Lecture 6 - Section 7.7 Inverse Trigonometric Functions Section 7.8

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