Download 5.6 INVERSE TRIGONOMETRIC FUNCTIONS

SECTION 5.6
5.6
1
3. cot
2. csc s2
(s3 )
4. sec1 2
5
5. arcsin sin
4
6. sin (2 sin
( ))
1 3
5
s3
2
7. cos1
■ Find the derivative of the function. Find the domains of
the function and its derivative.
36 – 41
1
1. sin 0.5
1
8. sin1 9. sinsin 1 0.7
14. secarctan 2
15. cos(2 sin
( ))
16. sin sin1 ( 3 ) sin1 ( 3 )
■
■
■
■
■
1
2
■
■
■
18. f x sin 2x 1
19. tx tan x 20. hx arcsin x ln x
21. f t cos1 tt
22. Ft s1 t 2 sin1 t
23. Gt cos1 s2t 1
24. tx tan1x 3
25. Gx sin1xa, a 0
1
1
■
1
1
32. y x cot 3x
2
1
34. y sin
Copyright © 2013, Cengage Learning. All rights reserved.
■
■
■
cos x
1 sin x
■
■
1
33. y tan sin x
s3
1
35. y tan x
■
■
■
■
■
arcsin x
tan x2
46. lim sin1
xl
■
■
x1
2x 1
■
■
■
■
xl
■
■
■
■
■
■
y s1 t
53.
55.
yx
57.
y 1 9x
0.5
dx
s1 x 2
50.
y
52.
ye
y x 4 ln x 54.
y 1x
x9
dx
2
9
56.
y 1 cos x dx
58.
y
1
t
4
dt
dx
2
dx
■
48. lim tan1x x 2
6
dx
1 x2
51.
■
■
■
Evaluate the integral.
y
2
31. tt sin 4t
1
■
xl1
■
■
1
30. f x arctan x ln x
■
44. lim
tan1 x
x
49.
3
29. y tan1e x 28. y sin x 2
■
Find the limit.
■
49 – 58
27. y sin x
26. Fx tan xa
■
4
xl
possible.
1
■
47. lim tan1x 2
2
Find the derivative of the function. Simplify where
1
■
■
xl
Find a similar simplification of sin2 cos1 x.
■
■
45. lim
]
17. It is a fact that cossin x s1 x for 1 x 1.
18 –35
41. Ut 2 arctan t
xl
1
■
40. Rt arcsin2 t 43. lim tan1 x 2
4
■
39. Sx sin1tan1 x
43– 48
13. sin(cos1 ( 5 ))
[
38. Fx ssin12x
42. If hx 3 tan x , find h3.
12. tancos 1 0.5
1 5
13
37. tx sin13x 1
1
4
3
10. tan1 tan
36. f x cos1sin1 x
■
1
s2
11. sin 1sin 1
■
1
S Click here for solutions.
Find the exact value of the expression.
■
■
INVERSE TRIGONOMETRIC FUNCTIONS
A Click here for answers.
1–16
INVERSE TRIGONOMETRIC FUNCTIONS
2
■
■
■
■
0
ex
dx
1
2x
tan1 x
2
dx
sin x
2
12
0
■
sin1x
dx
s1 x 2
■
■
2
■
SECTION 5.6
INVERSE TRIGONOMETRIC FUNCTIONS
5.6
ANSWERS
E Click here for exercises.
S Click here for solutions.
1. π
6
3.
2. π
4
5π
6
4.
5. − π
4
6. 24
25
7. π
6
8. − π
4
10. π
3
9. 0.7
√
12. 3
√
14. 5
√ √
16. 19
5+4 2
11. 1
13. 35
15. 119
169
√
17. 2x 1 − x2
1
x − x2
18. f (x) = √
19. g (x) =
3x2
1 + x6
ln x
arcsin x
20. h (x) = √
+
x
1 − x2
cos−1 t
1
− √
t2
t 1 − t2
1−t
1 − t2
22. F (t) = √
2 (−2t2
1
a2 − x2
a
26. F (x) = 2
a + x2
25. G (x) = √
arctan x
ln x
+
x
1 + x2
4
t4 − 16t2
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31. g (t) = − √
33. y =
[− sin 1, sin 1], (− sin 1, sin 1)
3
, − 23 , 0 , − 23 , 0
2
−9x − 6x
37. g (x) = √
38. F (x) = −
(3x) −
cos x
1 + sin2 x
1
, [2, ∞), (2, ∞)
(2/x) 1 − 4/x2
−1
39. S (x) =
1 − (tan−1 x)2 1 + x2
,
x2
−1
sin
2t ln 2
, (−∞, 0], (−∞, 0)
1 − 4t
arctan t
41. U (t) = 2
(ln 2) 1 + t2 , R, R
−1 3
42. 162
tan 3
5
40. R (t) = √
π2
4
44. 0
π
6
45. 0
46.
π
2
π
49.
2
48. −
50.
−1
(cos x) + C
−1
(3x) + C
56. − tan
ex
29. y =
1 + e2x
−1
1 − sin
1
2 · √1 − x2 ,
x
π
2
π
6
(ex ) + C
−1 1
53. 12 tan
ln x + C
2
−1 2
54. 12 tan
x +C
2
−1 1
55. 12 ln x + 9 + 9 · 13 tan
x +C
3
2x
1 − x4
−1
−1
28. y = √
32. y = 2x cot
1
52. tan
2 sin−1 x
1 − x2
2
−1
51. 12 sin
t +C
27. y = √
47.
3x2
24. g (x) =
1 + x6
30. f (x) =
1
(1 + x2 ) (tan−1 x)2
36. f (x) = − 43.
1
+ 3t − 1)
23. G (t) = − 35. y = −
−1
2 sin x + 2 sin2 x
[− tan 1, tan 1], (− tan 1, tan 1)
21. f (t) = −
34. y = π
3
3x2
1 + 9x2
57. 13 tan
58.
π2
72
SECTION 5.6
5.6
INVERSE TRIGONOMETRIC FUNCTIONS
■
SOLUTIONS
E Click here for exercises.
−1
(0.5) = π6 because sin π6 = 0.5.
√
√
−1
2. csc
2 = π4 because csc π4 = 2.
√
√ −1
3. cot
because cot 5π
= − 3.
− 3 = 5π
6
6
1. sin
−1
2 = π3 because sec π3 = 2.
arcsin sin 5π
= arcsin − √12 = − π4
4
Let θ = sin−1 35 . Then sin θ = 35
2
⇒ cos θ = 1 − 35 = 45 , so
.
sin 2 sin−1 35 = sin 2θ = 2 sin θ cos θ = 2 · 35 · 45 = 24
25
√ √
cos−1 23 = π6 because cos π6 = 23 .
sin−1 − √12 = − π4 because sin − π4 = − √12 .
sin sin−1 (0.7) = 0.7 since 0.7 is in [ 1, 1].
√
tan−1 tan 4π
= tan−1 3 = π3 since π3 is in − π2 , π2 .
3
4. sec
5.
6.
7.
8.
9.
10.
−1
(sin 1) = 1 since − π2 ≤ 1 ≤ π2 .
√
−1
12. tan cos
0.5 = tan π3 = 3
−1 4
13. Let θ = cos
, so cos θ = 45 . Then
5
−1 4 2 9
sin cos
1 − 45 = 25
= 35 .
=
sin
θ
=
5
11. sin
⇒
√
sec2 θ = 1 + tan2 θ = 1 + 4 = 5 ⇒ sec θ = 5 ⇒
√
sec (arctan 2) = sec θ = 5.
5
−1
5
15. Let θ = sin
, so
. Then sin θ = 13
13
5
cos 2 sin−1 13
= cos 2θ = 1 − 2 sin2 θ
5 2
= 1 − 2 13
= 119
.
169
−1 1
−1 2
16. Let x = sin
and y = sin
. Then
3
3
√
2
sin x = 13 , cos x = 1 − 13 = 2 3 2 , sin y = 23 ,
√
2
cos y = 1 − 23 = 35 , so
sin sin−1 13 + sin−1 23
= sin (x + y)
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14. Let θ = arctan 2, so tan θ = 2
= sin x cos y + cos x sin y
√ √ = 13 35 + 2 3 2 23
√
√ 5+4 2
= 19
−1
17. Let y = cos
x. Then cos y = x
√
⇒ sin y = 1 − x2 since 0 ≤ y ≤ π. So
√
sin 2 cos−1 x = sin 2y = 2 sin y cos y = 2x 1 − x2 .
−1
18. f (x) = sin
(2x − 1) ⇒
1
1
f (x) = (2) = √
2
x
− x2
1 − (2x − 1)
3
−1
19. g (x) = tan
x
⇒
1
3x2
2
g (x) =
=
2 3x
3
1 + x6
1 + (x )
20. h (x) = (arcsin x) ln x
⇒
ln
x
arcsin
x
h (x) = √
+
x
1 − x2
cos−1 t
⇒
t
√
t −1
1 − t2 − cos−1 t
f (t) =
t2
−1
cos t
1
=−
− √
t2
t 1 − t2
√
−1
22. F (t) = 1 − t2 + sin
t ⇒
−2t
1
1−t
F (t) = √
+√
= √
2 1 − t2
1 − t2
1 − t2
√
−1
23. G (t) = cos
2t − 1 ⇒
2
1
√
G (t) = − 2
2t
−1
1 − (2t − 1)
1
= −
2 (−2t2 + 3t − 1)
3
−1
24. g (x) = tan
x
⇒
1
3x2
g (x) =
3x2 =
2
1 + x6
1 + (x3 )
21. f (t) =
−1
25. G (x) = sin
(x/a) ⇒
1/a
1
1
= = √
G (x) = 2 − x2
2 /a2
2
a
a
1
−
x
1 − (x/a)
−1
26. F (x) = tan
(x/a) ⇒
1
1
a
F (x) =
· = 2
a + x2
1 + (x/a)2 a
−1 2
27. y = sin
x
⇒
d −1 2 sin−1 x
y = 2 sin−1 x
sin x = √
dx
1 − x2
−1
28. y = sin
x2
⇒
d 2
1
2x
y = x = √
dx
2
1
− x4
1 − (x2 )
−1
29. y = tan
(ex ) ⇒ y =
d x
1
ex
(e ) =
2
x
1 + e2x
1 + (e ) dx
3
4
■
SECTION 5.6
INVERSE TRIGONOMETRIC FUNCTIONS
30. f (x) = (arctan x) ln x
Dom (F ) = x | −1 ≤ 2/x ≤ 1 and sin−1 (2/x) ≥ 0
⇒
1
arctan x
1
ln x
=
+ ln x ·
+
x
1 + x2
x
1 + x2
4
−1
⇒
31. g (t) = sin
t
1
4
4
g (t) = − 2 = −√
4 − 16t2
t
2
t
1 − (4/t)
f (x) = arctan x ·
2
−1
32. y = x cot
(3x) ⇒
y = 2x cot−1 (3x) + x2
2x cot−1 (3x) −
−1
34. y = sin
y = cos x
1 + sin x
cos x
1 + sin2 x
⇒
2
1 − [cos x/ (1 + sin x)]
·
− sin x (1 + sin x) − cos2 x
(1 + sin x)2
1
− (1 + sin x)
·
2
(1 + sin x)2
1 + 2 sin x + sin2 x − cos2 x (1 + sin x)
−1
2 sin x + 2 sin2 x
−1 −1
35. y = tan
x
⇒
−2
y = − tan−1 x
= {x | sin (−1) ≤ x ≤ sin 1}
= [− sin 1, sin 1]
Dom (f ) = x | −1 < sin−1 x < 1 = (− sin 1, sin 1)
−1
(3x + 1) ⇒
3
3
g (x) = = √
,
2
−9x2 − 6x
1 − (3x + 1)
Copyright © 2013, Cengage Learning. All rights reserved.
37. g (x) = sin
Dom (g) = {x | −1 ≤ 3x + 1 ≤ 1}
= x | − 23 ≤ x ≤ 0 = − 23 , 0
Dom (g ) = {x | −1 < 3x + 1 < 1} = − 23 , 0
−1
38. F (x) =
sin (2/x) ⇒
1
1
2
F (x) = −1
·
− 2
x
2 sin (2/x)
1 − (2/x)2
x2
1
−1
sin
= {x | tan (−1) ≤ x ≤ tan 1} = [− tan 1, tan 1]
Dom (S ) = x | −1 < tan−1 x < 1 = (− tan 1, tan 1)
t
40. R (t) = arcsin 2
(2/x) 1 − 4/x2
⇒
2t ln 2
,
2t ln 2 = √
1 − 4t
1 − (2t )2
Dom (R) = t | −1 ≤ 2t ≤ 1 = {t | t ≤ 0} = (−∞, 0],
1
R (t) = Dom (R ) = (−∞, 0)
41. U (t) = 2
⇒ U (t) = 2arctan t (ln 2)
1 + t2 ,
Dom (U ) = Dom (U ) = R
4
−1
42. h (x) = 3 tan
x
⇒
3
3
⇒
h (x) = 4 3 tan−1 x
1 + x2
3 3 −1 3
h (3) = 4 3 tan−1 3
= 162
tan 3
10
5
2 π 2
2 π2
=
tan−1 x = lim tan−1 x =
x→∞
x→∞
2
4
arcsin x
44. lim
= 0 since lim arcsin x = π2 and
x→1− tan (πx/2)
x→1−
lim tan (πx/2) = ∞.
43. lim
1
1
=−
1 + x2
(1 + x2 ) (tan−1 x)2
−1 −1
36. f (x) = cos
sin x ⇒
1
1
f (x) = − −1 2 · √1 − x2 ,
1 − sin x
Dom (f ) = x | −1 ≤ sin−1 x ≤ 1
=−
Dom (F ) = {x | x > 2} = (2, ∞)
−1 −1
39. S (x) = sin
tan x ⇒
−1
S (x) =
1 − (tan−1 x)2 1 + x2
Dom (S) = x | −1 ≤ tan−1 x ≤ 1
arctan t
1
= = 3x2
1 + 9x2
(sin x) ⇒ y =
33. y = tan
−1
1
−
(3) =
1 + (3x)2
= {x | 0 < 2/x ≤ 1} = {x | x ≥ 2} = [2, ∞)
x→1−
tan−1 x
= 0 since lim tan−1 x =
x→∞
x→∞
x
lim x = ∞.
45. lim
π
2
and
x→∞
Or: Note that 0 <
Theorem.
−1
46. lim sin
x→∞
sin−1
1
2
=
π
6
x+1
2x + 1
−1
= sin
x+1
lim
x→∞ 2x + 1
−1
x2 =
−1
x − x2 = − π2 since
47. lim tan
x→∞
48. lim tan
x→∞
π/2
tan−1 x
<
and use the Squeeze
x
x
π
2
=
since x2 → ∞ as x → ∞.
x − x2 = x (1 − x) → −∞ as x → ∞.
√3
√3
6
−1
49.
dx
=
6
tan
x
1
1 + x2
1
−1 √
= 6 tan
3 − tan−1 1
π
= 6 3 − π4 = π2
0.5
0.5
dx
√
50.
= sin−1 x 0 = sin−1 12 − sin−1 0 =
2
1−x
0
π
6
SECTION 5.6
2
51. Let u = t . Then du = 2t dt, so
t
1
√
dt =
4
2
1−t
1
√
du =
1 − u2
= 12 sin−1 t2 + C.
x
1
2
sin−1 u + C
x
52. Let u = e . Then du = e dx, so
x
e dx
=
e2x + 1
du
= tan−1 u+C = tan−1 (ex )+C.
u2 + 1
53. Let u = ln x. Then du = (1/x) dx
⇒
u
du
= 12 tan−1
+C
2
4+u
2
= 12 tan−1 12 ln x + C.
dx
=
x 4 + (ln x)2
dx
x. Then du =
, so
1 + x2
2
tan−1 x
dx = u du = 12 u2 + C = 12 tan−1 x + C.
1 + x2
x+9
x
1
55.
dx
=
dx
+
9
dx
x2 + 9
x2 + 9
x2 + 9
= 12 ln x2 + 9 + 9 · 13 tan−1 13 x + C
−1
54. Let u = tan
[Let u = x2 + 9 in the first integral and use Formula 14 in
the second.]
56. Let u = cos x. Then du = − sin x dx, so
sin x
dx = −
1 + cos2 x
1
du = − tan−1 u + C
1 + u2
= − tan−1 (cos x) + C.
57. Let u = 3x. Then du = 3 dx, so
dx
1
=
1 + 9x2
3
=
1/2
Copyright © 2013, Cengage Learning. All rights reserved.
0
du
=
1 + u2
1
3
tan−1 u + C
tan−1 (3x) + C.
1
x. Then du = √
dx, so
1 − x2
π/6
π/6
sin−1 x
u2 √
dx =
u du =
2 0
1 − x2
0
1 π 2
π2
=
=
.
2 6
72
−1
58. Let u = sin
1
3
INVERSE TRIGONOMETRIC FUNCTIONS
■
5