integrals of inverse trig functions · integrals of inverse trig functions section 5.7 calculus...

INTEGRALS OF INVERSE TRIG FUNCTIONS

Section 5.7

Calculus AP/Dual, Revised ©2017

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12/6/2018 5:06 PM §5.7: Integrals of Inverse Trig Functions 1

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DERIVATIVE OF INVERSE TRIG FUNCTIONS (REVIEW)

A.𝒅

𝒅𝒙𝐬𝐢𝐧−𝟏 𝒖 =

𝒅𝒖

𝟏−𝒖𝟐

B.𝒅

𝒅𝒙𝐜𝐨𝐬−𝟏 𝒖 = −

𝒅𝒖

𝟏−𝒖𝟐

C.𝒅

𝒅𝒙𝐭𝐚𝐧−𝟏 𝒖 =

𝒅𝒖

𝟏+𝒖𝟐

D.𝒅

𝒅𝒙𝐜𝐨𝐭−𝟏 𝒖 = −

𝒅𝒖

𝟏+𝒖𝟐

E.𝒅

𝒅𝒙𝐬𝐞𝐜−𝟏 𝒖 =

𝒅𝒖

𝒖 𝒖𝟐−𝟏

F.𝒅

𝒅𝒙𝐜𝐬𝐜−𝟏 𝒖 = −

𝒅𝒖

𝒖 𝒖𝟐−𝟏


REVIEW

Solve 𝒇′ 𝒙 = 𝐚𝐫𝐜𝐬𝐢𝐧 𝒙𝟑


2

'arcsin

1

d uu

dx u

2

23

3arcsin

1

d xu

dxx

2

6

3

1

x

x

REVIEW

Solve 𝒇′ 𝒙 = 𝐚𝐫𝐜𝐜𝐬𝐜 𝟗𝒙


2

9arccsc

9 9 1

du

dx x x

2

'arccsc

1

d uu

dx u u

2

9

9 81 1x x

EQUATIONS

A. 𝒅𝒖

𝒂𝟐−𝒖𝟐= 𝐚𝐫𝐜𝐬𝐢𝐧

𝒖

𝒂+ 𝑪

B. −𝒅𝒖

𝒂𝟐−𝒖𝟐= 𝐚𝐫𝐜𝐜𝐨𝐬

𝒖

𝒂+ 𝑪

C. 𝒅𝒖

𝒂𝟐+𝒖𝟐=

𝟏

𝒂𝐚𝐫𝐜𝐭𝐚𝐧

𝒖

𝒂+ 𝑪

D. −𝒅𝒖

𝒂𝟐+𝒖𝟐=

𝟏

𝒂𝐚𝐫𝐜𝐜𝐨𝐭

𝒖

𝒂+ 𝑪

E. 𝒅𝒖

𝒖 𝒖𝟐−𝒂𝟐=

𝟏

𝒂𝐚𝐫𝐜𝐬𝐞𝐜

𝒖

𝒂+ 𝑪

F. −𝒅𝒖

𝒖 𝒖𝟐−𝒂𝟐=

𝟏

𝒂𝐚𝐫𝐜𝐜𝐬𝐜

𝒖

𝒂+ 𝑪


Number Minus Function

Number Function

Function Minus Number

Function Number

GRAPHIC ORGANIZER

𝐚𝐫𝐜 𝐬𝐢𝐧 𝐚𝐫𝐜 𝐜𝐨𝐬

𝐚𝐫𝐜 𝐭𝐚𝐧

𝐚𝐫𝐜 𝐜𝐨𝐭 𝐚𝐫𝐜 𝐬𝐞𝐜 𝐚𝐫𝐜 𝐜𝐬𝐜

𝒅𝒚

𝒅𝒙

න𝒇(𝒙)𝒅𝒙


2

'

1

u

u

2

'

1

u

u

2

'

1

u

u u

2 2

2 2

arcsin

arccos

du uC

aa u

du uC

aa u

2 2

2 2

1arctan

1arccot

du uC

a u a a

du uC

a u a a

2 2

2 2

1arcsec

1arccsc

uduC

a au u a

uduC

a au u a

Notice any patterns?

STEPS

A. Identify which equation does the rule apply

B. Identify the 𝒂 and the 𝒖 of the equation

1.The 𝒂 is the number

2.The 𝒖 is the function

C. Take the derivative 𝒅𝒖 of the equation

D. Simplify


EXAMPLE 1

Solve 𝟏

𝟏−𝟏𝟔𝒙𝟐𝒅𝒙


2

1

1 16dx

x

2 2arcsin

du uC

aa u

2 1

1

a

a

2 216

4

4

1

4

u x

u x

du dx

du dx

2 2

1

4

du

a u

2 2

2 2

2 2

2 2

2 2

2 2

arcsin

arccos

1arctan

1arccot

1arcsec

1arccsc

du uC

aa u

du uC

aa u

du uC

a u a a

du uC

a u a a

uduC

a au u a

uduC

a au u a

EXAMPLE 1

Solve 𝟏

𝟏−𝟏𝟔𝒙𝟐𝒅𝒙


2

1 1

4 1 16dx

x

2 1

1

a

a

2 216

4

4

1

4

u x

u x

du dx

du dx

41arcsin

4 1

xC

1arcsin 4

4x C

EXAMPLE 2

Solve 𝟑

𝟏−𝟒𝒙𝟐𝒅𝒙


2

3

1 4dx

x

2 2arcsin

du uC

aa u

2 1

1

a

a

2 24

2

2

1

2

u x

u x

du dx

du dx

2 2

13

2

du

a u

2 2

2 2

2 2

2 2

2 2

2 2

arcsin

arccos

1arctan

1arccot

1arcsec

1arccsc

du uC

aa u

du uC

aa u

du uC

a u a a

du uC

a u a a

uduC

a au u a

uduC

a au u a

EXAMPLE 2

Solve 𝟑

𝟏−𝟒𝒙𝟐𝒅𝒙


23arcsin

2 1

xC

2 2

3

2

du

a u

3arcsin 2

2x C

2 1

1

a

a

2 24

2

2

1

2

u x

u x

du dx

du dx

EXAMPLE 3

Solve 𝒅𝒙

𝟐+𝟗𝒙𝟐


2 2

2

a

a

2 29

3

3

1

3

u x

u x

du dx

du dx

22 9

dx

x

2 2

1arctan

du uC

a u a a

2 2

2 2

2 2

2 2

2 2

2 2

arcsin

arccos

1arctan

1arccot

1arcsec

1arccsc

du uC

aa u

du uC

aa u

du uC

a u a a

du uC

a u a a

uduC

a au u a

uduC

a au u a

EXAMPLE 3

Solve 𝒅𝒙

𝟐+𝟗𝒙𝟐


2 2

2

a

a

2 29

3

3

1

3

u x

u x

du dx

du dx

1 1 3arctan

3 2 2

xC

2

1 1arctan

3 2 9

dx uC

x a a

1 3arctan

3 2 2

xC

EXAMPLE 4

Solve 𝟏

𝟒+ 𝒙−𝟏 𝟐𝒅𝒙


1 1arctan

2 2

xC

YOUR TURN

Solve 𝟓𝒅𝒙

𝟗+𝟒𝒙𝟐


5 2arctan

6 3

xC

EXAMPLE 5

Solve 𝒅𝒙

𝒙 𝒙𝟒−𝟒


2 4

2

a

a

2 4

2

2

1

2

u x

u x

du xdx

du xdx

4 4

dx

x x

2 2

1arcsec

uduC

a au u a

2 2

2 2

2 2

2 2

2 2

2 2

arcsin

arccos

1arctan

1arccot

1arcsec

1arccsc

du uC

aa u

du uC

aa u

du uC

a u a a

du uC

a u a a

uduC

a au u a

uduC

a au u a

21arcsec

4 2

xC

1 1

2du dx

x

Can't take out a 1/x as a coefficient

EXAMPLE 6

Solve 𝒅𝒙

𝒙 𝟏𝟔𝒙𝟐−𝟏


41arcsec

1 1

xC

2 2

1arcsec

uduC

a au u a

2 2

1 1arcsec

44 1

4

udxC

u a ax

2 1

1

a

a

2 216

4

1

4

u x

u x

du dx

4

4

u x

ux

2 2

1 14 arcsec

4 4 1

udxC

a ax x

EXAMPLE 7

Solve 𝒅𝒙

𝒆𝟐𝒙−𝟏


2 1x

dx

e

1a

x

x

x

u e

du e dx

dudx

e

2 2 2

1x x

dx dudx dx

u u ae e

2

1x x

du

e e

pay attention to the order on the denominator

arcsec equation

EXAMPLE 7

Solve 𝒅𝒙

𝒆𝟐𝒙−𝟏


2 2 2

1x x

dx du

u u ae e

1sec

uarc C

a a

arcsec xe C

1a

1

x

x

x

u e

du e dx

du dxe

EXAMPLE 8

Solve 𝟎ൗ𝟑 𝟐 𝟏

𝟏+𝟒𝒙𝟐𝒅𝒙


32

20

1

1 4x

1a

2

2

1

2

u x

du dx

du dx

3

0

1 1arctan

2

u

a a

3

2 20

1

2

du

a u

0 0

33

2

u

u

EXAMPLE 8

Solve 𝟎ൗ𝟑 𝟐 𝟏

𝟏+𝟒𝒙𝟐𝒅𝒙


3

0

1 1arctan

2 1 1

u

10

2 3

1arctan 3 arctan 0

2

6

1a

2

2

1

2

u x

du dx

du dx

0 0

33

2

u

u

REVIEW

Use Completing the Square to write in Vertex form, 𝒙𝟐 − 𝟒𝒙 + 𝟕 = 𝟎


2 2

2 4 44 7

2 2x x

2 4 4 7 4x x

2

2 3y x

2 4 7x x

2

2 3x

EXAMPLE 9

Solve 𝒅𝒙

𝒙𝟐−𝟒𝒙+𝟕


2 4 7x x

2 2

2 2

2 2

2 2

2 2

2 2

arcsin

arccos

1arctan

1arccot

1arcsec

1arccsc

du uC

aa u

du uC

aa u

du uC

a u a a

du uC

a u a a

uduC

a au u a

uduC

a au u a

Completing The Square

2 2

2 4 44 7

2 2x x

2

2 3

dx

x

2

22

3

3

2

2

a

a

u x

u x

1 2arctan

3 3

xC

2 4 7 0x x

EXAMPLE 10

Solve 𝒅𝒙

𝟐𝒙𝟐−𝟖𝒙+𝟏𝟎


2

2

2 8 10

2 4 5

x x

x x

2 2

2 2

2 2

2 2

2 2

2 2

arcsin

arccos

1arctan

1arccot

1arcsec

1arccsc

du uC

aa u

du uC

aa u

du uC

a u a a

du uC

a u a a

uduC

a au u a

uduC

a au u a

Completing The Square

2

1

2 4 5

dx

x x

2

22

1

1

2

2

a

a

u x

u x

2

1

2 4 5

dx

x x

2 2

2

1

2 4 44 5

2 2

dx

x x

2

1

2 2 1

dx

x

1 2arctan

2 1

xC

YOUR TURN

Solve 𝒅𝒙

𝒙𝟐+𝟒𝒙+𝟏𝟑


1 2arctan

3 3

xC

AP MULTIPLE CHOICE PRACTICE QUESTION 1 (NON-CALCULATOR)

Solve 𝟏

𝟏𝟔+𝒙𝟐𝒅𝒙

(A) 𝟏

𝟏𝟔𝐚𝐫𝐜𝐭𝐚𝐧

𝒙

𝟒+ 𝑪

(B) 𝟏

𝟒𝐚𝐫𝐜𝐭𝐚𝐧

𝒙

𝟏𝟔+𝑪

(C) 𝟏

𝟒𝐚𝐫𝐜𝐭𝐚𝐧

𝒙

𝟒+ 𝑪

(D) 𝟏

𝟏𝟔𝐚𝐫𝐜𝐭𝐚𝐧

𝒙

𝟏𝟔+ 𝑪


Vocabulary Connections and Process Answer and Justifications

AP MULTIPLE CHOICE PRACTICE QUESTION 1 (NON-CALCULATOR)

Solve 𝟏

𝟏𝟔+𝒙𝟐𝒅𝒙


CIntegration of

Inverse Functions2 2

1 du

a u2 2u x

u x

du dx

2 2

1 1arctan

uC

a u a a

2 2

1 1arctan

4 44

xC

x

2 16

4

a

a

ASSIGNMENT

Page 380

1-17 odd, 21-29 EOO, 33-35 all


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