1. Lesson 7-3
Trig Substitution

2. Table of Trigonometric Substitutions
Expression
Substitution
Trig Identity
a² - x²
x = a sin θ
-π/2 ≤ θ ≤ π/2
1 - sin² θ = cos² θ
a² + x²
x = a tan θ
1 + tan² θ = sec² θ
x² - a²
x = a sec θ
0 ≤ θ ≤ π/2 or π ≤ θ ≤ 3π/2
sec²θ – 1 = tan² θ

3. Type 1: a²- x² Sub x = a sin θ and dx = a cos θ dθ
Square root reduces to a cos θ
Integrate
Sub back in x θ a x
a2 – x2

4. ∫ 4 - x² dx 7-3 Example 1 = ∫ 2cos² θ (2cos θ dθ)
x = 2 sin θ dx = 2 cos θ dθ
Use Trig id: sin² θ = 1 - cos² θ
= ∫ 2(cos θ) (2cos θ dθ)
= 4 ∫ cos² θ dθ
= 4( ½ θ + ¼ sin2θ ) + C
= 2θ + 2sin θ cos θ + C
Double Angle formula θ 2 x
4 – x2
= 2 sin-1 (x/2) + (x/2)4 – x2 + C 4

5. ∫ ∫ ∫ ∫ ∫ 7-3 Example 2 x2 dx (2sin θ)² 2 cos θ dθ -----------=
(4 – 4sin²θ)3/2 ∫
Let x = 2 sin θ
dx = 2 cos θ dθ
8 sin² θ cos θ dθ =
8 cos3θ ∫
sin² θ
= dθ = tan2 θ dθ
cos2θ ∫
Trig Reduction Formula
= tan2 θ dθ = tan θ - dθ ∫ θ 2 x
4 – x2
= tan θ - θ + C
x x
= sin C
4 – x 5

6. Type 2: a² + x² Sub x = a tan θ and dx = a sec² θ dθ
Square root reduces to a sec θ
Integrate trig function
Sub back in x θ a x
a2 – x2

7. ∫ ∫ ∫ ∫ 7-3 Example 3 dx (3sec² θ) dθ -----------=
(9 tan² θ + 9) ∫
Let x = 3 tan θ
dx = 3 sec² θ dθ
sec² θ dθ =
3 sec2 θ ∫ 1
= dθ 3 ∫
= (1/3) θ + C θ 3 x
9 – x2 x
= tan C 7

8. ∫ 4 + x² dx ∫ ∫ 7-3 Example 4 = ∫ 4 + 4tan² x (2 sec2 θ) dθ
Let x = 2 tan θ
and dx = 2 sec2 θ dθ
= 2 ∫ sec θ sec² θ dθ
= sec3-2 x tan x sec3-2 x dx ∫
Using Trig Reduction
Formula
= sec x tan x sec x dx ∫
From Table of Integrals
= ½ sec x tan x + ½ ln |sec x + tan x| + C θ 2 x
4 – x2
= ½(4+x² /2)(x/2) + ½ ln |(4+x² /2)+ (x/2)| + C 8

9. Type 3:  x² - a² Sub x = a sec θ and dx = a sec θ tan θ dθ
Square root reduces to a tan θ
Integrate trig function
Sub back in x θ a x
x2 – a2

10. ∫ ∫ ∫ ∫ ∫ 7-3 Example 5 dx 3 sec θ tan θ dθ ---------------- ==
x2 (x2 – 9)½
3 sec θ tan θ dθ
9 sec2 θ (9 sec2 θ – 9)½ ∫ ∫
let x = 3 sec θ and
dx = 3 sec θ tan θ dθ
3 sec θ tan θ dθ
9 sec2 θ 3 tan θ ∫ dθ
sec θ ∫ 1
cos θ dθ = (1/9) sin θ + C 9 ∫ θ 3 x
x2–9
(1/9) (x2 – 9)/x + C 10

11. ∫ ∫ ∫ ∫ ∫ 7-3 Example 6 (x2 – 16)½ (16sec² θ – 16)½ ----------- dx =
(4sec θ) (tan θ) dθ
4 sec θ ∫ ∫
let x = 4 sec θ and
dx = 4 sec θ tan θ dθ
4 tan θ (4sec θ) (tan θ) dθ
4 sec θ ∫
4 tan2 θ dθ ∫
4 (sec² θ - 1) dθ = (4) tan θ - 4θ + C ∫ θ 4 x
x2–16
4(x2 – 16)/4 – 4sec-1(x/4) + C 11

12. Summary & Homework Summary: Homework:
Trig Substitution can allow us to solve some hard integrals involving square roots
Basic steps the same (but different substitutions)
Substitute to eliminate square root
Evaluate the trigonometric integral
Convert back to original variable using triangle
All based on Geometric Right Triangle Trig Dfns
Homework:
pg , Day 1: 1, 2, 5, 9, Day 2: 3, 7, 11, 14, 17