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
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