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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