3 − 6
Z
1
du
u
=
ln |u| + C
=
ln |x3 − 6| + C .
Return to Exercise 2
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Solutions to exercises
14
Exercise 3.
Z
4x2
dx is of the slightly more general form
3
x −1
Z
Z 0
f 0 (x)
f (x)
k
dx = k
dx = k ln |f (x)| + C,
f (x)
f (x)
where k is a constant
i.e. the top line is a multiple of the derivative of the bottom line.
du
du
= 3x2 and
= x2 dx
dx
3
Z
Z
x2
1 du
∴ 4
dx = 4
x3 − 1
u 3
4
ln |u| + C
=
3
4
=
ln |x3 − 1| + C .
3
Return to Exercise 3
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Let u = x3 − 1 then
Solutions to exercises
15
Exercise 4.
Z
Z 0
2x − 1
f (x)
dx
is
of
the
form
dx = ln |f (x)| + C.
x2 − x − 7
f (x)
du
= 2x − 1 and du = (2x − 1)dx
dx
Z
Z
2x − 1
du
∴
dx
=
x2 − x − 7
u
Let u = x2 − x − 7 then
=
ln |u| + C
=
ln |x2 − x − 7| + C .
Return to Exercise 4
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Solutions to exercises
16
Exercise 5.
Z
x−1
dx is of the slightly more general form
2
x − 2x + 7
f 0 (x)
dx = k
k
f (x)
where k is a constant
Z
Z
f 0 (x)
dx = k ln |f (x)| + C,
f (x)
i.e. the top line is a multiple of the derivative of the bottom line.
Let u = x2 − 2x + 7 then
so that
du
= 2x − 2 and du = (2x − 2)dx
dx
du
= (x − 1)dx
2
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Solutions to exercises
Z
∴
17
x−1
dx =
2
x − 2x + 7
Z
1 du
u 2
Z
=
1
2
du
u
=
1
ln |u| + C
2
=
1
ln |x2 − 2x + 7| + C .
2
Return to Exercise 5
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Solutions to exercises
18
Exercise 6.
Z
8x
dx
2
x −4
du
= xdx
2
Z
1 du
8
u 2
Let u = x2 − 4 then du = 2x dx and
Z
∴
8x
dx =
−4
x2
=
4 ln |u| + C
=
4 ln |x2 − 4| + C .
Return to Exercise 6
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Solutions to exercises
19
Exercise 7.
Z
x2
dx
3
x +2
du
= x2 dx
3
Let u = x3 + 2 then du = 3x2 dx and
Z
∴
x2
dx =
x3 + 2
Z
1 du
u 3
=
1
ln |u| + C
3
=
1
ln |x3 + 2| + C .
3
Return to Exercise 7
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Solutions to exercises
20
Exercise 8.
R sin x
cos x dx
Let u = cos x then
du
dx
Z
∴
du
(−1)
= − sin x and
= sin x dx
Z
1 du
u (−1)
Z
du
= −
u
sin x
dx =
cos x
= − ln |u| + C
= − ln | cos x| + C.
Return to Exercise 8
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Solutions to exercises
21
Exercise 9.
R cos x
sin x dx
Let u = sin x then
du
dx
= cos x and du = cos x dx
Z
cos x
dx =
sin x
∴
Z
1
du
u
=
ln |u| + C
=
ln | sin x| + C.
Return to Exercise 9
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Solutions to exercises
22
Exercise 10.
Z
cos x
dx
5 + sin x
Let u = 5 + sin x then
Z
∴
Toc
du
= cos x and du = cos x dx
dx
cos x
dx =
5 + sin x
JJ
II
Z
du
u
=
ln |u| + C
=
ln |5 + sin x| + C .
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Solutions to exercises
23
Exercise 11.
Z
sinh x
dx
1 − cosh x
du
= − sinh x and −du = sinh x dx
dx
Z
Z
1
sinh x
dx =
· (−du)
∴
1 − cosh x
u
Let u = 1 − cosh x then
= − ln |u| + C
= − ln |1 − cosh x| + C .
Return to Exercise 11
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