Upload
others
View
2
Download
0
Embed Size (px)
Citation preview
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
[email protected]://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