# 1,5,9,13,...37 (hw link has all odds)

Transcription

1 February 8, 17 Goals: 1. Recognize trig functions and their integrals.. Learn trig identities useful for integration. 3. Understand which identities work and when. a) identities enable substitution by matching u to du. b) Half angle identities reduce even powers enabling direct integration or substitution. 4. Categorize trig function by powers. a) For sin m x and cos m x: i) If m is odd, rewrite as sinx sin (m 1) x or cosx cos (m 1) x and use pythagorean identity. ii) If m is even, rewrite using half angle identity. b) For sec m x and tan m x: i) If m is odd and tanx is a factor, rewrite as secx sec (m 1) x and use secx tanx as du. ii) If m is even, rewrite as sec x sec (m 1) x and use pythagorean identity. Study 7. # 1,5,9,13,...37 (hw link has all odds) Derivatives of Trig Functions d [sin x] = cos x d [tan x] = sec x d [sec x] = sec x tan x d [cos x] = sin x d [cot x] = csc x d [csc x] = csc x cot x Integrals of Trig Functions cos x = sin x + c sec x = tan x + c sec x tan x= sec x + c sin x = cos x + c csc x = cot x + c csc x cot x= csc x + c G. Battaly 17 1

2 February 8, 17 Trigonometric Identities sin x + cos x = 1 cot x + 1 = csc x Basic tan x = sin x / cos x cot x = cos x / sin x sec x = 1 / cos x csc x = 1 / sin x tan x = 1 / cot x Half angle (Power reducing) cos x = 1 + cos x sin x = 1 cos x Double angle sin x = sin x cos x cos x = cos x sin x = cos x 1 = 1 sin x cos 5 x sin x G. Battaly 17

3 February 8, 17 cos 5 x sin x cos 5 x ( )sin x u = cos x du = sin x u 5 du = u 6 + c = cos 6 x + c 6 6 easy! old stuff! sin 3 x G. Battaly 17 3

4 February 8, 17 sin 3 x sin x sinx (1 cos x) sinx sinx cos x sinx cos x ( ) cos x ( )sinx cos x + cos 3 x 3 keep one factor of sinx for the du sin x + cos x = 1 cot x + 1 = csc x u = cos x du = sin x cos + 1/3 (cos ) 3 [ cos + 1/3 (cos ) 3 ] + 1/3 () 3 [ 1 + 1/3 (1) 3 ] = ( /3 ) = /3 cos 5 x sin x sin x + cos x = 1 cot x + 1 = csc x G. Battaly 17 4

5 February 8, 17 cos 5 x sin x sin x cos 4 x cosx sin x (cos x) cosx sin x (1 sin x) cosx sin x (1 sin x+sin 4 x) cosx (sin x sin 4 x + sin 6 x) cosx sin x + cos x = 1 cot x + 1 = csc x sin x cosx sin 4 x cosx + sin 6 x cosx sin 3 x sin 5 x + sin 7 x + c even # of sinx factors, odd # of cosx keep one factor of cosx for the du sec (x 1) sin x + cos x = 1 cot x + 1 = csc x G. Battaly 17 5

6 February 8, 17 sec (x 1) sin x + cos x = 1 cot x + 1 = csc x ½ sec (x 1) ½ tan (x 1) + c old stuff! sec 4 (5x) sin x + cos x = 1 cot x + 1 = csc x G. Battaly 17 6

7 sec 4 (5x) sec (5x) sec (5x) keep sec x for the du [ tan (5x) + 1] sec (5x) sin x + cos x = 1 cot x + 1 = csc x February 8, 17 tan (5x) sec (5x) + sec (5x) u = tan 5x du = 5 sec 5x u = 5 x du = 5 1/5 tan (5x) 5sec (5x) + 1/5 sec (5x) 5 1/5 u du + 1/5 sec (u) du 1/5 u 3 + 1/5 tan (u) + c = 1/15 tan 3 5x + 1/5 tan (5x) + c 3 sin x sin x + cos x = 1 sin x = sin x sin x cot x + 1 = csc x (1 cos x) neither approach helps directly BUT, could try IP: u = sin x dv= sinx du=cosx v = cosx sin x = sinx cosx cosx cosx sin x = sinx cosx + cos x sin x = sinx cosx + (1 sin x) sin x = sinx cosx + x sin x sin x = sinx cosx + x sin x = ½ sinx cosx + ½ x + c G. Battaly 17 7

8 February 8, 17 sin x OR try half angle for simpler solution 1 cos x Half angle cos x = 1 + cos x sin x = 1 cos x ½ ½ cos x ½ x ½½ cos x ½ x ¼ sin x + c sin x = ½ sinx cosx + ½ x + c x sin x Half angle cos x = 1 + cos x sin x = 1 cos x G. Battaly 17 8

9 x sin x ½ x (1 cos x) Half angle cos x = 1 + cos x sin x = 1 cos x February 8, 17 ½ x ½ x cos x u = x dv= ½ cos x du= v = ½ sinx ½ ½ x ½ [½ x sin x ½ sinx ] ¼ x ¼ x sin x + ¼ ½ sinx ¼ x ¼ x sin x ¼ ½ cosx + c ¼ x ¼ x sin x 1/8 cosx + c Practice: ( sin θ) dθ tan x cos 3 x π/4 tan 4 x sin θ dθ cos 3 θ (tan x + tan 4 x) 1 θ G. Battaly 17 9

10 February 8, 17 Practice: ( sin θ) dθ tan x cos 3 x π/4 tan 4 x sin θ dθ cos 3 θ (tan x + tan 4 x) (9π 16)/4 1/3 sin 3 x + c π/4 /3 ½ tan θ + c 1/3 tan 3 x + c 1 θ Practice: ( sin θ) dθ (9π 16)/4 tan x cos 3 x 1/3 sin 3 x + c 1 θ G. Battaly 17 1

11 February 8, 17 Practice: π/4 tan 4 x π/4 /3 sin θ dθ cos 3 θ ½ tan θ + c 1 θ Practice: (tan x + tan 4 x) 1/3 tan 3 x + c G. Battaly 17 11