Chapter 7

Chapter 7

Exercise 7A

1. I will use the "intelligent guess" method for this

question, but my preference is for the "rearrang-

ing" method, so I will use that for most of the

questions where one of these approaches is suit-

able.

Guess

y

=

(2x+5)5 5

d (2x+5)5 dx 5

=

(2x + 5)4(2)

Adjust

by

a

factor

of

1 2

:

(2x + 5)5 +c

10

2. dy = (3x + 1)3 dx

=

1 (3(3x +

1)3)

3

1 (3x + 1)4

y=

+c

34

(3x + 1)4

=

+c

12

3. This can not be rearranged to the form f (x)(f(x))n so we expand:

dy = x(3x + 4)

dx = 3x2 + 4x

3x3 4x2 y= + +c

32 = x3 + 2x2 + c

4. dy = 50(1 + 5x)4 dx = 10(5(1 + 5x)4)

(1 + 5x)5

y = 10

+c

5

= 2(1 + 5x)5 + c

5. dy = 24x(2 - x2)3 dx = -12(-2x(2 - x2)3)

(2 - x2)4

y = -12

+c

4

= -3(2 - x2)4 + c

6. This can not be rearranged to the form f (x)(f(x))n so we expand:

dy = x(1 + 4x)2 dx

= x(1 + 8x + 16x2)

= x + 8x2 + 16x3

x2 8x3 16x4

y= + +

+c

23

4

= x2 + 8x3 + 4x4 + c 23

7. dy = 30x(x2 - 3)2 dx = 15(2x(x2 - 3)2)

(x2 - 3)3

y = 15

+c

3

= 5(x2 - 3)3 + c

8. This can not be rearranged to the form f (x)(f(x))n so we expand:

dy = x(2x - 3)2 dx

= x(4x2 - 12x + 9)

= 4x3 - 12x2 + 9x

4x4 12x3 9x2

y= -

+ +c

4

3

2

= x4 - 4x3 + 9x2 + c 2

9. dy = 12(2x + 1)3 dx = 6(2(2x + 1)3)

(2x + 1)4

y=6

+c

4

3(2x + 1)4

=

+c

2

10. dy = 2(3x + 1)4 dx

=

2 (3(3x

+

1)4)

3

2 (3x + 1)5

y=

+c

3

5

2(3x + 1)5

=

+c

15

11. dy = (2x - 3)4 dx

=

1 (2(2x

-

3)4)

2

1 (2x - 3)5

y=

+c

2

5

(2x - 3)5

=

+c

10

12. dy = 5x2(3 - x3)4 dx

= 5 (-3x2(3 - x3)4) -3

5 (3 - x3)5

y=-

+c

3

5

(3 - x3)5

=-

+c

3

(x3 - 3)5

=

+c

3

(Note that the last step is valid because (-1)5 =

1

Exercise 7A

Solutions to A.J. Sadler's

-1. Such a simplification would not be possible if we were raising to an even power.)

13. dy = (1 + x)4 dx

(1 + x)5

y=

+c

5

14. dy = 4x(2 + x2)4 dx = 2(2x(2 + x2)4)

(2 + x2)5

y=2

+c

5

2(2 + x2)5

=

+c

5

15. dy = (1 + 2x)4 dx

=

1 (2(1

+ 2x)4)

2

1 (1 + 2x)5

y=

+c

2

5

(1 + 2x)5

=

+c

10

16. dy = 4x(1 + x2) dx = 2(2x(1 + x2))

(1 + x2)2

y=2

+c

2

= (1 + x2)2 + c This question is probably just as easy to do by first expanding:

dy = 4x(1 + x2) dx

= 4x + 4x3

x2 x4 y=4 +4 +c

23 = 2x2 + x4 + c

Although these two solutions may not look quite the same, you should satisfy yourself that they are, in fact, both correct. (Remember, c is an arbitrary constant.)

17. dy = 60(2x - 3)5 dx = 30(2(2x - 3)5

(2x - 3)6

y = 30

+c

6

= 5(2x - 3)6 + c

18. dy = 60(3 - 2x)5 dx = -30(-2(3 - 2x)5

(3 - 2x)6

y = -30

+c

6

= -5(3 - 2x)6 + c

dy

1

19. dx = (x + 2)4

= (x + 2)-4

(x + 2)-3

y=

+c

-3

1

=

- 3(x

+

2)3

+

c

dy

1

20. dx = (2x + 1)4

= (2x + 1)-4

=

1 (2(2x

+

1)-4

2

1 (2x + 1)-3

y=

+c

2 -3

1

=

- 6(2x

+

1)3

+

c

dy

25x

21. dx = - (x2 + 1)5

= -25x(x2 + 1)-5

= -25 (2x(x2 + 1)-5) 2

-25 (x2 + 1)-4

y=

+c

2 -4

25 = 8(x2 + 1)4 + c

dy

6

22. dx = (3x + 5)3

= 6(3x + 5)-3

= 2(3(3x + 5)-3)

(3x + 5)-2

y=2

+c

-2

1

=

- (3x

+

5)2

+

c

dy

18x

23. dx = (3x2 + 5)3

= 3(6x)(3x2 + 5)-3

(3x2 + 5)-2

y=3

+c

-2

3 = - 2(3x2 + 5)2 + c

24.

dy

= 12 3 3x - 2

dx

1

= 4 3(3x - 2) 3

(3x

-

2)

4 3

y=4

4

+c

3

43

= 4(3x - 2) 3

+c

4

4

= 3(3x - 2) 3 + c

2

Unit 3C Specialist Mathematics

dy

25. = 12 3x + 5

dx

1

= 4 3(3x + 5) 2

(3x

+

5)

3 2

y=4

3

+c

2

32

= 4(3x + 5) 2

+c

3

8(3x

+

5)

3 2

=

+c

3

dy

12

26. =

dx 3x + 5

=4

3(3x

+

5)-

1 2

(3x

+

5)

1 2

y=4

1

+c

2

1

= 8(3x + 5) 2 + c

= 8 3x + 5 + c

27. dy = 3 - 12(-3x2(1 - x3)2) dx

(1 - x3)3

y = 3x - 12

+c

3

= 3x - 4(1 - x3)3 + c

28. dP = 12(3(2 + 3t)3) dt

(2 + 3t)4

P = 12

+c

4

= 3(2 + 3t)4 + c

50 = 3(2 + 3(0))4 + c

50 = 3(24) + c

c = 50 - 3 ? 16

=2

P = 3(2 + 3t)4 + 2

29. dP = 12(2t(t2 - 5)3 dt

(t2 - 5)4

P = 12

+c

4

= 3(t2 - 5)4 + c

10 = 3(22 - 5)4 + c

10 = 3(-1)4 + c

10 = 3 + c

c=7

P = 3(t2 - 5)4 + 7

Exercise 7B

Exercise 7B

1. 5 cos x dx = 5 sin x + c

2. 2 sin x dx = -2 cos x + c

3. -10 sin x dx = 10 cos x + c

4. -2 cos x dx = -2 sin x + c

5. 6 cos 2x dx = 3 2 cos 2x dx

= 3 sin 2x + c

1 6. 2 cos 6x dx = 6 cos 6x dx

3

sin 6x

=

+c

3

7. 12 sin 4x dx = -3 -4 sin 4x dx = -3 cos 4x + c

8. 9. 10. 11. 12. 3

1 - sin 3x dx = -3 sin 3x dx

3

cos 3x

=

+c

3

4 -8 cos 10x dx = - 10 cos 10x dx

5

4 sin 10x

=-

+c

5

x

1x

sin dx = -2 - sin dx

2

22

x = -2 cos + c

2

3x

2 3 3x

cos dx =

cos dx

2

32 2

2 3x = sin + c

32

2x

3

-6 sin dx = 6 ?

3

2

2 2x - sin dx

33

2x = 9 cos + c

3

Exercise 7B

Solutions to A.J. Sadler's

13. For this question and the next, you should note

that sin

x

+

2

= cos x and cos

x

-

2

= sin x.

You may make the substitution at the beginning

and then antidifferentiate, or first antidifferenti-

ate and then substitute.

14. See previous question.

2 15. cos 2x + dx

3

1

2

= 2 cos 2x +

dx

2

3

2 = sin 2x + + c

3

16. sin(-x) dx = - sin x dx

= cos x + c

4 17. cos2 x dx = 4 tan x + c

1

1

4

18. cos2 4x dx = 4 cos2 4x dx

1 = tan(4x) + c

4

1 19. cos2(x - 1) dx = tan(x - 1) + c

20. 6 cos2x + 6 sin 3x dx

= 3 2 cos 2x dx - 2 -3 sin 3x dx = 3 sin 2x - 2 cos 3x + c

21. cos8x - 4 sin 2x dx 1

= 8 cos 8x dx + 2 -2 sin 2x dx 8 1

= sin 8x + 2 cos 2x + c 8

22. 2x + 4 cos x + 6 cos 2x dx

= 2x dx + 4 cos x dx + 3 2 cos 2x dx

= x2 + 4 sin x + 3 sin 2x + c 23. Although this looks long, you should by now be

able to antidifferentiate it in a single step, simply working term by term.

3 + 4x - 6x2 dx = 3x + 2x2 - 2x3 + c1

so the overall antiderivative is 3x + 2x2 - 2x3 + 2 sin 5x + cos 4x + c 2

24.

cos3

x

sin

x

dx

=

cos4 -

x

+

c

4

25.

30

cos5

x

sin

x

dx

=

30 -

cos6

x

+

c

6

= -5 cos6 x + c

26. sin4 x cos x dx = sin5 x + c 5

27. 6 sin3 x cos x dx = 6 sin4 x + c 4

3 sin4 x

=

+c

2

28. -2 cos4 x sin x dx = 2 cos5 x + c 5

29.

-2

sin7

x

cos

x

dx

=

2 -

sin8

x

+

c

8

sin8 x

=-

+c

4

30. 32 sin3 4x cos 4x dx = 8 sin3 4x(4 cos 4x) dx

8 sin4 4x

=

+c

4

= 2 sin4 4x + c

31. -24 sin3 2x cos2x dx

= -12 sin3 2x(2 cos 2x) dx

-12 sin4 2x

=

+c

4

= -3 sin4 2x + c

32. 20 sin4 2x cos 2x dx

= 10 sin4 2x(2 cos 2x) dx

10 sin5 2x

=

+c

5

= 2 sin5 2x + c

10 cos 5x dx = 2 5 cos 5x dx

= 2 sin 5x + c2 1

-2 sin 4x dx = -4 sin 4x dx 2 cos 4x

= 2 + c3

33. 4

-6 cos2 4x sin4x dx

3 =

cos2 4x(-4 sin 4x) dx

2

3 cos3 4x

=

+c

2?3

cos3 4x

=

+c

2

Unit 3C Specialist Mathematics

Exercise 7B

34. sin3 x dx = sin x(sin2 x) dx

= sin x(1 - cos2 x) dx

= sin x - sin x cos2 x dx

cos3 x

= - cos x +

+c

3

35. cos3 x dx = cos x(cos2 x) dx

= cos x(1 - sin2 x) dx

= cos x - cos x sin2 x dx

sin3 x

= sin x -

+c

3

39. 8 sin4 x dx = 8 (sin2 x)2 dx

-2 sin2 x 2

=8

dx

-2

(-2 sin2 x)2

=8

dx

4

= 2 (-2 sin2 x)2 dx

= 2 (1 - 2 sin2 x - 1)2 dx

= 2 (cos 2x - 1)2 dx

= 2 cos2 2x - 2 cos 2x + 1 dx

= 2 cos2 2x - 4 cos 2x + 2 dx

= 2 cos2 2x - 1 - 4 cos 2x + 3 dx

36. cos5x dx = cos x(cos4 x) dx = cos x(cos2 x)2 dx

= cos 4x - 4 cos 2x + 3 dx

sin 4x 4 sin 2x

=

-

+ 3x + c

4

2

sin 4x

=

- 2 sin 2x + 3x + c

4

= cos x(1 - sin2 x)2 dx

40. cos2 x + sin2 x dx = 1 dx

= cos x(1 - 2 sin2 x + sin4 x) dx = cos x - 2 cos x sin2 x + cos x sin4 x dx

=x+c (Compare this with your answers for questions 37 and 38.)

2 sin3 x sin5 x

= sin x -

+

+c

3

5

37. cos2 x dx = 1 2 cos2 x dx 2

1 =

2 cos2 x - 1 + 1 dx

2

1 = cos 2x + 1 dx

2

1 sin 2x

=

+x +c

22

sin 2x x

=

+ +c

42

38. sin2 x dx = - 1 -2 sin2 x dx 2

1 =-

1 - 2 sin2 x - 1 dx

2

1 = - cos 2x - 1 dx

2

1 sin 2x

=-

-x +c

22

sin 2x x

=-

+ +c

42

41. cos2 x - sin2 x dx = cos 2x dx

sin 2x

=

+c

2

(Compare this with your answers for questions

37 and 38.)

42. sin 5x cos 2x + cos 5x sin 2x dx

= sin(5x + 2x) dx

= sin 7x dx

cos 7x

=-

+c

7

(You need to know the angle sum trig identities

well enough to recognise them.)

43. sin 3x cos x - cos 3x sin x dx

= sin(3x - x) dx

= sin 2x dx

cos 2x

=-

+c

2

5

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