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 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 θ a x a2 – x2

∫ 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

∫ ∫ ∫ ∫ ∫ 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-1 --- + C 4 – x2 2 5

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

∫ ∫ ∫ ∫ 7-3 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 1 x = ---- tan-1 --- + C 3 3 7

∫ 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 1 3 - 2 = ------- sec3-2 x tan x + ------- sec3-2 x dx 3 - 1 3 - 1 ∫ Using Trig Reduction Formula 1 1 = --- sec x tan x + ----- sec x dx 2 2 ∫ 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

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

∫ ∫ ∫ ∫ ∫ 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 ∫ 1 dθ --- ----------- 9 sec θ ∫ 1 --- cos θ dθ = (1/9) sin θ + C 9 ∫ θ 3 x x2–9 (1/9) (x2 – 9)/x + C 10

∫ ∫ ∫ ∫ ∫ 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

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 488-489, Day 1: 1, 2, 5, 9, 10 Day 2: 3, 7, 11, 14, 17