M1120 Class 7 Dan Barbasch September 13, 2011 - PowerPoint PPT Presentation

M1120 Class 7 Dan Barbasch September 13, 2011 http://www.math.cornell.edu/ ˜ web1120/index.html Dan Barbasch () M1120 Class 7 September 13, 2011 1 / 26

Integration by Parts � � Basic Formula: u dv = uv − vdu . d ( uv ) = udv + vdu ⇔ udv = d ( uv ) − vdu ⇔ � � � udv = d ( uv ) − vdu . Dan Barbasch () M1120 Class 7 September 13, 2011 2 / 26

x sec 2 x dx . � Dan Barbasch () M1120 Class 7 September 13, 2011 3 / 26

x sec 2 x dx . � Example x sec 2 x dx set u = x and dv = sec 2 x dx . Then � For dv = sec 2 x dx u = x du = dx v = tan x . The integration by parts formula gives � � x sec 2 x dx = x tan x − tan x dx = x tan x − ln | cos x | + C . Dan Barbasch () M1120 Class 7 September 13, 2011 4 / 26

Further Examples � (a) x sin x dx . xe 3 x dx . � (b) x 4 ln x dx . � (c) Dan Barbasch () M1120 Class 7 September 13, 2011 5 / 26

x 4 ln x dx � Dan Barbasch () M1120 Class 7 September 13, 2011 6 / 26

Example (c) Set dv = x 4 dx u = ln x v = x 5 du = 1 x dx 5 . So � 1 5 dx = x 5 ln x x 4 dx = x 5 ln x x 4 ln x dx = x 5 x 5 − x 5 − 1 � � 5 ln x − 25+ C . 5 5 5 x Dan Barbasch () M1120 Class 7 September 13, 2011 7 / 26

Harder Examples e ax sin ( bx ) dx . � (d) � (e) ln x dx . Dan Barbasch () M1120 Class 7 September 13, 2011 8 / 26

e x sin x dx � Dan Barbasch () M1120 Class 7 September 13, 2011 9 / 26

e 2 x cos x dx � u = e 2 x dv = cos x dx du = 2 e 2 x dx v = sin x . Then � � � e 2 x cos x dx = e 2 x sin x − (sin x )(2 e 2 x dx ) = e 2 x sin x − 2 e 2 x sin x dx . For the second integral, u = e 2 x dv = sin x dx du = 2 e 2 x dx v = − cos x . So � � e 2 x sin x dx = − e 2 x cos x − ( − cos x )(2 e 2 x dx ) = � = − e 2 x cos x + 2 e 2 x cos x dx . Dan Barbasch () M1120 Class 7 September 13, 2011 10 / 26

e 2 x cos x dx � Plugging into the first equation, � � � � e 2 x cos x dx = e 2 x sin x − 2 − e 2 x cos x + 2 e 2 x cos x dx = � e 2 x sin x + 2 e 2 x cos x e 2 x cos x dx . � � = − 4 This is an equation, the integral we want appears on both sides. So we move it to the left and solve: � e 2 x cos x dx = e 2 x sin x + 2 e 2 x cos x (1 + 4) e 2 x cos x dx = e 2 x sin x + 2 e 2 x cos x � + C . 5 Dan Barbasch () M1120 Class 7 September 13, 2011 10 / 26

Reduction formulas The expression to be integrated depends on some integer n . We write the integral in terms of similar integrals, but for smaller n . sec n x dx . � (f) x n e x dx . � (g) Dan Barbasch () M1120 Class 7 September 13, 2011 11 / 26

Example (f) Dan Barbasch () M1120 Class 7 September 13, 2011 12 / 26

Example (f) u = sec n − 2 x dv = sec 2 x dx du = ( n − 2) sec n − 3 x sec x tan x dx v = tan x . � � sec n x dx = sec n − 2 x tan x − ( n − 2) sec n − 2 x tan 2 x dx = � = sec n − 2 x tan x − ( n − 2) sec n − 2 x − 1 + sec 2 x � � dx = � � = sec n − 2 x tan x + ( n − 2) sec n − 2 x dx − ( n − 2) sec n x dx sec n x dx to the other side of the equation: � Now move the integral � � sec 2 x dx = sec n − 2 x tan x + ( n − 2) sec n − 2 x dx (1 + n − 2) sec n x dx = sec n − 2 x tan x � + n − 2 � sec n − 2 x dx . n − 1 n − 1 Dan Barbasch () M1120 Class 7 September 13, 2011 13 / 26

Trigonometric integrals � sin m x cos n x dx , � sin ax cos bx dx . sin 4 x cos 2 x dx . � Example. The other type will arise as we compute the example. Dan Barbasch () M1120 Class 7 September 13, 2011 14 / 26

sin 4 x · cos 2 x dx . � Double Angle Formulas: sin 2 θ = 1 − cos 2 θ cos 2 θ = 1 + cos 2 θ 2 2 � 2 � � 1 − cos 2 x � 1 + cos 2 x � � � � 2 cos 2 x dx = sin 2 x · dx . 2 2 The powers have decreased, but instead the angles have doubled. Dan Barbasch () M1120 Class 7 September 13, 2011 15 / 26

sin 4 x cos 2 x dx � � sin 4 x cos 2 x dx = 1 � � 1 − 2 cos 2 x + cos 2 2 x � (1 + cos 2 x ) dx = 8 =1 � � 1 − cos 2 x − cos 2 2 x + cos 3 2 x � dx = 8 8 − 1 cos 2 x dx − 1 cos 2 2 x dx + 1 = x � � � cos 3 2 x dx . 8 8 8 Then � cos 2 x dx = 1 2 sin 2 x + C , and � � � 1 + cos 4 x � dx = 1 2 x + 1 cos 2 2 x dx = 8 sin 4 x + C . 2 � cos 3 2 x dx . Remains to compute Dan Barbasch () M1120 Class 7 September 13, 2011 16 / 26

cos 3 2 x dx . � Change variables u = 2 x , du = 2 dx : � � 2 = 1 � cos 3 u du cos 3 2 x dx = cos 3 u du . 2 Dan Barbasch () M1120 Class 7 September 13, 2011 17 / 26

cos 3 x dx , odd powers � Dan Barbasch () M1120 Class 7 September 13, 2011 18 / 26

cos 3 x dx , odd powers � Apply the identity sin 2 u + cos 2 u = 1 : � � � � cos 3 u du = cos 2 u · cos u du = 1 − sin 2 u � · (cos u du ) . Change variables w = sin u so dw = cos u du : 3 + C = sin u − sin 3 u dw = w − w 3 � � � cos 3 u du = 1 − w 2 � + C = 3 = sin 2 x − sin 3 2 x + C . 3 Dan Barbasch () M1120 Class 7 September 13, 2011 18 / 26

General Strategy 1 When at least one of sin x or cos x appears to an odd power say sin x , change variables u = cos θ , and use sin 2 θ + cos 2 θ = 1. This will convert � sin n x cos m x dx to the integral of a polynomial, assuming n , m are positive integers. 2 For even powers, half angle formulas (or reduction formulas) reduce the powers. Dan Barbasch () M1120 Class 7 September 13, 2011 19 / 26

cos 3 x dx , products of cosines � cos 3 x = cos x · cos 2 x = cos x · 1 + cos 2 x = 1 2 cos x + 1 2 cos x · cos 2 x , 2 So � cos 3 x dx = 1 � cos x dx + 1 � cos x cos 2 x dx = 2 2 =1 2 sin x + 1 � cos x · cos 2 x dx . 2 We need to compute � cos x · cos 2 x dx . Dan Barbasch () M1120 Class 7 September 13, 2011 20 / 26

Trigonometry Formulas (1) cos ( u + v ) = cos u cos v − sin u sin v . (2) cos ( u − v ) = cos u cos v + sin u sin v . (3) sin ( u + v ) = sin u cos v + cos u sin v . (4) sin ( u − v ) = sin u cos v − cos u sin v . Solving, (5) cos u cos v = 1 2 (cos ( u − v ) + cos ( u + v )) . (6) sin u sin v = 1 2 (cos ( u − v ) − cos ( u + v )) . (7) sin u cos v = 1 2 (sin ( u + v ) + sin ( u − v )) . Dan Barbasch () M1120 Class 7 September 13, 2011 21 / 26

� cos x cos 2 x dx cos u cos v = 1 2 (cos ( u + v ) + cos ( u − v )) . Dan Barbasch () M1120 Class 7 September 13, 2011 22 / 26

� cos x · cos 2 x dx cos u cos v = 1 2 cos( u + v ) + 1 2 cos( u − v ) . � cos x · cos 2 x dx = 1 � cos( x − 2 x ) dx + 1 � cos( x + 2 x ) dx = 2 2 =1 � cos( − x ) dx + 1 � cos 3 x dx = 1 2 sin x + 1 6 sin 3 x + C . 2 2 We used the relation cos( − x ) = cos x . Plug this answer into the formula on a previous page. Question: This answer does not seem to coincide with the previous one.Why?! Dan Barbasch () M1120 Class 7 September 13, 2011 23 / 26

cos 3 x dx , recursion formula � Dan Barbasch () M1120 Class 7 September 13, 2011 24 / 26

cos 3 x dx , recursion formula � Use integration by parts: u = cos 2 x dv = cos x dx du = 2 cos x ( − sin x dx ) v = sin x � � cos 3 x = sin x cos 2 x − ( − 2 sin x cos x ) sin x dx = � = sin x cos 2 x + 2 sin 2 x cos x dx = � � = sin x cos 2 x + 2 1 − cos 2 x � cos x dx = � � = sin x cos 2 x + 2 cos 3 dx = cos x dx − 2 � = sin x cos 2 x + 2 sin x − 2 cos 3 dx . Dan Barbasch () M1120 Class 7 September 13, 2011 24 / 26

cos 3 x dx , recursion formula � cos n x dx . Solving for � This is the derivation of the recursion formula for cos 3 x dx , � � � cos 3 x dx = sin x cos 2 x + 2 sin x − 2 cos 3 dx . � cos 3 x dx = sin x cos 2 x + 2 sin x + C . 3 cos 3 x dx = sin x cos 2 x + 2 sin x � + C . 3 Dan Barbasch () M1120 Class 7 September 13, 2011 24 / 26

sin 4 x cos 2 x dx � Remark: There are a lot of steps and calculations, the answer is supposed to be 16 x − 1 1 64 sin 2 x − 1 1 64 sin 4 x + 192 sin 6 x + C . With this in mind, recall sin 2 x = 2 sin x cos x . So we can write � 2 � sin 2 x � 1 − cos 2 x � sin 4 x cos 2 x = (sin x cos x ) 2 · sin 2 x = · = 2 2 � 1 − cos 4 x � =1 · (1 − cos 2 x ) = 8 2 = 1 16 (1 − cos 2 x − cos 4 x + cos 2 x cos 4 x ) . Now use cos 2 x cos 4 x = 1 2 cos( − 2 x ) + 1 2 cos 6 x to get the answer above. Check the arithmetic carefully. Dan Barbasch () M1120 Class 7 September 13, 2011 25 / 26

Exercises for next time � � � sin 5 x cos x dx sin 4 x cos 3 x dx sin 8 x cos 7 x dx ( a ) ( b ) ( c ) � � � sin 2 x dx sin 4 x dx sin 8 x cos 2 x dx ( d ) ( e ) ( f ) � 2 π � ( g ) sin 5 x cos 2 x dx ( h ) sin 2 x sin 4 x dx 0 Dan Barbasch () M1120 Class 7 September 13, 2011 26 / 26