Integration
SUBSTITUTION III .. [f (x)]n · f 0 (x)
Graham S McDonald
and Silvia C Dalla
A self-contained Tutorial Module for practising the integration of expressions of the form
[f (x)]n · f 0 (x), where n 6= −1
● 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
[f (x)]n f 0 (x)dx
Letting u = f (x) gives
Z
∴
du
= f 0 (x) and du = f 0 (x)dx
dx
[f (x)]n f 0 (x)dx =
Z
un du =
un+1
[f (x)]n+1
+C =
+C
n+1
n+1
For example, when n = 1,
Z
f (x)f 0 (x)dx =
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Z
u du =
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u2
[f (x)]2
+C =
+C
2
2
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Section 2: Exercises
4
2. Exercises
Click on EXERCISE links for full worked solutions (10 exercises in
total).
Perform the following integrations:
Exercise 1.
Z
sin x cos x dx
Exercise 2.
Z
sinh x cosh x dx
Exercise 3.
Z
tan x sec2 x dx
● Theory ● Standard integrals ● Answers ● Tips
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Section 2: Exercises
5
Exercise 4.
Z
sin 2x cos 2x dx
Exercise 5.
Z
sinh 3x cosh 3x dx
Exercise 6.
Z
1
ln x dx , x > 0
x
Exercise 7.
Z
sin4 x cos x dx
Exercise 8.
Z
sinh3 x cosh x dx
● Theory ● Standard integrals ● Answers ● Tips
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Section 2: Exercises
6
Exercise 9.
Z
cos3 x sin x dx
Exercise 10.
Z
2x
dx
(x2 − 4)2
● Theory ● Standard integrals ● Answers ● Tips
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Section 3: Answers
7
3. Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
1
2
1
2
1
2
1
4
1
6
1
2
sin2 x + C ,
sinh2 x + C ,
tan2 x + C ,
sin2 2x + C ,
sinh2 3x + C ,
ln2 x + C
sin5 x
+C,
5
4
1
4 sinh x +
C,
4
− cos4 x + C,
− x21−4 + C .
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Section 4: Standard integrals
8
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
II
J
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
9
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
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h
h
sinh−1
− cosh−1
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x
a
i
√
x a2 +x2
2
a
i
√
2
2
+ x xa2−a
+
x
a
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Section 5: Tips
10
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
11
Full worked solutions
Exercise 1.
Z
Z
1
2
sin x cos x dx is of the form
f (x)f 0 (x)dx = [f (x)] + C
2
To see this, set u = sin x, to find
du
= cos x and du = cos x dx
dx
Z
∴
Z
sin x cos x dx =
=
=
u du
1 2
u +C
2
1
sin2 x + C.
2
Return to Exercise 1
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Solutions to exercises
12
Exercise 2.
Z
Z
2
sinh x cosh x dx is of the form
f (x)f 0 (x)dx = [f (x)] + C
To see this, set u = sinh x then
du
= cosh x and du = cosh x dx
dx
Z
∴
Z
sinh x cosh x dx =
=
=
u du
1 2
u +C
2
1
sinh2 x + C.
2
Return to Exercise 2
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Solutions to exercises
13
Exercise 3.
Z
Z
2
tan x sec2 x dx is of the form
f (x)f 0 (x)dx = [f (x)] + C
Let u = tan x then
Z
∴
du
= sec2 x and du = sec2 x dx
dx
tan x sec2 x dx =
=
=
Z
u du
1 2
u +C
2
1
tan2 x + C.
2
Return to Exercise 3
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Solutions to exercises
14
Exercise 4.
Z
Z
2
sin 2x cos 2x dx is close to the form f (x)f 0 (x)dx = [f (x)] +C
Let u = sin 2x then
du
du
= 2 cos 2x and
= cos 2x dx
dx
2
Z
∴
Z
sin 2x cos 2x dx =
=
=
=
du
u
2
Z
1
u du
2
11 2
u +C
22
1
sin2 2x + C.
4
Return to Exercise 4
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Solutions to exercises
15
Exercise 5.
Z
Z
2
sinh 3x cosh 3x dx is close to the form f (x)f 0 (x)dx = [f (x)] + C
Let u = sinh 3x then
du
du
= 3 cosh 3x and
= cosh 3x dx
dx
3
Z
∴
Z
sinh 3x cosh 3x dx =
=
=
=
du
u
3
Z
1
u du
3
11 2
u +C
32
1
sinh2 3x + C.
6
Return to Exercise 5
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Solutions to exercises
16
Exercise 6.
Z
Z
1
2
ln x dx is of the form
f (x)f 0 (x)dx = [f (x)] + C
x
Let u = ln x then
1
1
du
=
and du = dx
dx
x
x
Z
Z
1
∴
ln x dx =
u du
x
1 2
=
u +C
2
1 2
=
ln x + C .
2
Return to Exercise 6
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Solutions to exercises
17
Exercise 7.
Z
Z
n+1
[f (x)]
n
4
sin x cos x dx is of the form [f (x)] f 0 (x)dx =
+C
n+1
To see this, set u = sin x to find du = cos x dx
Z
∴
sin4 x cos x dx =
=
=
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Z
u4 du
u5
+C
5
sin5 x
+ C.
5
Return to Exercise 7
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Solutions to exercises
18
Exercise 8.
Z
Z
n+1
[f (x)]
n
3
sinh x cosh x dx is of the form [f (x)] f 0 (x)dx =
+C
n+1
To see this, set u = sinh x then du = cosh x dx
Z
∴
sinh3 x cosh x dx =
=
=
Z
u3 du
1 4
u +C
4
1
sinh4 x + C.
4
Return to Exercise 8
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Solutions to exercises
19
Exercise 9.
Z
Z
n+1
[f (x)]
n
3
cos x sin x dx is close to the form [f (x)] f 0 (x)dx =
+C
n+1
Let u = cos x then du = − sin x dx
Z
∴
Z
3
cos x sin x dx =
Z
3
u · (−du) = −
u3 du
u4
+C
4
cos4 x
+ C.
= −
4
= −
Return to Exercise 9
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Solutions to exercises
20
Exercise 10.
Z
Z
n+1
2x
[f (x)]
n 0
+C
dx
is
of
the
form
[f
(x)]
f
(x)dx
=
(x2 − 4)2
n+1
Let u = x2 − 4 then du = 2x dx
Z
∴
2x
dx =
(x2 − 4)2
Z
1
du
u2
Z
=
u−2 du
= −u−1 + C = −
= −
x2
1
+C
u
1
+C.
−4
Return to Exercise 10
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