1 Inverse Trig Functions

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  • Lecture 6Section 7.7 Inverse Trigonometric Functions Section

    7.8 Hyperbolic Sine and Cosine

    Jiwen He

    1 Inverse Trig Functions

    1.1 Inverse Sine

    Inverse Since sin1 x (or arcsin x)

    1

  • domain:[ 12, 1

    2] range:[1, 1]

    2

  • sin(sin1 x) = x

    3

  • 4

  • domain:[1, 1] range:[ 12, 1

    2]

    Trigonometric Properties

    5

  • sin(sin1 x) = x cos(sin1 x) =

    1 x2

    tan(sin1 x) =x

    1 x2cot(sin1 x) =

    1 x2

    x

    sec(sin1 x) =1

    1 x2csc(sin1 x) =

    1x

    Differentiation

    Theorem 1.d

    dxsin1 x =

    11 x2

    .

    Proof.Let y = sin1 x. Then x = sin y,

    d

    dxsin1 x =

    1ddy sin y

    =1

    cos y=

    1cos(sin1 x)

    =1

    1 x2.

    Theorem 2.

    d

    dxsin1 u =

    11 u2

    du

    dx,

    1

    1 u2du = sin1 u + C

    Integration: u-Substitution

    6

  • Theorem 3. g(x)

    1 (g(x))2dx = sin1(g(x)) + C

    ProofLet u = g(x). Then du = g(x) dx,

    g(x)1 (g(x))2

    dx =

    11 u2

    du = sin1 u + C = sin1(g(x)) + C

    Examples 4.

    14 x2

    dx =

    11 u2

    du = sin1 u+C = sin1x

    2+C. Note

    that 4 x2 = 4(1

    (x2

    )2). Let u = x2 . Then du = 12dx. 12x x2 dx =1

    1 u2du = sin1 u+C = sin1(x 1)+C. Note that 2xx2 = 1 (x2

    2x + 1) = 1 (x 1)2 (complete the square). Let u = x 1. Then du = dx.

    1.2 Inverse Tangent

    Inverse Tangent tan1 x (or arctanx)

    7

  • y = tan x domain:( 12, 1

    2) range:(,)

    8

  • Trigonometric Properties

    tan(tan1 x) = x cot(tan1 x) =1x

    sin(tan1 x) =x

    1 + x2cos(tan1 x) =

    11 + x2

    sec(tan1 x) =

    1 + x2 csc(tan1 x) =

    1 + x2

    x

    Differentiation

    Theorem 5.d

    dxtan1 x =

    11 + x2

    .

    Proof.Let y = tan1 x. Then x = tan y,

    d

    dxtan1 x =

    1ddy tan y

    =1

    (sec y)2=

    1(sec(tan1 x)

    )2 = 11 + x2 .Theorem 6.

    d

    dxtan1 u =

    11 + u2

    du

    dx,

    1

    1 + u2du = tan1 u + C

    9

  • Integration: u-Substitution

    Theorem 7. g(x)

    1 + (g(x))2dx = tan1(g(x)) + C

    ProofLet u = g(x). Then du = g(x) dx,

    g(x)1 + (g(x))2

    dx =

    11 + u2

    du = tan1 u + C = tan1(g(x)) + C

    Examples 8.

    14 + x2

    dx =12

    1

    1 + u2du =

    12

    tan1 u + C =12

    tan1x

    2+ C.

    Note that 4+x2 = 4(1 +

    (x2

    )2). Let u = x2 . Then du = 12dx. 12 + 2x + x2 dx =1

    1 + u2du = tan1(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.

    ex

    1 + e2xdx =

    11 + u2

    du = tan1(ex) + C. Note that 1 + e2x = 1 + (ex)2 (complete

    the square). Let u = ex. Then du = exdx.

    Quiz

    Quiz

    10

  • Let f (t) = kf(t).

    1. For f(0) = 4, f(t) =: (a) kt + 4, (b) 4ekt, (c) 4ekt.

    2. For k > 0, double time T =: (a)4k

    , (b)ln 2k

    (c) ln 2k

    .

    1.3 Inverse Secant

    Inverse Secant sec1 x

    11

  • y = sec x domain:[0, 12)( 1

    2, ] range:(,1][1,)

    12

  • Trigonometric Properties

    sec(sec1 x) = x csc(sec1 x) =x

    x2 1

    sin(sec1 x) =

    x2 1x

    cos(sec1 x) =1x

    tan(sec1 x) =

    x2 1 cot(sec1 x) = 1x2 1

    Differentiation

    Theorem 9.d

    dxsec1 x =

    1|x|

    x2 1.

    Proof.Let y = sec1 x. Then x = sec y,

    d

    dxsec1 x =

    1ddy sec y

    =1

    (sec y tan y)2=

    1|x|

    x2 1.

    Theorem 10.

    d

    dxsec1 u =

    1|u|

    u2 1du

    dx,

    1

    u

    u2 1du = sec1 |u|+ C

    13

  • Integration: u-Substitution

    Theorem 11. g(x)

    g(x)

    (g(x))2 1dx = sec1(|g(x)|) + C

    ProofLet u = g(x). Then du = g(x) dx,

    g(x)g(x)

    (g(x))2 1

    dx =

    1u

    u2 1du = sec1(|g(x)|) + C

    Examples 12.

    1x

    x 1dx = 2

    1

    u

    u2 1du =

    12

    sec1

    x + C. Note that

    x 1 = (

    x)2 1. Let u =

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

    1.4 Other Trig Inverses

    Other Trigonometric Inverses

    Other Trigonometric Inverses

    14

  • sin1 x + cos1 x =

    2or cos1 x =

    2 sin1 x

    tan1 x + cot1 x =

    2or cot1 x =

    2 tan1 x

    sec1 x + csc1 x =

    2or csc1 x =

    2 sec1 x

    Differentiation

    Theorem 13.

    d

    dxcos1 x = d

    dxsin1 x = 1

    1 x2d

    dxcot1 x = d

    dxtan1 x = 1

    1 + x2d

    dxcsc1 x = d

    dxsec1 x = 1

    |x|

    x2 1

    Quiz (cont.)The value, at the end of the 4 years, of a principle of $100 invested at 4%

    compounded

    3. annually: (a) 400(1 + 0.04), (b) 100(1 + 0.04)4, (c) 100(1 + 0.16).

    4. continuously: (a) 100e0.04, (b) 100e0.16, (c) 100(1 + 0.04)4.

    2 Hyperbolic Sine and Cosine

    2.1 Definition

    Hyperbolic Sine and Cosine

    15

  • Definition 14.

    sinhx =12(ex ex

    ), coshx =

    12(ex + ex

    )Theorem 15.

    d

    dxsinhx = cosh,

    d

    dxcoshx = sinh,

    Identities

    16

  • 17

  • cosh2 x sinh2 x = 1sinh(x + y) = sinhx cosh y + coshx sinh ycosh(x + y) = coshx cosh y + sinhx sinh y

    cos2 x + sin2 x = 1sin(x + y) = sin x cos y + cos x sin ycos(x + y) = cos x cos y sinx sin y

    Outline

    Contents

    1 Inverse Trig Functions 1

    18

  • 1.1 Inverse Sine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Inverse Tangent . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 Inverse Secant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.4 Other Trig Inverses . . . . . . . . . . . . . . . . . . . . . . . . . . 14

    2 Hyperbolic Sine and Cosine 152.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

    19