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Lesson 7-2

Lesson 7-2. Hard Trig Integrals. Strategies for Hard Trig Integrals. Type I: sin n x or cos n x , n is odd. Keep one sin x or cos x or for du Convert remainder with sin ² x + cos² x = 1 Using U substitution to get power rules. 7-2 Example 1. ∫ sin³ x dx.

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Lesson 7-2

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  1. Lesson 7-2 Hard Trig Integrals

  2. Strategies for Hard Trig Integrals

  3. Type I: sinnx or cosnx, n is odd • Keep one sin x or cos x or for du • Convert remainder with sin²x + cos² x = 1 • Using U substitution to get power rules

  4. 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 du = - cos x + ⅓ cos³ x + C Let u = cos x then du = -sin x

  5. 7-2 Example 2 ∫ cos5 xdx = ∫ cos4 x (cos x dx) 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 =∫(1- sin2 x)2 (cos x dx) =∫(1- 2sin2 x + sin4 x) (cos x dx) let u = sin x and du = cos x dx mostly un du form = ∫(cos x dx) - 2 ∫ sin2 x (cos x dx) + ∫ sin4 x (cos x dx) = sin x – 2/3 sin3 x + 1/5 sin5 x + C

  6. Type 2: sinnx or cosnx, n is even • Use half angle formulas:  • sin² x = ½(1 - cos 2x) • cos² x = ½(1 + cos 2x) • Use form of cos u du

  7. 7-2 Example 3 ∫ sin² x dx = ∫ ½ (1 - cos 2x) dx Use double angle formulas: Sin2 x = ½(1 – cos 2x) Then use u = 2x and du = 2dx, so you need an extra ½ out front = ½∫ dx - ½∫ cos 2x dx = ½ x - ½(½ sin 2x)+ C = (¼) (2x – sin 2x) + C

  8. 7-2 Example 4 = ∫ cos4 x dx = ∫ (½(1 + cos 2x))² ∫ cos4 x dx Use double angle formulas: cos2 x = ½(1 + cos 2x) Twice on last term! Then use cos u du forms = ¼ ∫ (1 + 2cos 2x + cos2 2x)dx = ¼( ∫ dx + 2 ∫ cos 2x dx + ∫ cos2 2x dx) = ¼( ∫ dx + 2 ∫ cos 2x dx + ∫ ½(1 + cos 4x) dx) = ¼x + ¼sin 2x + (1/8)x + (1/8)(1/4) sin 4x + C = (3/8)x + ¼sin 2x + (1/32) sin 4x + 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

  9. Type 3:sinmx • cosnx, n or m is odd • From odd power, keep one sin x or cosx, for du • Use identities to substitute • Convert remainder with sin²x + cos² x = 1 • With U-substitutions, use power rule

  10. 7-2 Example 5 ∫ sin³ x cos4 x dx = ∫ (1 – cos² x) (cos4 x) (sin x)dx = = -1 ∫ (cos4 x) (-sin x) dx – (- ∫(cos6 x) (-sin x) dx) let u = cos x and du = -sin x dx = - ∫ u4 du + ∫u6 du = (-1/5) u5 + (1/7) u7 + C = (-1/5) (cos5 x) + (1/7) (cos7 x) + C

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

  12. 7-2 Example 7 = ∫ (1/2) (1 – cos 2x) (1/2) (1 + cos 2x) dx ∫ sin² x cos² x dx Use ½ angle formulas = (1/4) ∫ (1 – cos2 2x) dx Have to use ½ angle formula again = (1/4) ∫ dx – (1/4)∫ (1/2)(1 - cos 4x) dx = (1/4) x – (1/8) x + (1/8) ∫ cos 4x dx = (1/8) x + (1/32) sin 4x + C (not similar to calculator answer!)

  13. Type V: tann x or cotn x • From power pull out tan2x or cot2 x and substitute cot2x = csc2x - 1or tan2x = sec2x – 1 • Sometimes it converts directly into u-substitution and the power rule;other times, this may have to be repeated several times

  14. 7-2 Example 7 = ∫ cot2 x (csc2 x – 1)dx ∫ cot4 x dx Use trig id to convert cot2 = ∫ cot2 x (csc2 x) dx – ∫ cot2 x dx First  is a u-sub power rule and second, we reapply step 1 = - ∫ u²du - ∫ (csc2 x – 1)dx = (-1/3)(cot3 x) + cot x + x + C

  15. 7-2 Example 8 ∫ tan5 x dx = ∫ tan3 x (sec2 x – 1)dx Use trig id to convert cot2 = ∫ tan3 x (sec2 x) dx – ∫ tan3 x dx First  is a u-sub power rule and second, we reapply step 1 = ∫ u3du - ∫ tan x(sec2 x – 1)dx = ∫ u3du - ∫ u du + ∫tan x dx = (1/4)(tan4 x) - (1/2)tan2 x - ln |cos x| + C

  16. Type VI:tanmx•secnx or cotmx •cscnx , where n is even • Pull out sec2x or csc2x for du • Convert rest using trig ids: • csc2x = cot2x+ 1 • sec2x = tan2x+ 1 • Use u-substitution and power rules

  17. 7-2 Example 9 = ∫ (tan-3/2 x) (tan2 + 1) (sec2 x) dx ∫ tan-3/2 x sec4 x dx = ∫ (tan1/2 x + tan-3/2 ) (sec2 x) dx Keep a sec2 for du and convert other using trig id = ∫ u1/2 du + ∫ u-3/2 du = (2/3)u3/2 – (2) u-1/2 + C = (2/3)tan3/2 x – (2)tan-1/2 x + C

  18. Trigonometric Reduction Formulas ∫ ∫ ∫ ∫ 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

  19. 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.

  20. 7-2 Example 11 ∫ tan5 x dx = (1/5-1)tan5-1 x + ∫ tan5-2 x dx Use trig reduction formula = (1/4) tan4 x + ∫ tan3 x dx Use trig reduction formula again = (1/4) tan4 x + (1/3-1)tan3-1 x + ∫ tan3-2 x dx = (1/4) tan4 x + (1/2)tan2 x + ∫ tan x dx = (1/4) tan4 x + (1/2)tan2 x + ln|sec x| + C

  21. 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

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