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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

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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

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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

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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

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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

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Definition 14.

sinhx =12(ex − e−x

), coshx =

12(ex + e−x

)Theorem 15.

d

dxsinhx = cosh,

d

dxcoshx = sinh,

Identities

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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

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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

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