Lesson 7 2 Hard Trig Integrals Strategies for

Lesson 7 -2 Hard Trig Integrals

Strategies for Hard Trig Integrals Expression Substitution Trig Identity sinn x or cosn x, n is odd Keep one sin x or cos x or for du Convert remainder using Trig ID sin² x + cos² x = 1 sinn x or cosn x, n is even Use half angle formulas: sin² x = ½(1 – cos 2 x) cos² x = ½(1 + cos 2 x) sinm x • cosn x, n or m is odd From odd power, keep one sin x or cos x, for du Use identities to substitute sin² x + cos² x = 1 sinm x • cosn x, n & m are even Use half angle identities sin² x = ½(1 – cos 2 x) cos² x = ½(1 + cos 2 x) tann x or cotn x From power pull out tan 2 x or cot 2 x and substitute using Trig ID cot 2 x = csc 2 x - 1 or tan 2 x = sec 2 x – 1 tanm x • secn x or cotm x • cscn x , where n is even Pull out sec 2 x or csc 2 x for du Convert rest to tan or cot using the Trig ID sec² θ – 1 = tan² θ

Type I: sinn x or cosn x, n is odd • Keep one sin x or cos x or for du • Convert remainder with sin² x + cos² x = 1 • Using U substitution to get power rules

7 -2 Example 1 ∫ sin³ x dx = ∫ sin² x (sin x dx) Remove one sin x and combine with dx to form du Use Trig id: sin² x = 1 - cos² x to get the uⁿ du form = ∫ (1 – cos² x) (sin x dx) = ∫ sin x dx – ∫ cos² x (sin x dx) ) = ∫ sin x dx – (- 1) ∫ u² Let u = cos x then du = -sin x = - cos x + ⅓ cos³ x + C du

7 -2 Example 2 ∫ cos 5 xdx Remove one cos x and combine with dx to form du Use Trig id: sin² x = 1 - cos² x to get the uⁿ du form let u = sin x and du = cos x dx mostly un du form ∫ = (cos x dx) - 2 ∫ cos = 4 x (cos x dx) ∫ = (1 - sin 2 x)2 (cos x dx) ∫ = (1 - 2 sin 2 x + sin 4 x) (cos x dx) ∫ sin 2 x (cos x dx) + = sin x – 2/3 sin 3 x + 1/5 sin 5 x + C ∫ sin 4 x (cos x dx)

Type 2: sinn x or cosn x, n is even • Use half angle formulas: – sin² x = ½(1 - cos 2 x) – cos² x = ½(1 + cos 2 x) • Use form of cos u du

7 -2 Example 3 ∫ sin² x dx = ∫ ½ (1 - cos 2 x) dx Use double angle formulas: ∫ ∫ cos 2 x dx Sin 2 x = ½(1 – cos 2 x) =½ Then use u = 2 x and du = 2 dx, so you need an extra ½ out front = ½ x - ½(½ sin 2 x) + C dx - ½ = (¼) (2 x – sin 2 x) + C

7 -2 Example 4 ∫ cos 4 x dx = ∫ (½(1 + cos 2 x))² Use double angle formulas: cos 2 x = ½(1 + cos 2 x) Twice on last term! Then use cos u du forms ∫ = ¼ (1 + 2 cos 2 x + cos 2 2 x) dx = ¼( ∫ dx + 2 ∫ cos 2 x dx + ∫ cos 2 2 x dx) = ¼( ∫ dx + 2 ∫ cos 2 x dx + ∫ ½(1 + cos 4 x) dx) = ¼x + ¼sin 2 x + (1/8)(1/4) sin 4 x + C = (3/8)x + ¼sin 2 x + (1/32) sin 4 x + C = ¼ (sin x cos³ x) + (3/8) sin x cos x + 3/8(x) + C Note: Calculators will use other trig IDs to simplify into a different form

Type 3: sinm x • cosn x, n or m is odd • From odd power, keep one sin x or cos x, for du • Use identities to substitute – Convert remainder with sin² x + cos² x = 1 • With U-substitutions, use power rule

7 -2 Example 5 ∫ sin³ x cos 4 let u = cos x and du = -sin x dx ∫ (1 – cos² x) (cos x dx = = -1 =- ∫ (cos ∫u 4 4 4 x) (sin x) dx = ∫ x) (-sin x) dx – (- (cos 6 x) (-sin x) dx) ∫ du + u 6 du = (-1/5) u 5 + (1/7) u 7 + C = (-1/5) (cos 5 x) + (1/7) (cos 7 x) + C

Type IV: sinm x • cosn x, n and m are even. • Use half angle identities – sin² x = ½(1 - cos 2 x) – cos² x = ½(1 + cos 2 x)

7 -2 Example 7 ∫ sin² x cos² x dx ∫ = (1/2) (1 – cos 2 x) (1/2) (1 + cos 2 x) dx Use ½ angle formulas ∫ = (1/4) (1 – cos 2 2 x) dx Have to use ½ angle formula again ∫ ∫ = (1/4) dx – (1/4) (1/2)(1 - cos 4 x) dx = (1/4) x – (1/8) x + (1/8) ∫ cos 4 x dx = (1/8) x + (1/32) sin 4 x + C (not similar to calculator answer!)

Type V: tann x or cotn x • From power pull out tan 2 x or cot 2 x and substitute cot 2 x = csc 2 x - 1 or tan 2 x = sec 2 x – 1 • Sometimes it converts directly into usubstitution and the power rule; other times, this may have to be repeated several times

7 -2 Example 7 ∫ cot 4 x dx ∫ = cot 2 x (csc 2 x – 1) dx Use trig id to convert cot 2 ∫ ∫ = cot 2 x (csc 2 x) dx – cot 2 x dx First is a u-sub power rule and second, we reapply step 1 =- ∫ u² du - ∫ (csc 2 x – 1) dx = (-1/3)(cot 3 x) + cot x + C

7 -2 Example 8 ∫ tan 5 x dx ∫ = tan 3 x (sec 2 x – 1) dx Use trig id to convert cot 2 ∫ ∫ = tan 3 x (sec 2 x) dx – tan 3 x dx First is a u-sub power rule and second, we reapply step 1 = ∫u 3 du - tan x(sec 2 x – 1) dx ∫ = ∫u 3 du - u du + tan x dx ∫ ∫ = (1/4)(tan 4 x) - (1/2)tan 2 x - ln |cos x| + C

Type VI: tanm x • secn x or cotm x • cscn x , where n is even • Pull out sec 2 x or csc 2 x for du • Convert rest using trig ids: – csc 2 x = cot 2 x + 1 – sec 2 x = tan 2 x + 1 • Use u-substitution and power rules

7 -2 Example 9 ∫ tan -3/2 x sec 4 x dx Keep a sec 2 for du and convert other using trig id ∫ = (tan-3/2 x) (tan 2 + 1) (sec 2 x) dx ∫ = (tan 1/2 x + tan-3/2 ) (sec 2 x) dx = ∫u 1/2 du + ∫u -3/2 du = (2/3)u 3/2 – (2) u-1/2 + C = (2/3)tan 3/2 x – (2)tan-1/2 x + C

Trigonometric Reduction Formulas Expression ∫ sinn x dx ∫ cosn x dx ∫ tann x dx ∫ secn x dx Reduction Formula 1 n – 1 n-2 x dx = - --- sinn-1 x cos x + ------- sin ∫ n 1 n – 1 n-2 x dx = --- cosn-1 x sin x + ------- cos ∫ n 1 n-2 x dx = ------- tann-1 x - tan ∫ n - 1 n - 2 n-2 x dx = ------- secn-2 x tan x + ------- sec ∫ n - 1 n - 1 Remember the following integrals: (when n=1 in the above) ∫ tan x dx = ln |sec x| + C ∫ sec x dx = ln |sec x + tan x| + C

7 -2 Example 10 ∫ sin² x dx ∫ = -(1/2) sin x cos x + (1/2) dx = Using reduction formulas = (-1/2) sin x cos x + (1/2) x + C Use your calculator to check. Calculator uses the reduction formulas.

7 -2 Example 11 ∫ tan 5 x dx ∫ = (1/5 -1)tan 5 -1 x + tan 5 -2 x dx Use trig reduction formula ∫ = (1/4) tan 4 x + tan 3 x dx Use trig reduction formula again ∫ = (1/4) tan 4 x + (1/3 -1)tan 3 -1 x + tan 3 -2 x dx ∫ = (1/4) tan 4 x + (1/2)tan 2 x + tan x dx = (1/4) tan 4 x + (1/2)tan 2 x + ln|sec x| + C

Summary & Homework • Summary: – Hard Trig integrals can be solved • Homework: – pg 488 -489, Day 1: 1, 2, 5, 9, 10 Day 2: 3, 7, 11, 14, 17