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)
● Table of contents
● Begin Tutorial
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Table of contents
1. Theory 2. Exercises 3. Answers 4. Standard integrals 5. TipsSection 1: Theory 3
1. Theory
Consider an integral of the form R f (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 then du
dx = a , and this gives dx = du
a This allows us to change the integration variable from x to u
Z
f (ax + b)dx = Z
f (u)du a
Section 1: Theory 4
The final result is
R f (ax + b)dx = 1
aR f (u) du
where u = ax + b
This is a general result for integrating functions of a linear function of x
Each application of this result involves dividing by the coefficient of x and then integrating
Section 2: Exercises 5
2. Exercises
Click on EXERCISE links for full worked solutions (10 exercises in total).
Perform the following integrations:
Exercise 1. Z (2x − 1)3dx Exercise 2. Z cos(3x + 5)dx Exercise 3. Z e5x+2dx
Section 2: Exercises 6 Exercise 4. Z sinh 3x dx Exercise 5. Z dx 2x − 1 Exercise 6. Z dx 1 + (5x)2 Exercise 7. Z sec2(7x + 1)dx
Section 2: Exercises 7 Exercise 8. Z sin(3x − 1)dx Exercise 9. Z cosh(1 + 2x)dx Exercise 10. Z tan(9x − 1)dx
Section 3: Answers 8
3. Answers
1. 1 8(2x − 1) 4+ C , 2. 13sin(3x + 5) + C , 3. 15e5x+2+ C , 4. 13cosh 3x + C , 5. 12ln |2x − 1| + C , 6. 15tan−15x + C , 7. 17tan(7x + 1) + C , 8. −13cos(3x − 1) + C , 9. 12sinh(1 + 2x) + C , 10. −19ln | cos(9x − 1)| + C .Section 4: Standard integrals 9
4. Standard integrals
f (x) R f (x)dx f (x) R f (x)dx xn xn+1n+1 (n 6= −1) [g (x)]ng0(x) [g(x)]n+1n+1 (n 6= −1) 1 x ln |x| g0(x) g(x) ln |g (x)| ex ex ax ln aax (a > 0)sin x − cos x sinh x cosh x
cos x sin x cosh x sinh x
tan x − ln |cos x| tanh x ln cosh x cosec x ln tanx2 cosech x ln
tanhx2 sec x ln |sec x + tan x| sech x 2 tan−1ex
sec2x tan x sech2
x tanh x
cot x ln |sin x| coth x ln |sinh x| sin2x x2 −sin 2x 4 sinh 2 x sinh 2x4 −x 2 cos2x x 2 + sin 2x 4 cosh 2x sinh 2x 4 + x 2
Section 4: Standard integrals 10 f (x) R f (x) dx f (x) R f (x) dx 1 a2+x2 1 atan −1 x a 1 a2−x2 1 2aln a+x a−x (0 < |x| < a) (a > 0) x2−a1 2 1 2aln x−a x+a (|x| > a > 0) 1 √ a2−x2 sin −1 x a 1 √ a2+x2 ln x+√a2+x2 a (a > 0) (−a < x < a) √ 1 x2−a2 ln x+√x2−a2 a (x > a > 0) √ a2− x2 a2 2sin −1 x a √ a2+x2 a2 2 h sinh−1 xa +x√a2+x2 a2 i +x √ a2−x2 a2 i √ x2−a2 a2 2 h − cosh−1 x a + x√x2−a2 a2 i
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, AN-SWERS 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
Solutions to exercises 12
Full worked solutions
Exercise 1. Z (2x − 1)3dx Let u = 2x − 1 then du dx = 2 and dx = du 2 ∴ Z (2x − 1)3dx = Z u3du 2 = 1 2 Z u3du = 1 2 1 4u 4+ C = 1 8(2x − 1) 4+ C .
Note. The final result can also be obtained using the general pattern: Z
f (ax + b) dx = 1 a
Z
f (u) du where a = 2 andR f (u)du is R u3du.
Solutions to exercises 13 Exercise 2. Z cos(3x + 5)dx Let u = 3x + 5 then du dx = 3 and dx = du 3 ∴ Z cos(3x + 5)dx = Z cos u ·du 3 = 1 3 Z cos u du = 1 3sin u + C = 1 3sin(3x + 5) + C . Note. The final result can also be obtained using the general pattern:
Z
f (ax + b) dx = 1 a
Z
f (u) du
Solutions to exercises 14 Exercise 3. Z e5x+2dx Let u = 5x + 2 then du dx = 5 and dx = du 5 ∴ Z e5x+2dx = Z eudu 5 = 1 5 Z eudu = 1 5e u+ C = 1 5e 5x+2+ C .
Note. The final result can also be obtained using the general pattern: Z
f (ax + b) dx = 1 a
Z
f (u) du
Solutions to exercises 15 Exercise 4. Z sinh 3x dx Let u = 3x then du dx = 3 and dx = du 3 ∴ Z sinh 3x dx = Z sinh u ·du 3 = 1 3 Z sinh u du = 1 3cosh u + C = 1 3cosh 3x + C .
Note. The final result can also be obtained using the general pattern: Z
f (ax + b) dx = 1 a
Z
f (u) du
Solutions to exercises 16 Exercise 5. Z dx 2x − 1 Let u = 2x − 1 then du dx = 2 and dx = du 2 ∴ Z dx 2x − 1 = Z du u · 1 2 = 1 2 Z du u = 1 2ln |u| + C = 1 2ln |2x − 1| + C .
Note. The final result can also be obtained using the general pattern: Z
f (ax + b) dx = 1 a
Z
f (u) du where a = 2 andR f (u)du is R du
u.
Solutions to exercises 17 Exercise 6. Z dx 1 + (5x)2 Let u = 5x then du dx = 5 and dx = du 5 ∴ Z dx 1 + (5x)2 = Z du 1 + u2 · 1 5 = 1 5 Z du 1 + u2 = 1 5tan −1u + C = = 1 5tan −15x + C .
Note. The final result can also be obtained using the general pattern: Z
f (ax + b) dx = 1 a
Z
f (u) du where a = 5 andR f (u)du is R du
Solutions to exercises 18 Exercise 7. Z sec2(7x + 1)dx Let u = 7x + 1 then du dx = 7 and dx = du 7 ∴ Z sec2(7x + 1)dx = 1 7 Z sec2u du = 1 7tan u + C = 1 7tan(7x + 1) + C .
Note. The final result can also be obtained using the general pattern: Z
f (ax + b) dx = 1 a
Z
f (u) du where a = 7 andR f (u)du is R sec2udu.
Solutions to exercises 19 Exercise 8. Z sin(3x − 1)dx Let u = 3x − 1 then du dx = 3 and dx = du 3 ∴ Z sin(3x − 1)dx = 1 3 Z sin u du = −1 3cos u + C = −1 3cos(3x − 1) + C .
Note. The final result can also be obtained using the general pattern: Z
f (ax + b) dx = 1 a
Z
f (u) du where a = 3 andR f (u)du is R sin udu.
Solutions to exercises 20 Exercise 9. Z cosh(1 + 2x)dx Let u = 1 + 2x then du dx = 2 and dx = du 2 ∴ Z cosh(1 + 2x)dx = Z cosh udu 2 = 1 2sinh u + C = 1 2sinh(1 + 2x) + C .
Note. The final result can also be obtained using the general pattern: Z
f (ax + b) dx = 1 a
Z
f (u) du where a = 2 andR f (u)du is R cosh u du.
Solutions to exercises 21 Exercise 10. Z tan(9x − 1)dx Let u = 9x − 1 then du dx = 9 and dx = du 9 ∴ Z tan(9x − 1) dx = Z tan u ·du 9 = 1 9 Z tan u du = −1 9ln | cos u| + C = −1 9ln | cos(9x − 1)| + C . Note. The final result can also be obtained using the general pattern:
Z
f (ax + b) dx = 1 a
Z
f (u) du where a = 9 andR f (u)du is R tan u du.
