1. Lesson 7-2 Hard Trig Integrals

2. Strategies for Hard Trig Integrals ExpressionSubstitutionTrig Identity sin n x or cos n x, n is odd Keep one sin x or cos x or for du Convert remainder using Trig ID sin² x + cos² x = 1 sin n x or cos n x, n is even Use half angle formulas: sin² x = ½(1 – cos 2x) cos² x = ½(1 + cos 2x) sin m x cos n 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 sin m x cos n x, n & m are even Use half angle identities sin² x = ½(1 – cos 2x) cos² x = ½(1 + cos 2x) tan n x or cot n 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 tan m x sec n x or cot m x csc n 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

3. 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 sin² x + cos² x = 1 Using U substitution to get power rules

4. 7-2 Example 1 ∫ sin³ 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 = ∫ sin² x (sin x dx) = ∫ (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 ∫ cos 5 xdx let u = sin x and du = cos x dx mostly u n du form = ∫ cos 4 x (cos x dx) = ∫ (cos x dx) - 2 ∫ sin 2 x (cos x dx) + ∫ sin 4 x (cos x dx) = sin x – 2/3 sin 3 x + 1/5 sin 5 x + C = ∫ (1- sin 2 x) 2 (cos x dx) = ∫ (1- 2sin 2 x + sin 4 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

6. Type 2: sin n x or cos n x, 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 Use double angle formulas: Sin 2 x = ½(1 – cos 2x) Then use u = 2x and du = 2dx, so you need an extra ½ out front = ∫ ½ (1 - cos 2x) dx = ½ x - ½(½ sin 2x) + C = (¼) (2x – sin 2x) + C = ½ ∫ dx - ½ ∫ cos 2x dx

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

9. Type 3: sin m x cos n 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

10. 7-2 Example 5 ∫ sin³ x cos 4 x dx = let u = cos x and du = -sin x dx ∫ (1 – cos² x) (cos 4 x) (sin x) dx = = -1 ∫ (cos 4 x) (-sin x) dx – (- ∫ (cos 6 x) (-sin x) dx) = (-1/5) u 5 + (1/7) u 7 + C = ( -1 /5) ( cos 5 x) + (1/7) (cos 7 x) + C = - ∫ u 4 du + ∫ u 6 du

11. Type IV: sin m x cos n 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 ∫ sin² x cos² x dx Use ½ angle formulas = ∫ (1/2) (1 – cos 2x) (1/2) (1 + cos 2x) dx Have to use ½ angle formula again = (1/4) ∫ (1 – cos 2 2x) 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!) = (1/4) ∫ dx – (1/4) ∫ (1/2)(1 - cos 4x) dx

13. Type V: tan n x or cot n 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 u- substitution and the power rule; other times, this may have to be repeated several times

14. 7-2 Example 7 ∫ cot 4 x dx Use trig id to convert cot 2 = ∫ cot 2 x (csc 2 x – 1) dx First  is a u-sub power rule and second, we reapply step 1 = ∫ cot 2 x (csc 2 x) dx – ∫ cot 2 x dx = - ∫ u² du - ∫ (csc 2 x – 1) dx = (-1/3)(cot 3 x) + cot x + x + C

15. 7-2 Example 8 Use trig id to convert cot 2 = ∫ tan 3 x (sec 2 x – 1) dx First  is a u-sub power rule and second, we reapply step 1 = ∫ tan 3 x (sec 2 x) dx – ∫ tan 3 x dx = ∫ 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 ∫ tan 5 x dx

16. Type VI: tan m x sec n x or cot m x csc n 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

17. 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

18. Trigonometric Reduction Formulas ExpressionReduction Formula ∫ sin n x dx 1 n – 1 = - --- sin n-1 x cos x + ------- sin n-2 x dx n n ∫ cos n x dx 1 n – 1 = --- cos n-1 x sin x + ------- cos n-2 x dx n n ∫ tan n x dx 1 = ------- tan n-1 x - tan n-2 x dx n - 1 ∫ sec n x dx 1 n - 2 = ------- sec n-2 x tan x + ------- sec n-2 x dx 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

19. 7-2 Example 10 ∫ sin² x dx Using reduction formulas = -(1/2) sin x cos x + (1/2) ∫ dx = = (-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 Use trig reduction formula = (1/5-1)tan 5-1 x + ∫ tan 5-2 x dx Use trig reduction formula again = (1/4) tan 4 x + ∫ tan 3 x dx ∫ tan 5 x dx = (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

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