1 inverse trig functions
TRANSCRIPT
Lecture 6Section 7.7 Inverse Trigonometric Functions Section
7.8 Hyperbolic Sine and Cosine
Jiwen He
1 Inverse Trig Functions
1.1 Inverse Sine
Inverse Since sin−1 x (or arcsin x)
1
domain:[− 12π, 1
2π] range:[−1, 1]
2
sin(sin−1 x) = x
3
4
domain:[−1, 1] range:[− 12π, 1
2π]
Trigonometric Properties
5
sin(sin−1 x) = x cos(sin−1 x) =√
1− x2
tan(sin−1 x) =x√
1− x2cot(sin−1 x) =
√1− x2
x
sec(sin−1 x) =1√
1− x2csc(sin−1 x) =
1x
Differentiation
Theorem 1.d
dxsin−1 x =
1√1− x2
.
Proof.Let y = sin−1 x. Then x = sin y,
d
dxsin−1 x =
1ddy sin y
=1
cos y=
1cos(sin−1 x)
=1√
1− x2.
Theorem 2.
d
dxsin−1 u =
1√1− u2
du
dx,
∫1√
1− u2du = sin−1 u + C
Integration: u-Substitution
6
Theorem 3. ∫g′(x)√
1− (g(x))2dx = sin−1(g(x)) + C
ProofLet u = g(x). Then du = g′(x) dx,∫
g′(x)√1− (g(x))2
dx =∫
1√1− u2
du = sin−1 u + C = sin−1(g(x)) + C
Examples 4.∫
1√4− x2
dx =∫
1√1− u2
du = sin−1 u+C = sin−1 x
2+C. Note
that 4− x2 = 4(1−
(x2
)2). Let u = x
2 . Then du = 12dx.
∫1√
2x− x2dx =∫
1√1− u2
du = sin−1 u+C = sin−1(x− 1)+C. Note that 2x−x2 = 1− (x2−
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 arctanx)
7
y = tan x domain:(− 12π, 1
2π) range:(−∞,∞)
8
Trigonometric Properties
tan(tan−1 x) = x cot(tan−1 x) =1x
sin(tan−1 x) =x√
1 + x2cos(tan−1 x) =
1√1 + x2
sec(tan−1 x) =√
1 + x2 csc(tan−1 x) =√
1 + x2
x
Differentiation
Theorem 5.d
dxtan−1 x =
11 + x2
.
Proof.Let y = tan−1 x. Then x = tan y,
d
dxtan−1 x =
1ddy tan y
=1
(sec y)2=
1(sec(tan−1 x)
)2 =1
1 + x2.
Theorem 6.
d
dxtan−1 u =
11 + u2
du
dx,
∫1
1 + u2du = tan−1 u + C
9
Integration: u-Substitution
Theorem 7. ∫g′(x)
1 + (g(x))2dx = tan−1(g(x)) + C
ProofLet u = g(x). Then du = g′(x) dx,∫
g′(x)1 + (g(x))2
dx =∫
11 + u2
du = tan−1 u + C = tan−1(g(x)) + C
Examples 8.∫
14 + x2
dx =12
∫1
1 + u2du =
12
tan−1 u + C =12
tan−1 x
2+ C.
Note that 4+x2 = 4(1 +
(x2
)2). Let u = x
2 . Then du = 12dx.
∫1
2 + 2x + x2dx =∫
11 + u2
du = tan−1(x+1)+C. Note that 2+2x+x2 = 1+(x2+2x+1) = 1+(x+
1)2 (complete the square). Let u = x + 1. Then du = dx.∫
e−x
1 + e−2xdx =
−∫
11 + u2
du = − tan−1(e−x) + C. Note that 1 + e−2x = 1 + (e−x)2 (complete
the square). Let u = e−x. Then du = −e−xdx.
Quiz
Quiz
10
Let f ′(t) = kf(t).
1. For f(0) = 4, f(t) =: (a) kt + 4, (b) 4ekt, (c) 4e−kt.
2. For k > 0, double time T =: (a)4k
, (b)ln 2k
(c) − ln 2k
.
1.3 Inverse Secant
Inverse Secant sec−1 x
11
y = sec x domain:[0, 12π)∪( 1
2π, π] range:(−∞,−1]∪[1,∞)
12
Trigonometric Properties
sec(sec−1 x) = x csc(sec−1 x) =x√
x2 − 1
sin(sec−1 x) =√
x2 − 1x
cos(sec−1 x) =1x
tan(sec−1 x) =√
x2 − 1 cot(sec−1 x) =1√
x2 − 1
Differentiation
Theorem 9.d
dxsec−1 x =
1|x|√
x2 − 1.
Proof.Let y = sec−1 x. Then x = sec y,
d
dxsec−1 x =
1ddy sec y
=1
(sec y tan y)2=
1|x|√
x2 − 1.
Theorem 10.
d
dxsec−1 u =
1|u|√
u2 − 1du
dx,
∫1
u√
u2 − 1du = sec−1 |u|+ C
13
Integration: u-Substitution
Theorem 11. ∫g′(x)
g(x)√
(g(x))2 − 1dx = sec−1(|g(x)|) + C
ProofLet u = g(x). Then du = g′(x) dx,∫
g′(x)g(x)
√(g(x))2 − 1
dx =∫
1u√
u2 − 1du = sec−1(|g(x)|) + C
Examples 12.∫
1x√
x− 1dx = 2
∫1
u√
u2 − 1du =
12
sec−1√
x + C. Note that
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
sin−1 x + cos−1 x =π
2or cos−1 x =
π
2− sin−1 x
tan−1 x + cot−1 x =π
2or cot−1 x =
π
2− tan−1 x
sec−1 x + csc−1 x =π
2or csc−1 x =
π
2− sec−1 x
Differentiation
Theorem 13.
d
dxcos−1 x = − d
dxsin−1 x = − 1√
1− x2
d
dxcot−1 x = − d
dxtan−1 x = − 1
1 + x2
d
dxcsc−1 x = − d
dxsec−1 x = − 1
|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 Hyperbolic Sine and Cosine
2.1 Definition
Hyperbolic Sine and Cosine
15
Definition 14.
sinhx =12(ex − e−x
), coshx =
12(ex + e−x
)Theorem 15.
d
dxsinhx = cosh,
d
dxcoshx = sinh,
Identities
16
17
cosh2 x− sinh2 x = 1sinh(x + y) = sinhx cosh y + coshx sinh y
cosh(x + y) = coshx cosh y + sinhx sinh y
cos2 x + sin2 x = 1sin(x + y) = sin x cos y + cos x sin y
cos(x + y) = cos x cos y − sinx sin y
Outline
Contents
1 Inverse Trig Functions 1
18
1.1 Inverse Sine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Inverse Tangent . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 Inverse Secant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.4 Other Trig Inverses . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2 Hyperbolic Sine and Cosine 152.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
19