Lesson 7 3 Trig Substitution Table of Trigonometric

Lesson 7 -3 Trig Substitution

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 -π/2 ≤ θ ≤ π/2 1 + tan² θ = sec² θ x² - a² x = a sec θ 0 ≤ θ ≤ π/2 or π ≤ θ ≤ 3π/2 sec²θ – 1 = tan² θ

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 a x θ a 2 – x 2

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

7 -3 Example 2 ∫ x 2 dx -----(4 - x 2)3/2 Let x = 2 sin θ dx = 2 cos θ dθ θ x 4 – x 2 ∫ 8 sin² θ cos θ dθ --------------8 cos 3θ = ∫ sin² θ -----dθ = cos 2θ = ∫tan = Trig Reduction Formula 2 ∫ (2 sin θ)² 2 cos θ dθ --------------(4 – 4 sin²θ)3/2 = 2 ∫ tan 2 θ dθ ∫ θ dθ = tan θ - θ + C x x -1 = ------ - sin --- + C 4 – x 2 2

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 a 2 – x 2 x θ a

7 -3 Example 4 ∫ 4 + x² dx Let x = 2 tan θ and dx = 2 sec 2 θ dθ = ∫ 4 + 4 tan² x (2 sec 2 θ) dθ ∫ = 2 sec θ sec² θ dθ Using Trig Reduction Formula = 1 3 -2 ------- sec x tan x + ------3 -1 ∫ sec 3 -2 x dx 1 1 = --- sec x tan x + ----- sec x dx 2 2 = ½ sec x tan x + ½ ln |sec x + tan x| + C ∫ From Table of Integrals = 4 – x 2 θ x 2 ½( 4+x² /2)(x/2) + ½ ln |( 4+x² /2)+ (x/2)| + C

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 x x 2 – a 2 θ a

7 -3 Example 5 ∫ dx -------- = x 2 (x 2 – 9)½ ∫ let x = 3 sec θ and dx = 3 sec θ tan θ dθ 1 --9 x θ x 2– 9 3 3 sec θ tan θ dθ -------------9 sec 2 θ (9 sec 2 θ – 9)½ ∫ 3 sec θ tan θ dθ -------------9 sec 2 θ 3 tan θ ∫ dθ -----sec θ ∫ cos θ dθ = (1/9) sin θ + C (1/9) (x 2 – 9)/x + C

7 -3 Example 6 ∫ (x 2 – 16)½ ------ dx = x ∫ (16 sec² θ – 16)½ ------------ (4 sec θ) (tan θ) dθ 4 sec θ ∫ 4 tan θ (4 sec θ) (tan θ) ---------------- dθ 4 sec θ 4 ∫ tan 2 θ dθ 4 ∫ (sec² θ let x = 4 sec θ and dx = 4 sec θ tan θ dθ x θ x 2– 16 4 - 1) dθ = (4) tan θ - 4θ + C 4 (x 2 – 16)/4 – 4 sec-1(x/4) + C

Summary & Homework • Summary: – 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 488 -489, Day 1: 1, 2, 5, 9, 10 Day 2: 3, 7, 11, 14, 17