Trigonometric Substitution

Lesson 8.4

New Patterns for the Integrand

• Now we will look for a different set of patterns

• And we will use them in the context of a right triangle

• Draw and label the other two triangles which show the relationships of a and x 2

2 2 2 2 2 2a x a x x a

a

x

2 2a x

Example

• Given

• Consider the labeled triangle Let x = 3 tan θ (Why?) And dx = 3 sec2 θ dθ

• Then we have

3

2 9

dx

x 3

x

2 23 xθ

2

2

3sec

9 tan 9

d

Use identity

tan2x + 1 = sec2x

Use identitytan2x + 1 = sec2x

23sec3 sec ln sec tan

3sec

dd C

Finishing Up

• Our results are in terms of θ We must un-substitute back into x

Use the triangle relationships

4

ln sec tan C 3

x

2 23 xθ

29ln

3 3

x xC

Knowing Which Substitution

5

u

u

2 2u a

Try It!!

• For each problem, identify which substitution and which triangle should be used

6

3 2 9x x dx

2

2

1 xdx

x

2 2 5x x dx

24 1x dx

Keep Going!

• Now finish the integration

7

3 2 9x x dx

2

2

1 xdx

x

2 2 5x x dx

24 1x dx

Application

• Find the arc length of the portion of the parabola y = 10x – x2 that is above the x-axis

• Recall the arc length formula

8

21 '( )b

i

a

L f x dx

Special Integration Formulas

• Useful formulas from Theorem 8.2

• Look for these patterns and plug in thea2 and u2 found in your particular integral

2 2 2 2

2 2 2 2 2 2 2

2 2 2 2 2 2 2

11. arcsin

2

12. ln ,

21

3. ln2

ua u du a u a u C

a

u a du u u a a u u a C u a

a u du u u a a u u a C

Assignment

• Lesson 8.4

• Page 550

• Exercises 1 – 45 EOOAlso 67, 69, 73, and 77

10