lecture 6 - section 7.7 inverse trigonometric functions section 7… · 2008-02-05 · inverse trig...
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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