Lecture 6Section 7.7 Inverse Trigonometric Functions Section
7.8 Hyperbolic Sine and Cosine
Jiwen He
1
1.1
Inverse Trig Functions
Inverse Sine
Inverse Since sin−1 x (or arcsin x)
1
domain:[− 12 π, 12 π]
range:[−1, 1]
2
sin(sin−1 x) = x
3
4
domain:[−1, 1]
range:[− 12 π, 12 π]
Trigonometric Properties
5
sin(sin−1 x) = x
cos(sin−1 x) =
x
1 − x2
1
sec(sin−1 x) = √
1 − x2
cot(sin−1 x) =
tan(sin−1 x) = √
csc(sin−1 x) =
p
√
1 − x2
1 − x2
x
1
x
Differentiation
Theorem 1.
1
d
sin−1 x = √
.
dx
1 − x2
Proof.
Let y = sin−1 x. Then x = sin y,
d
sin−1 x =
dx
d
dy
1
1
1
1
=
.
=
=√
cos y
cos(sin−1 x)
sin y
1 − x2
Theorem 2.
Z
1
d
du
sin−1 u = √
,
dx
1 − u2 dx
Integration: u-Substitution
6
√
1
du = sin−1 u + C
1 − u2
Theorem 3.
g 0 (x)
Z
p
1 − (g(x))2
dx = sin−1 (g(x)) + C
Proof
Let u = g(x). Then du = g 0 (x) dx,
Z
Z
g 0 (x)
1
p
dx = √
du = sin−1 u + C = sin−1 (g(x)) + C
2
1 − u2
1 − (g(x))
Z
Z
1
1
x
√
Examples 4.
dx = √
du = sin−1 u+C = sin−1 +C. Note
2
2
2
1−u
Z
4 − x 1
2
√
that 4 − x2 = 4 1 − x2
. Let u = x2 . Then du = 12 dx.
dx =
2x − x2
Z
1
√
du = sin−1 u + C = sin−1 (x − 1) + C. Note that 2x − x2 = 1 − (x2 −
1 − u2
2x + 1) = 1 − (x − 1)2 (complete the square). Let u = x − 1. Then du = dx.
1.2
Inverse Tangent
Inverse Tangent tan−1 x (or arctan x)
7
8
y = tan x domain:(− 12 π, 12 π) range:(−∞, ∞)
Trigonometric Properties
1
x
tan(tan−1 x) = x
cot(tan−1 x) =
x
1 + x2
p
sec(tan−1 x) = 1 + x2
1
cos(tan−1 x) = √
1 + x2
√
1 + x2
csc(tan−1 x) =
x
sin(tan−1 x) = √
Differentiation
Theorem 5.
d
1
tan−1 x =
.
dx
1 + x2
Proof.
Let y = tan−1 x. Then x = tan y,
d
tan−1 x =
dx
d
dy
1
1
1
1
=
=
.
2 =
−1
(sec y)2
1
+
x2
tan y
sec(tan x)
Theorem 6.
Z
d
1
du
tan−1 u =
,
dx
1 + u2 dx
9
1
du = tan−1 u + C
1 + u2
Integration: u-Substitution
Theorem 7.
Z
g 0 (x)
dx = tan−1 (g(x)) + C
1 + (g(x))2
Proof
Let u = g(x). Then du = g 0 (x) dx,
Z
Z
g 0 (x)
1
dx =
du = tan−1 u + C = tan−1 (g(x)) + C
1 + (g(x))2
1 + u2
Z
Z
1
1
1
x
1
1
dx
=
du = tan−1 u + C = tan−1 + C.
2
2
4+x
2 1+u
2
2
Z2
1
x 2
x
1
2
dx =
Note that 4+x = 4 1 + 2
. Let u = 2 . Then du = 2 dx.
2 + 2x + x2
Z
1
du = tan−1 (x+1)+C. Note that 2+2x+x2 = 1+(x2 +2x+1) = 1+(x+
1 + u2
Z
e−x
2
dx =
1) (complete the square). Let u = x + 1. Then du = dx.
1 + e−2x
Z
1
−
du = − tan−1 (e−x ) + C. Note that 1 + e−2x = 1 + (e−x )2 (complete
1 + u2
the square). Let u = e−x . Then du = −e−x dx.
Examples 8.
Quiz
Quiz
10
Let f 0 (t) = kf (t).
1. For f (0) = 4, f (t) =:
(a) kt + 4,
2. For k > 0, double time T =:
(a)
1.3
Inverse Secant
Inverse Secant sec−1 x
11
4
,
k
(b) 4ekt ,
(b)
ln 2
k
(c) 4e−kt .
(c) −
ln 2
.
k
y = sec x domain:[0, 21 π)∪( 12 π, π] range:(−∞, −1]∪[1, ∞)
12
Trigonometric Properties
sec(sec−1 x) = x
√
csc(sec−1 x) = √
x2 − 1
sin(sec−1 x) =
x
p
−1
tan(sec x) = x2 − 1
x
x2
−1
1
cos(sec−1 x) =
x
cot(sec−1 x) = √
1
x2
−1
Differentiation
Theorem 9.
d
1
√
sec−1 x =
.
dx
|x| x2 − 1
Proof.
Let y = sec−1 x. Then x = sec y,
d
sec−1 x =
dx
d
dy
1
1
1
√
=
=
.
2
(sec y tan y)
sec y
|x| x2 − 1
Theorem 10.
Z
d
1
du
√
sec−1 u =
,
2
dx
|u| u − 1 dx
13
1
du = sec−1 |u| + C
u u2 − 1
√
Integration: u-Substitution
Theorem 11.
Z
g 0 (x)
p
dx = sec−1 (|g(x)|) + C
g(x) (g(x))2 − 1
Proof
Let u = g(x). Then du = g 0 (x) dx,
Z
Z
1
g 0 (x)
p
√
dx =
du = sec−1 (|g(x)|) + C
2
u u2 − 1
g(x) (g(x)) − 1
Z
Z
√
1
1
1
√
√
Examples 12.
du = sec−1 x + C. Note that
dx = 2
2
2
x x−1 √
u u −1
√
x − 1 = ( x)2 − 1. Let u = x. Then x = u2 , dx = 2udu.
1.4
Other Trig Inverses
Other Trigonometric Inverses
Other Trigonometric Inverses
14
π
2
π
−1
−1
tan x + cot x =
2
π
sec−1 x + csc−1 x =
2
sin−1 x + cos−1 x =
or
or
or
π
− sin−1 x
2
π
cot−1 x = − tan−1 x
2
π
csc−1 x = − sec−1 x
2
cos−1 x =
Differentiation
Theorem 13.
d
d
1
cos−1 x = − sin−1 x = − √
dx
dx
1 − x2
d
d
1
cot−1 x = − tan−1 x = −
dx
dx
1 + x2
d
d
1
csc−1 x = − sec−1 x = − √
dx
dx
|x| x2 − 1
Quiz (cont.)
The value, at the end of the 4 years, of a principle of $100 invested at 4%
compounded
3. annually: (a) 400(1 + 0.04), (b) 100(1 + 0.04)4 , (c) 100(1 + 0.16).
4. continuously: (a) 100e0.04 , (b) 100e0.16 , (c) 100(1 + 0.04)4 .
2
2.1
Hyperbolic Sine and Cosine
Definition
Hyperbolic Sine and Cosine
15
Definition 14.
sinh x =
Theorem 15.
1 x
e − e−x ,
2
cosh x =
1 x
e + e−x
2
d
cosh x = sinh,
dx
d
sinh x = cosh,
dx
Identities
16
17
cosh2 x − sinh2 x = 1
sinh(x + y) = sinh x cosh y + cosh x sinh y
cosh(x + y) = cosh x cosh y + sinh x sinh y
cos2 x + sin2 x = 1
sin(x + y) = sin x cos y + cos x sin y
cos(x + y) = cos x cos y − sin x sin y
Outline
Contents
1 Inverse Trig Functions
1
18
1.1
1.2
1.3
1.4
Inverse Sine . . . .
Inverse Tangent . .
Inverse Secant . . .
Other Trig Inverses
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1
7
11
14
2 Hyperbolic Sine and Cosine
15
2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
19