Download inverse trigonometric functions

INVERSE TRIGONOMETRIC FUNCTIONS
y = cos −1 x
π
y = sin− 1 x
π
2
y = tan−1 x
π
2
π
2
−
π
2
−
π
2
y = sec −1 x
y = csc −1 x
3π
2
π
π
π
2
π
2
3π
2
y = cot − 1 x
π
π
2
The graphs of the inverse trigonometric functions are shown above. There has been disagreement about the graphs of
the inverses of the cosecant, secant and cotangent functions. Our textbook accepts the graphs as shown above.
Know the following formulas.
f (x)
Additional formulas are given below.
f′( x )
f ′( x)
1
sin
−1
u
cos− 1 u
du
1 − u2 dx
1
du
−
1 − u2 dx
f (x)
csc
−1
−
u
cot − 1 u
1
du
u u − 1 dx
1 du
−
1 + u2 dx
2
1
tan
−1
u
sec − 1 u
du
1 + u dx
1
du
2
u u − 1 dx
2
PROBLEM SET
Find f ′ ( x ) if f ( x ) is the given expression.
f (x)
1. sin−1 ( 4x )
ANSWER
4
1.
1 − 16x 2
⎛ 3⎞
2. tan− 1 ⎜ e x ⎟
⎝
⎠
2.
3. cos−1 ( 6x )
3. −
( )
4. sec − 1 x 5
5. sin−1 ( tan x )
6. tan− 1 ( cos3x )
⎛2⎞
7. cos− 1 ⎜ ⎟
⎝x⎠
8. sec − 1 x
3x2 e x
3
1 + e2x
4.
6
1 − 36x 2
5
x x10 − 1
sec 2 x
5.
1 − tan2 x
6. −
3 sin ( 3x )
1 + cos2 ( 3x )
2
7.
x
8.
3
x2 − 4
1
2x x − 1
THE HYPERBOLIC FUNCTIONS
In this handout we will study special combinations of e x and e − x . These combinations are called “hyperbolic
functions.” These functions, which arise in certain engineering applications, have many properties in common
with the trigonometric functions. The name “hyperbolic” comes from the fact that the point ( cosh t, sinh t ) is on the
hyperbola x 2 − y2 = 1.
Know the following definitions.
sinh x =
Hyperbolic Sine
Hyperbolic Cosine
Hyperbolic Tangent
Hyperbolic Cosecant
Hyperbolic Secant
cosh x =
x
e −e
2
x
Know the starred identities.
sinh ( x + y ) = sinh x cosh y + cosh x sinh y
* cosh x + sinh x = e x
−x
e +e
2
−x
* cosh x − sinh x = e− x
sinh ( x − y ) = sinh x cosh y − cosh x sinh y
* coth2 x − 1 = csc h2 x
cosh ( − x ) = cosh x
sinh2x = 2sinh x cosh x
* cosh2 x − sinh2 x = 1 cosh ( x + y ) = cosh x cosh y + sinh x sinh y
*1 − tanh2 x = sec h2 x cosh ( x − y ) = cosh x cosh y − sinh x sinh y
sinh x e x − e− x
=
cosh x e x + e− x
1
2
csc h x =
=
sinh x e x − e− x
1
2
sec h x =
=
x
cosh x e + e− x
tanh x =
sinh ( − x ) = − sinh x
tanh ( − x ) = − tanh x
cosh2x = cosh2 x + sinh2 x
cosh2x = 2sinh2 x + 1
cosh2x = 2cosh2 x − 1
y = cosh x
y = sinh x
cosh x e x + e− x
coth x =
=
Hyperbolic Cotangent
sinh x e x − e− x
Know the following formulas.
f (x)
f ′( x )
f (x)
f ′( x )
sinh ( u )
du
cosh ( u )
dx
du
sinh ( u )
dx
du
2
sec h ( u )
dx
csc h ( u )
− csc h ( u ) coth ( u )
cosh ( u )
tanh ( u )
sec h ( u )
coth ( u )
y = tanh x
du
dx
du
− sec h ( u ) tanh ( u )
dx
du
− csc h2 ( u )
dx
PROBLEM SET
1. Prove the starred identities listed in the chart of identities above.
2. Find f ′ ( x ) if f ( x ) is the given expression.
f (x)
Answer
1. sinh7x
1. 7cosh7x
2. cosh x5
2. 5x 4 sinh x5
3. sinh2 ( 4x )
3. 8 sinh 4x cosh 4x or 4 sinh8x
4. cosh ( sinh ( x ) )
4. sinh ( sinh x ) cosh x
5. sinh5x cosh5x
5. 5 sinh2 5x + cosh2 5x
6. sec h 4x
6. − 4 sec h 4x tanh 4x
7. esec h3x
7. − 3esec h3x sec h3x tanh3x
( )
(
( )
)
or 5cosh10x
( )
8. sec h x 4
8. − 4x 3 sec h x 4 tanh x 4
9. tanh 8x
9. 8 sec h2 8x
10. tanh3 7x
10. 21tanh2 7x sec h2 7x
4
11.
5
tanh x
11.
−
1
( tanh x ) 5 sec h2 x
5
12. coth8x
12. − 8csc h2 8x
13. coth6 4x
13. − 24coth5 4x csc h2 4x
14. csc h5x7
14. − 35x6 csc h5x7 coth5x7
15.
16.
sinh x
x
csc h x
e
4x
15.
16.
x cosh x − sinh x
x2
− csc h x ( coth x + 4 )
e 4x
INDEFINITE INTEGRALS THAT YIELD INVERSE TRIGONOMETRIC FUNCTIONS
Assume that a > 0 in the following formulas.
du
∫
1 − u2
du
∫ 1 + u2
∫
= sin − 1 u + C
∫
u u2 − 1
a 2 − u2
du
= tan − 1 u + C
du
du
∫ a2 + u2
= sec − 1 u + C
∫
=
= sin − 1
u
+C
a
1
u
tan − 1 + C
a
a
du
=
u u2 − a 2
1
u
sec − 1 + C
a
a
INDEFINITE INTEGRALS OF HYPERBOLIC FUNCTIONS
∫ sinhudu = cosh u + C
∫ csch udu = − coth u + C
∫ coshudu = sinh u + C
∫ sechu tanhu du = − sech u + C
∫ sech udu = tanh u + C
∫ cschu cothu du = − csch u + C
2
2
INVERSE HYPERBOLIC FUNCTIONS
y = sinh − 1x
y = tanh − 1x
y = cosh − 1x
DERIVATIVES OF INVERSE HYPERBOLIC FUNCTIONS
f (x)
sinh−1u
cosh−1u
tanh−1u
f (x)
f′( x )
du
u + 1 dx
1
du
, u >1
2
u − 1 dx
csch−1u
1
coth−1u
1
2
sech−1u
du
, u <1
1 − u dx
2
f ′( x )
−1
2
u 1+ u
du
, u ≠0
dx
−1
du
, 0 <u <1
u 1 − u2 dx
1 du
, u >1
1 − u2 dx
LOGARITHMIC IDENTITIES
⎛1
1 + x2
csch−1x = ln ⎜ +
⎜x
x
⎝
sinh−1x = ln ⎛⎜ x + x 2 + 1 ⎞⎟
⎝
⎠
⎛1
1 − x2
cosh−1x = ln ⎛⎜ x + x 2 − 1 ⎞⎟ , x ≥ 1 sech−1x = ln ⎜ +
⎜x
x
⎝
⎠
⎝
1 ⎛ 1+ x ⎞
1 ⎛ x + 1⎞
tanh−1x = ln ⎜
, x <1
coth−1x = ln ⎜
, x
⎟
2 ⎝ 1− x ⎠
2 ⎝ x − 1 ⎟⎠
⎞
⎟, x ≠ 0
⎟
⎠
⎞
⎟, 0 < x ≤ 1
⎟
⎠
>1
INDEFINITE INTEGRALS THAT YIELD INVERSE HYPERBOLIC FUNCTIONS
∫
∫
du
2
u +a
2
du
2
u −a
2
= sinh−1
u
+ C = ln ⎛⎜ u + u2 + a2
a
⎝
= cosh−1
⎞+C
⎟
⎠
u
+ C, u > a > 0; = ln ⎛⎜ u + u2 − a2
a
⎝
⎞ + C, u > a > 0
⎟
⎠
⎧1
1
a+u
−1 u
⎪ tanh a + C, u < a; = 2a ln a − u + C, u ≠ a
⎪a
∫ a2 − u2 = ⎨ 1 −1 u
1
a+u
⎪ coth
ln
+ C, u > a; =
+ C, u ≠ a
⎪⎩ a
a
2a a − u
du
INVERSE TRIGONOMETRIC FUNCTIONS AND COMPLETING THE SQUARE
KNOW ENTIRE PAGE! (I will give you the integration formulas.)
EXAMPLE 1
dx
∫ x2 − 4x + 7
dx
=
∫ x2 − 4x +
=
∫
=
∫
(x
4 − 4 +7
dx
2
)
− 4x + 4 + 3
dx
( x − 2)
2
( 3)
+
2
du
=
∫ u2 + a2 , where u = x − 2 and a =
=
1
u
1
x−2
tan− 1 + C =
tan− 1
+C
a
a
3
3
3
PROBLEMS: Find the following indefinite integrals.
dx
1.
∫ x2 − 8x + 21
2.
∫ x2 + 10x + 29
dx
=
∫
=
=
dx
− x 2 + 6x
(
∫
=
∫
5
)
dx
(
∫
=
∫
=
)
2
dx
(
9 − x 2 − 6x + 9
)
dx
(3)
2
− ( x − 3)
du
2
a −u
= sin− 1
1
x+5
tan− 1
+C
2
2
2
2
, where a = 3 and u = x − 3
u
x−3
+ C = sin− 1
+C
a
3
=
∫
∫
=
∫
=
∫
dx
− x 2 + 12x + 13
dx
(
)
− x 2 − 12x + 13
dx
(
2
)
− x − 12x + 36 + 36 + 13
(
dx
49 − x 2 − 12x + 36
∫
2.
∫
dx
8x − x
2
dx
11 + 10x − x
2
)
dx
(7)
2
− ( x − 6)
du
2
a − u2
= sin− 1
2
, where a = 7 and u = x − 6
u
x−6
+ C = sin− 1
+C
a
7
PROBLEMS: Find the following indefinite integrals.
1.
+C
13 + 12x − x 2
=
− x − 6x + 9 + 9
∫
=
dx
− x 2 − 6x
∫
x−4
dx
∫
6x − x 2
∫
tan− 1
EXAMPLE 3
dx
=
5
Answer:
EXAMPLE 2
∫
1
Answer:
Answer: sin− 1
x−4
+C
4
Answer: sin− 1
x−5
+C
6