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8/8/2019 Integration by Substitution 1
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Integration
SUBSTITUTION I .. f(ax + b)
Graham S McDonaldand Silvia C Dalla
A Tutorial Module for practising the integra-tion of expressions of the form f(ax + b)
q Table of contents
q Begin Tutorial
c 2004 [email protected]
http://www.cse.salford.ac.uk/http://www.cse.salford.ac.uk/mailto:[email protected]:[email protected]://www.cse.salford.ac.uk/http://www.cse.salford.ac.uk/8/8/2019 Integration by Substitution 1
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Table of contents
1. Theory
2. Exercises
3. Answers
4. Standard integrals
5. Tips
Full worked solutions
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Section 1: Theory 3
1. Theory
Consider an integral of the formf(ax + b)dx
where a and b are constants. We have here an unspecified function f
of a linear function of x
Letting u = ax + b thendu
dx= a , and this gives dx =
du
a
This allows us to change the integration variable from x to u
f(ax + b)dx =
f(u)
du
a
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Section 1: Theory 4
The final result is
f(ax + b)dx = 1
a
f(u) du
where u = ax + b
This is a general result for integrating functions of a linear functionof x
Each application of this result involves dividing by the coefficient
of x and then integrating
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Section 2: Exercises 5
2. Exercises
Click on EXERCISE links for full worked solutions (10 exercises in
total).
Perform the following integrations:
Exercise 1.
(2x 1)3dx
Exercise 2.cos(3x + 5)dx
Exercise 3.e5x+2dx
q Theory q Standard integrals q Answers q Tips
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Section 2: Exercises 6
Exercise 4.
sinh3x dxExercise 5.
dx
2x
1
Exercise 6.dx
1 + (5x)2
Exercise 7.sec2(7x + 1)dx
q Theory q Standard integrals q Answers q Tips
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Section 2: Exercises 7
Exercise 8.
sin(3x 1)dxExercise 9.
cosh(1 + 2x)dx
Exercise 10.tan(9x 1)dx
q Theory q Standard integrals q Answers q Tips
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Section 3: Answers 8
3. Answers
1. 1
8
(2x
1)4 + C ,
2. 13 sin(3x + 5) + C ,
3. 15e5x+2 + C ,
4. 13 cosh3x + C ,
5.12 ln |2x 1| + C ,
6. 15 tan1 5x + C ,
7. 17 tan(7x + 1) + C ,
8. 13 cos(3x 1) + C ,9. 12 sinh(1 + 2x) + C ,
10. 19 ln | cos(9x 1)| + C .
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Section 4: Standard integrals 9
4. Standard integrals
f(x)
f(x)dx f(x)
f(x)dxxn x
n+1
n+1 (n = 1) [g (x)]n g (x) [g(x)]n+1
n+1 (n = 1)1x
ln |x| g(x)g(x) ln |g (x)|
ex ex ax ax
ln a (a > 0)
sin x cos x sinh x cosh xcos x sin x cosh x sinh xtan x ln |cos x| tanh x ln cosh xcosec x ln
tan x2 cosech x ln tanh x2
sec x ln
|sec x + tan x
|sech x 2tan1 ex
sec2 x tan x sech2 x tanh xcot x ln |sin x| coth x ln |sinh x|sin2 x x2 sin 2x4 sinh2 x sinh2x4 x2cos2 x x2 +
sin 2x4 cosh
2 x sinh2x4 +x2
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Section 4: Standard integrals 10
f(x)
f(x) dx f(x)
f(x) dx
1
a2+x21
a tan1 x
a
1
a2x21
2a lna+xax (0 < |x|< a)
(a > 0) 1x2a2
12a ln
xax+a (|x| > a > 0)
1a2x2 sin
1 xa
1a2+x2 ln
x+a2+x2a
(a > 0)
(a < x < a) 1x2a2 ln
x+x2a2a (x > a > 0)
a2 x2 a22
sin1xa
a2+x2 a22 sinh1 xa + xa2+x2a2
+xa2x2a2
x2a2 a22
cosh1 x
a
+ x
x2a2a2
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Section 5: Tips 11
5. Tips
q
STANDARD INTEGRALS are provided. Do not forget to use thesetables when you need to
q When looking at the THEORY, STANDARD INTEGRALS, AN-SWERS or TIPS pages, use the Back button (at the bottom of thepage) to return to the exercises
q Use the solutions intelligently. For example, they can help you getstarted on an exercise, or they can allow you to check whether your
intermediate results are correct
q Try to make less use of the full solutions as you work your waythrough the Tutorial
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Solutions to exercises 12
Full worked solutions
Exercise 1.(2x 1)3dx
Let u = 2x 1 then dudx
= 2 and dx =du
2
(2x 1)3dx = u3 du2 = 12 u3du=
1
2
1
4u4 + C
=1
8(2x
1)4 + C .
Note. The final result can also be obtained using the general pattern:f(ax + b) dx =
1
a
f(u) du
where a = 2 and
f(u)du is
u3
du. Return to Exercise 1
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Solutions to exercises 13
Exercise 2.
cos(3
x+ 5)
dx
Let u = 3x + 5 thendu
dx= 3 and dx =
du
3
cos(3x + 5)dx =
cos u
du
3 =
1
3
cos u du
=1
3sin u + C
=1
3sin(3x + 5) + C .
Note. The final result can also be obtained using the general pattern:f(ax + b) dx =
1
a
f(u) du
where a = 3 and
f(u)du is
cos udu. Return to Exercise 2
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Solutions to exercises 14
Exercise 3.
e5x+2dx
Let u = 5x + 2 thendu
dx= 5 and dx =
du
5
e5x+2dx = eudu
5
=1
5 eudu
=1
5eu + C
=1
5e5x+2 + C .
Note. The final result can also be obtained using the general pattern:f(ax + b) dx =
1
a
f(u) du
where a = 5 and
f(u)du is
eu
du. Return to Exercise 3
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Solutions to exercises 15
Exercise 4.
sinh 3x dx
Let u = 3x thendu
dx= 3 and dx =
du
3
sinh 3x dx =
sinh u
du
3 =
1
3
sinh u du
=1
3cosh u + C
=1
3cosh 3x + C .
Note. The final result can also be obtained using the general pattern:f(ax + b) dx =
1
a
f(u) du
where a = 3 and
f(u)du is
sinh u du. Return to Exercise 4
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Solutions to exercises 16
Exercise 5.
dx2x 1
Let u = 2x 1 then dudx
= 2 and dx =du
2
dx
2x 1=
du
u 1
2
=1
2
du
u
=1
2ln |u| + C
=1
2ln |2x 1| + C .
Note. The final result can also be obtained using the general pattern:f(ax + b) dx =
1
a
f(u) du
where a = 2 and f(u)du is duu
.
Return to Exercise 5
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Solutions to exercises 17
Exercise 6.
dx1 + (5x)2
Let u = 5x thendu
dx= 5 and dx =
du
5
dx
1 + (5x)2
= du
1 + u2
1
5
=1
5
du
1 + u2
=1
5tan1 u + C
= =1
5tan1 5x + C .
Note. The final result can also be obtained using the general pattern:f(ax + b) dx =
1
a
f(u) du
where a = 5 and f(u)du is du
1+u2 du. Return to Exercise 6
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Solutions to exercises 18
Exercise 7.
sec2(7x + 1)dx
Let u = 7x + 1 thendu
dx= 7 and dx =
du
7
sec2(7x + 1)dx =
1
7 sec
2 u du
=1
7tan u + C
=1
7tan(7x + 1) + C .
Note. The final result can also be obtained using the general pattern:f(ax + b) dx =
1
a
f(u) du
where a = 7 and
f(u)du is
sec2 udu.
Return to Exercise 7
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S l ti t i 19
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Solutions to exercises 19
Exercise 8.
sin(3x 1)dx
Let u = 3x 1 then dudx
= 3 and dx =du
3
sin(3x 1)dx =1
3 sin u du
= 13
cos u + C
= 13
cos(3x 1) + C .Note. The final result can also be obtained using the general pattern:
f(ax + b) dx =1
a
f(u) du
where a = 3 and f(u)du is sin udu.Return to Exercise 8
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Solutions to exercises 20
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Solutions to exercises 20
Exercise 9.
cosh(1 + 2x)dx
Let u = 1 + 2x thendu
dx= 2 and dx =
du
2
cosh(1 + 2x)dx = cosh u du2
=1
2sinh u + C
=1
2sinh(1 + 2x) + C .
Note. The final result can also be obtained using the general pattern:f(ax + b) dx =
1
a
f(u) du
where a = 2 and f(u)du is cosh u du.Return to Exercise 9
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Solutions to exercises 21
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Solutions to exercises 21
Exercise 10.
tan(9x 1)dx
Let u = 9x 1 then dudx
= 9 and dx =du
9
tan(9x 1) dx = tan u du
9=
1
9 tan u du
= 19
ln | cos u| + C
= 19
ln | cos(9x 1)| + C .Note. The final result can also be obtained using the general pattern:
f(ax + b) dx =1
a
f(u) du
where a = 9 and f(u)du is tan u du.Return to Exercise 10
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