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Warm-up: 2 4 4 2 dx x x 2 4 4 2 x x 2 4 4 11 x x 2 2 1 1 x 2 2 1 1 dx x Let 2 1 u x 2 du dx 2 1 2 1 du u 1 1 tan 2 u C 1 1 tan 2 1 2 x C 1 2 du dx

# Warm-up:. 7.4 Trigonometric Substitutions Integrals involving Inverse Trig Functions When listing the Antiderivative that corresponds to each of the

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### Text of Warm-up:. 7.4 Trigonometric Substitutions Integrals involving Inverse Trig Functions When listing...

24 4 2

dx

x x Warm-up:

24 4 2x x 24 4 1 1x x

22 1 1x

22 1 1

dx

x

Let 2 1u x 2 du dx

2

1

2 1

du

u

11 tan

2u C

11 tan 2 1

2x C

1

2du dx

7.4 Trigonometric Substitutions

Identities you need to know for this section:

2 2

2 2

2 2

Pythagorean Identities:

sin cos 1

tan 1 sec

1 cot csc

2 2

2

2

2

Double Angle Identities:

sin 2 2sin cos

cos 2 cos sin

1 2sin

2cos 1

2 tantan 2

1 tan

Integrals involving Inverse Trig Functions

2 2arcsin

du uC

aa u

2 2

1arctan

du uC

a u a a

2 2

1arcsec

uduC

a au u a

When listing the Antiderivative that corresponds to each of the inverse trigonometric functions, only use one member from each pair.

a is the number.u is the variable.

We can use right triangles and the pythagorean theorem to simplify some problems.

a

x

2 2a x1

24

dx

x

These are in the same form.

2

24 x

24sec

2

x

22sec 4 x

tan2

x

2 tan x

22sec d dx

Write a 'simple' relationship

that involves xKeeping Pyth. Ident.

in mind, write

another 'simple'

relationship

for the other

terms in problem.

We can use right triangles and the pythagorean theorem to simplify some problems.

a

x

2 2a x1

24

dx

x

These are in the same form.

2

24 x

24sec

2

x

22sec 4 x

tan2

x

2 tan x

22sec d dx

22sec

2sec

d

sec d ln sec tan C

24ln

2 2

x xC

We can use right triangles and the pythagorean theorem to simplify some problems.

1

24

dx

x

22sec

2sec

d

sec d ln sec tan C

24ln

2 2

x xC

24ln

2

x xC

2ln 4 ln 2x x C This is a constant.

2ln 4 x x C

a

x

2 2a x

This method is called Trigonometric Substitution.

If the integral contains ,

we use the triangle at right.

2 2a x

If we need , we

move a to the hypotenuse.

2 2a x If we need , we

move x to the hypotenuse.

2 2x a

a

x

2 2a x

a

x2 2x a

2 2

29

x dx

x 3

x

29 x

sin3

x

3sin x

3cos d dx

29cos

3

x

23cos 9 x 29sin 3cos

3cos

d

1 cos 29

2d

91 cos 2

2d

9 9 1sin 2

2 2 2C

sin3

x

1sin3

x

19 9sin 2sin cos

2 3 4

xC

double angle formula

219 9 9

sin2 3 2 3 3

x x xC

2 2

29

x dx

x 3

x

29 x

sin3

x

3sin x

29cos

3

x

23cos 9 x

sin3

x

1sin3

x

19 9sin 2sin cos

2 3 4

xC

double angle formula

219 9 9

sin2 3 2 3 3

x x xC

1 29sin 9

2 3 2

x xx C

3cos d dx

522

dx

x x We can get into the necessary

form by completing the square.

22x x

22x x

2 2 x x

2 2 1 1x x

21 1x

21 1x

21 1

dx

x

Let 1u x

du dx

21

du

u 1

u

21 ucos

cos

d

d C 1 sin u C

1 sin 1x C

21cos

1

u 21 u

sin u cos d du

624 4 2

dx

x x Complete the square:24 4 2x x

24 4 1 1x x

22 1 1x

22 1 1

dx

x

Let 2 1u x 2 du dx

2

1

2 1

du

u

1

u

2 1u

2

2

1 sec

2 sec

d

1

2d

1

2C 11

tan2

u C

11 tan 2 1

2x C

tan u

2sec 1u 1

2du dx

2sec d du

2 2sec 1u

Here are a couple of shortcuts that are result from Trigonometric Substitution:

12 2

1tan

du uC

u a a a

1

2 2sin

du uC

aa u

These are on your list of formulas. They are not really new.

p

HW Day 1: p. 512 #’s 1-4, 5-17 odd, 41-45 odd

In Class/HW Day 2: p. 512 #’s 19-37 odd, 47, 49 odd

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