Inverse Trig Functions Hyperbolic Sine and Cosine

Lecture 6Section 7.7 Inverse Trigonometric Functions

Section 7.8 Hyperbolic Sine and Cosine

Jiwen He

Department of Mathematics, University of Houston

jiwenhe@math.uh.eduhttp://math.uh.edu/∼jiwenhe/Math1432

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 1 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Since sin−1 x (or arcsin x)

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 2 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Since sin−1 x (or arcsin x)

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 2 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Since sin−1 x (or arcsin x)

domain:[− 12π, 1

2π]

range:[−1, 1]

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 2 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Since sin−1 x (or arcsin x)

domain:[− 12π, 1

2π]

range:[−1, 1]

sin(sin−1 x) = x

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 2 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Since sin−1 x (or arcsin x)

domain:[− 12π, 1

2π]

range:[−1, 1]

sin(sin−1 x) = x

domain:[−1, 1]

range:[− 12π, 1

2π]

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 2 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Trigonometric Properties

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) =

1

x

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 3 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Trigonometric Properties

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) =

1

x

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 3 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Trigonometric Properties

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) =

1

x

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 3 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Trigonometric Properties

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) =

1

x

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 3 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Trigonometric Properties

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) =

1

x

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 3 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Trigonometric Properties

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) =

1

x

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 3 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Differentiation

Theorem

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=

1

cos(sin−1 x)=

1√1− x2

.

Theorem

d

dxsin−1 u =

1√1− u2

du

dx,

∫1√

1− u2du = sin−1 u + C

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 4 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Differentiation

Theorem

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=

1

cos(sin−1 x)=

1√1− x2

.

Theorem

d

dxsin−1 u =

1√1− u2

du

dx,

∫1√

1− u2du = sin−1 u + C

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 4 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Differentiation

Theorem

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=

1

cos(sin−1 x)=

1√1− x2

.

Theorem

d

dxsin−1 u =

1√1− u2

du

dx,

∫1√

1− u2du = sin−1 u + C

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 4 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Differentiation

Theorem

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=

1

cos(sin−1 x)=

1√1− x2

.

Theorem

d

dxsin−1 u =

1√1− u2

du

dx,

∫1√

1− u2du = sin−1 u + C

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 4 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)√

1− (g(x))2dx = sin−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)√

1− (g(x))2dx =

∫1√

1− u2du = sin−1 u+C = sin−1(g(x))+C

Examples∫1√

4− x2dx =

∫1√

1− u2du = sin−1 u + C = sin−1 x

2+ C .

Note that 4− x2 = 4(1−

(x2

)2). Let u = x

2 . Then du = 12dx .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 5 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)√

1− (g(x))2dx = sin−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)√

1− (g(x))2dx =

∫1√

1− u2du = sin−1 u+C = sin−1(g(x))+C

Examples∫1√

4− x2dx =

∫1√

1− u2du = sin−1 u + C = sin−1 x

2+ C .

Note that 4− x2 = 4(1−

(x2

)2). Let u = x

2 . Then du = 12dx .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 5 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)√

1− (g(x))2dx = sin−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)√

1− (g(x))2dx =

∫1√

1− u2du = sin−1 u+C = sin−1(g(x))+C

Examples∫1√

4− x2dx =

∫1√

1− u2du = sin−1 u + C = sin−1 x

2+ C .

Note that 4− x2 = 4(1−

(x2

)2). Let u = x

2 . Then du = 12dx .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 5 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)√

1− (g(x))2dx = sin−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)√

1− (g(x))2dx =

∫1√

1− u2du = sin−1 u+C = sin−1(g(x))+C

Examples∫1√

4− x2dx =

∫1√

1− u2du = sin−1 u + C = sin−1 x

2+ C .

Note that 4− x2 = 4(1−

(x2

)2). Let u = x

2 . Then du = 12dx .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 5 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)√

1− (g(x))2dx = sin−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)√

1− (g(x))2dx =

∫1√

1− u2du = sin−1 u+C = sin−1(g(x))+C

Examples∫1√

4− x2dx =

∫1√

1− u2du = sin−1 u + C = sin−1 x

2+ C .

Note that 4− x2 = 4(1−

(x2

)2). Let u = x

2 . Then du = 12dx .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 5 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)√

1− (g(x))2dx = sin−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)√

1− (g(x))2dx =

∫1√

1− u2du = sin−1 u+C = sin−1(g(x))+C

Examples∫1√

2x − x2dx =

∫1√

1− u2du = sin−1 u+C = sin−1(x−1)+C .

Note that 2x − x2 = 1− (x2 − 2x + 1) = 1− (x − 1)2 (completethe square). Let u = x − 1. Then du = dx .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 5 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)√

1− (g(x))2dx = sin−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)√

1− (g(x))2dx =

∫1√

1− u2du = sin−1 u+C = sin−1(g(x))+C

Examples∫1√

2x − x2dx =

∫1√

1− u2du = sin−1 u+C = sin−1(x−1)+C .

Note that 2x − x2 = 1− (x2 − 2x + 1) = 1− (x − 1)2 (completethe square). Let u = x − 1. Then du = dx .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 5 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)√

1− (g(x))2dx = sin−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)√

1− (g(x))2dx =

∫1√

1− u2du = sin−1 u+C = sin−1(g(x))+C

Examples∫1√

2x − x2dx =

∫1√

1− u2du = sin−1 u+C = sin−1(x−1)+C .

Note that 2x − x2 = 1− (x2 − 2x + 1) = 1− (x − 1)2 (completethe square). Let u = x − 1. Then du = dx .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 5 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Tangent tan−1 x (or arctan x)

y = tan x

domain:(− 12π, 1

2π)

range:(−∞,∞)

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 6 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Tangent tan−1 x (or arctan x)

y = tan x

domain:(− 12π, 1

2π)

range:(−∞,∞)

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 6 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Tangent tan−1 x (or arctan x)

y = tan x

domain:(− 12π, 1

2π)

range:(−∞,∞)

Trigonometric Properties

tan(tan−1 x) = x cot(tan−1 x) =1

x

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 6 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Tangent tan−1 x (or arctan x)

y = tan x

domain:(− 12π, 1

2π)

range:(−∞,∞)

Trigonometric Properties

tan(tan−1 x) = x cot(tan−1 x) =1

x

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 6 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Tangent tan−1 x (or arctan x)

y = tan x

domain:(− 12π, 1

2π)

range:(−∞,∞)

Trigonometric Properties

tan(tan−1 x) = x cot(tan−1 x) =1

x

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 6 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Tangent tan−1 x (or arctan x)

y = tan x

domain:(− 12π, 1

2π)

range:(−∞,∞)

Trigonometric Properties

tan(tan−1 x) = x cot(tan−1 x) =1

x

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 6 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Tangent tan−1 x (or arctan x)

y = tan x

domain:(− 12π, 1

2π)

range:(−∞,∞)

Trigonometric Properties

tan(tan−1 x) = x cot(tan−1 x) =1

x

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 6 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Tangent tan−1 x (or arctan x)

y = tan x

domain:(− 12π, 1

2π)

range:(−∞,∞)

Trigonometric Properties

tan(tan−1 x) = x cot(tan−1 x) =1

x

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 6 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Tangent tan−1 x (or arctan x)

y = tan x

domain:(− 12π, 1

2π)

range:(−∞,∞)

Trigonometric Properties

tan(tan−1 x) = x cot(tan−1 x) =1

x

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 6 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Differentiation

Theorem

d

dxtan−1 x =

1

1 + 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

d

dxtan−1 u =

1

1 + u2

du

dx,

∫1

1 + u2du = tan−1 u + C

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 7 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Differentiation

Theorem

d

dxtan−1 x =

1

1 + 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

d

dxtan−1 u =

1

1 + u2

du

dx,

∫1

1 + u2du = tan−1 u + C

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 7 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Differentiation

Theorem

d

dxtan−1 x =

1

1 + 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

d

dxtan−1 u =

1

1 + u2

du

dx,

∫1

1 + u2du = tan−1 u + C

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 7 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Differentiation

Theorem

d

dxtan−1 x =

1

1 + 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

d

dxtan−1 u =

1

1 + u2

du

dx,

∫1

1 + u2du = tan−1 u + C

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 7 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)

1 + (g(x))2dx = tan−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)

1 + (g(x))2dx =

∫1

1 + u2du = tan−1 u + C = tan−1(g(x)) + C

Examples∫1

4 + x2dx =

1

2

∫1

1 + u2du =

1

2tan−1 u + C =

1

2tan−1 x

2+ C .

Note that 4 + x2 = 4(1 +

(x2

)2). Let u = x

2 . Then du = 12dx .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 8 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)

1 + (g(x))2dx = tan−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)

1 + (g(x))2dx =

∫1

1 + u2du = tan−1 u + C = tan−1(g(x)) + C

Examples∫1

4 + x2dx =

1

2

∫1

1 + u2du =

1

2tan−1 u + C =

1

2tan−1 x

2+ C .

Note that 4 + x2 = 4(1 +

(x2

)2). Let u = x

2 . Then du = 12dx .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 8 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)

1 + (g(x))2dx = tan−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)

1 + (g(x))2dx =

∫1

1 + u2du = tan−1 u + C = tan−1(g(x)) + C

Examples∫1

4 + x2dx =

1

2

∫1

1 + u2du =

1

2tan−1 u + C =

1

2tan−1 x

2+ C .

Note that 4 + x2 = 4(1 +

(x2

)2). Let u = x

2 . Then du = 12dx .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 8 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)

1 + (g(x))2dx = tan−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)

1 + (g(x))2dx =

∫1

1 + u2du = tan−1 u + C = tan−1(g(x)) + C

Examples∫1

4 + x2dx =

1

2

∫1

1 + u2du =

1

2tan−1 u + C =

1

2tan−1 x

2+ C .

Note that 4 + x2 = 4(1 +

(x2

)2). Let u = x

2 . Then du = 12dx .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 8 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)

1 + (g(x))2dx = tan−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)

1 + (g(x))2dx =

∫1

1 + u2du = tan−1 u + C = tan−1(g(x)) + C

Examples∫1

4 + x2dx =

1

2

∫1

1 + u2du =

1

2tan−1 u + C =

1

2tan−1 x

2+ C .

Note that 4 + x2 = 4(1 +

(x2

)2). Let u = x

2 . Then du = 12dx .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 8 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)

1 + (g(x))2dx = tan−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)

1 + (g(x))2dx =

∫1

1 + u2du = tan−1 u + C = tan−1(g(x)) + C

Examples∫1

2 + 2x + x2dx =

∫1

1 + u2du = 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 .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 8 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)

1 + (g(x))2dx = tan−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)

1 + (g(x))2dx =

∫1

1 + u2du = tan−1 u + C = tan−1(g(x)) + C

Examples∫1

2 + 2x + x2dx =

∫1

1 + u2du = 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 .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 8 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)

1 + (g(x))2dx = tan−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)

1 + (g(x))2dx =

∫1

1 + u2du = tan−1 u + C = tan−1(g(x)) + C

Examples∫1

2 + 2x + x2dx =

∫1

1 + u2du = 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 .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 8 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)

1 + (g(x))2dx = tan−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)

1 + (g(x))2dx =

∫1

1 + u2du = tan−1 u + C = tan−1(g(x)) + C

Examples∫e−x

1 + e−2xdx = −

∫1

1 + u2du = − tan−1(e−x) + C .

Note that 1 + e−2x = 1 + (e−x)2 (complete the square). Letu = e−x . Then du = −e−xdx .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 8 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)

1 + (g(x))2dx = tan−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)

1 + (g(x))2dx =

∫1

1 + u2du = tan−1 u + C = tan−1(g(x)) + C

Examples∫e−x

1 + e−2xdx = −

∫1

1 + u2du = − tan−1(e−x) + C .

Note that 1 + e−2x = 1 + (e−x)2 (complete the square). Letu = e−x . Then du = −e−xdx .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 8 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)

1 + (g(x))2dx = tan−1(g(x)) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)

1 + (g(x))2dx =

∫1

1 + u2du = tan−1 u + C = tan−1(g(x)) + C

Examples∫e−x

1 + e−2xdx = −

∫1

1 + u2du = − tan−1(e−x) + C .

Note that 1 + e−2x = 1 + (e−x)2 (complete the square). Letu = e−x . Then du = −e−xdx .

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 8 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Quiz

Quiz

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)4

k, (b)

ln 2

k(c) − ln 2

k.

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 9 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Secant sec−1 x

y = sec x

domain:[0, 12π)∪

( 12π, π]

range:(−∞,−1]∪[1,∞)

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 10 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Secant sec−1 x

y = sec x

domain:[0, 12π)∪

( 12π, π]

range:(−∞,−1]∪[1,∞)

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 10 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Secant sec−1 x

y = sec x

domain:[0, 12π)∪

( 12π, π]

range:(−∞,−1]∪[1,∞)

Trigonometric Properties

sec(sec−1 x) = x csc(sec−1 x) =x√

x2 − 1

sin(sec−1 x) =

√x2 − 1

xcos(sec−1 x) =

1

x

tan(sec−1 x) =√

x2 − 1 cot(sec−1 x) =1√

x2 − 1

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 10 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Secant sec−1 x

y = sec x

domain:[0, 12π)∪

( 12π, π]

range:(−∞,−1]∪[1,∞)

Trigonometric Properties

sec(sec−1 x) = x csc(sec−1 x) =x√

x2 − 1

sin(sec−1 x) =

√x2 − 1

xcos(sec−1 x) =

1

x

tan(sec−1 x) =√

x2 − 1 cot(sec−1 x) =1√

x2 − 1

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 10 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Secant sec−1 x

y = sec x

domain:[0, 12π)∪

( 12π, π]

range:(−∞,−1]∪[1,∞)

Trigonometric Properties

sec(sec−1 x) = x csc(sec−1 x) =x√

x2 − 1

sin(sec−1 x) =

√x2 − 1

xcos(sec−1 x) =

1

x

tan(sec−1 x) =√

x2 − 1 cot(sec−1 x) =1√

x2 − 1

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 10 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Secant sec−1 x

y = sec x

domain:[0, 12π)∪

( 12π, π]

range:(−∞,−1]∪[1,∞)

Trigonometric Properties

sec(sec−1 x) = x csc(sec−1 x) =x√

x2 − 1

sin(sec−1 x) =

√x2 − 1

xcos(sec−1 x) =

1

x

tan(sec−1 x) =√

x2 − 1 cot(sec−1 x) =1√

x2 − 1

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 10 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Secant sec−1 x

y = sec x

domain:[0, 12π)∪

( 12π, π]

range:(−∞,−1]∪[1,∞)

Trigonometric Properties

sec(sec−1 x) = x csc(sec−1 x) =x√

x2 − 1

sin(sec−1 x) =

√x2 − 1

xcos(sec−1 x) =

1

x

tan(sec−1 x) =√

x2 − 1 cot(sec−1 x) =1√

x2 − 1

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 10 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Secant sec−1 x

y = sec x

domain:[0, 12π)∪

( 12π, π]

range:(−∞,−1]∪[1,∞)

Trigonometric Properties

sec(sec−1 x) = x csc(sec−1 x) =x√

x2 − 1

sin(sec−1 x) =

√x2 − 1

xcos(sec−1 x) =

1

x

tan(sec−1 x) =√

x2 − 1 cot(sec−1 x) =1√

x2 − 1

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 10 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Inverse Secant sec−1 x

y = sec x

domain:[0, 12π)∪

( 12π, π]

range:(−∞,−1]∪[1,∞)

Trigonometric Properties

sec(sec−1 x) = x csc(sec−1 x) =x√

x2 − 1

sin(sec−1 x) =

√x2 − 1

xcos(sec−1 x) =

1

x

tan(sec−1 x) =√

x2 − 1 cot(sec−1 x) =1√

x2 − 1

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 10 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Differentiation

Theorem

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

d

dxsec−1 u =

1

|u|√

u2 − 1

du

dx,

∫1

u√

u2 − 1du = sec−1 |u|+C

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 11 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Differentiation

Theorem

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

d

dxsec−1 u =

1

|u|√

u2 − 1

du

dx,

∫1

u√

u2 − 1du = sec−1 |u|+C

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 11 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Differentiation

Theorem

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

d

dxsec−1 u =

1

|u|√

u2 − 1

du

dx,

∫1

u√

u2 − 1du = sec−1 |u|+C

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 11 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Differentiation

Theorem

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

d

dxsec−1 u =

1

|u|√

u2 − 1

du

dx,

∫1

u√

u2 − 1du = sec−1 |u|+C

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 11 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)

g(x)√

(g(x))2 − 1dx = sec−1(|g(x)|) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)

g(x)√

(g(x))2 − 1dx =

∫1

u√

u2 − 1du = sec−1(|g(x)|) + C

Examples∫1

x√

x − 1dx = 2

∫1

u√

u2 − 1du =

1

2sec−1√x + C .

Note that x − 1 = (√

x)2 − 1. Let u =√

x . Then x = u2,dx = 2udu.

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 12 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)

g(x)√

(g(x))2 − 1dx = sec−1(|g(x)|) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)

g(x)√

(g(x))2 − 1dx =

∫1

u√

u2 − 1du = sec−1(|g(x)|) + C

Examples∫1

x√

x − 1dx = 2

∫1

u√

u2 − 1du =

1

2sec−1√x + C .

Note that x − 1 = (√

x)2 − 1. Let u =√

x . Then x = u2,dx = 2udu.

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 12 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)

g(x)√

(g(x))2 − 1dx = sec−1(|g(x)|) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)

g(x)√

(g(x))2 − 1dx =

∫1

u√

u2 − 1du = sec−1(|g(x)|) + C

Examples∫1

x√

x − 1dx = 2

∫1

u√

u2 − 1du =

1

2sec−1√x + C .

Note that x − 1 = (√

x)2 − 1. Let u =√

x . Then x = u2,dx = 2udu.

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 12 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)

g(x)√

(g(x))2 − 1dx = sec−1(|g(x)|) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)

g(x)√

(g(x))2 − 1dx =

∫1

u√

u2 − 1du = sec−1(|g(x)|) + C

Examples∫1

x√

x − 1dx = 2

∫1

u√

u2 − 1du =

1

2sec−1√x + C .

Note that x − 1 = (√

x)2 − 1. Let u =√

x . Then x = u2,dx = 2udu.

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 12 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Integration: u-Substitution

Theorem

∫g ′(x)

g(x)√

(g(x))2 − 1dx = sec−1(|g(x)|) + C

Proof

Let u = g(x). Then du = g ′(x) dx ,∫g ′(x)

g(x)√

(g(x))2 − 1dx =

∫1

u√

u2 − 1du = sec−1(|g(x)|) + C

Examples∫1

x√

x − 1dx = 2

∫1

u√

u2 − 1du =

1

2sec−1√x + C .

Note that x − 1 = (√

x)2 − 1. Let u =√

x . Then x = u2,dx = 2udu.

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 12 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Other Trigonometric Inverses

Other Trigonometric Inverses

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 13 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Other Trigonometric Inverses

Other Trigonometric Inverses

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 13 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Other Trigonometric Inverses

Other Trigonometric Inverses

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 13 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Other Trigonometric Inverses

Other Trigonometric Inverses

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 13 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Other Trigonometric Inverses

Other Trigonometric Inverses

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 13 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Other Trigonometric Inverses

Other Trigonometric Inverses

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 13 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Other Trigonometric Inverses

Other Trigonometric Inverses

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 13 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Differentiation

Theorem

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 14 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Differentiation

Theorem

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 14 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Differentiation

Theorem

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 14 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Differentiation

Theorem

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 14 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Inverse Sine Inverse Tangent Inverse Secant Other Trig Inverses

Quiz (cont.)

The value, at the end of the 4 years, of a principle of $100 investedat 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.

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 15 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Definition

Hyperbolic Sine and Cosine

Definition

sinh x =1

2

(ex − e−x

), cosh x =

1

2

(ex + e−x

)Theorem

d

dxsinh x = cosh,

d

dxcosh x = sinh,

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 16 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Definition

Identities

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 17 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Definition

Identities

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 17 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Definition

Identities

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 17 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Definition

Identities

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 17 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Definition

Identities

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 17 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Definition

Identities

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 17 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Definition

Identities

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

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 17 / 18

Inverse Trig Functions Hyperbolic Sine and Cosine Definition

Outline

Inverse Trig FunctionsInverse SineInverse TangentInverse SecantOther Trig Inverses

Hyperbolic Sine and CosineDefinition

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 6 January 31, 2008 18 / 18