Study 5.5, # 1 5, 9, 13 27, 35, 39, 49 59, 63, 69, 71, 81. Class Notes: Prof. G. Battaly, Westchester Community College, NY Homework.

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1 Goals: 1. Recognize an integrand that is the derivative of a composite function. 2. Generalize the Basic Integration Rules to include composite functions. 3. Use substitution to simplify the process of integration of composite functions. Study 5.5, # 1 5, 9, 13 27, 35, 39, 49 59, 63, 69, 71, Chain Rule: Derivative of Composite Functions Revisit, from 2.4: Chain Rule If: 1) y = f(u) is a differentiable function of u, and 2) u=g(x) is a differentiable function of x, then y = f(g(x)) is a differentiable function of x, and or 217 1

2 Use the Chain Rule G: y = (x 2 + 2) 1 F: dy/dx Use the Chain Rule G: y = (x 2 + 2) 1 F: dy/dx dy/dx = 1 (x 2 + 2) 9 (2x) dy/dx = 2x (x 2 + 2)

3 Remember the Chain Rule y = (x 2 + 2) 1 dy/dx = 2x (x 2 + 2) 9 y Complete the table. dy/dx y = (2x + 1) 3 y = sin(x 2 ) dy/dx = 2 cos(2x+1) y = tan(x 2 ) dy/dx = 3x 2 sec 2 (x 3 ) Remember the Chain Rule y = (x 2 + 2) 1 dy/dx = 2x (x 2 + 2) 9 y Complete the table. dy/dx y = (2x + 1) 3 y = sin(x 2 ) dy/dx = 6 (2x+1) 2 dy/dx = 2x cos(x 2 ) dy/dx = 2 cos(2x+1) y = tan(x 2 ) dy/dx = 2x sec 2 (x 2 ) dy/dx = 3x 2 sec 2 (x 3 ) 1. All derivatives here use the Chain Rule to find the derivative of composite functions

4 Remember the Chain Rule y = (x 2 + 2) 1 dy/dx = 2x (x 2 + 2) 9 y Complete the table. dy/dx y = (2x + 1) 3 y = sin(x 2 ) dy/dx = 6 (2x+1) 2 dy/dx = 2x cos(x 2 ) dy/dx = 2 cos(2x+1) y = tan(x 2 ) dy/dx = 2x sec 2 (x 2 ) dy/dx = 3x 2 sec 2 (x 3 ) Derivative of the primary function. Derivative of the nested function. Complete the table. Remember the Chain Rule y = (2x + 1) 3 y = sin(x 2 ) y = sin(2x+1) + c y = tan(x 2 ) y = tan(x 3 ) + c dy/dx = 6 (2x+1) 2 dy/dx = 2x cos(x 2 ) dy/dx = 2 cos(2x+1) dy/dx = 2x sec 2 (x 2 ) dy/dx = 3x 2 sec 2 (x 3 ) Derivative of the primary function. Derivative of the nested function. 2. To find the integrals of functions that are the derivatives of composite functions, the integrand requires the presence of the derivative of the nested function as a factor. This is the reverse process of the Chain Rule

5 (x 2 9) 3 (2x)dx u = du = 2x (x 2 9) 3 dx (x 2 9) 3 (2x)dx u 3 du u c (x 2 9) 4 + c 4 u = x 2 9 du = 2x dx u substitution 2x (x 2 9) 3 dx same as original 217 5

6 x 2 x 3 +1 dx u = du = 1 3 x 2 x 3 +1 dx u = x 3 +1 (x 3 +1) ½ 3x 2 dx du = 3x 2 dx 1 3 u½ du 1 3 u + c = (x3 +1) + c 3 2 3/2 3/

7 2 sec 2x tan 2x dx u = du = 2 sec 2x tan 2x dx u = 2x du = 2 dx sec u tan u du sec u + c sec 2x + c 217 7

8 sec 2x tan 2x dx u = du = sec 2x tan 2x dx u = 2x du = 2 dx ½ sec 2x tan 2x 2dx ½ sec u tan u du ½ sec u + c ½ sec 2x + c 217 8

9 sin x cos x dx u = du = sin x cos x dx u du u = sin x du = cos x dx u 2 + c 2 ½ (sin x) 2 + c 217 9

10 sin x cos x dx u = cos x du = sin x cos x dx sin x cos x dx u = cos x du = sin x dx u du u 2 + c 2 ½ (cos x) 2 + c 217 1

11 sin x cos x dx ½ 2sin x cos x dx ½ sin 2x dx Trig Identity: sin 2x = 2 sin x cos x u = 2x du = sin x cos x dx Trig Identity: sin 2x = 2 sin x cos x ½ 2sin x cos x dx ½ sin 2x dx ½½ sin 2x 2dx u = 2x du = 2 dx ¼ sin u du ¼ cos u + c ¼ cos 2x + c

12 sin x cos x dx Trig Identity: sin 2x = 2 sin x cos x 3 solutions look different, but they are equivalent, different only by constants ½ (sin x) 2 + c 1 ½ (cos x) 2 + c 2 ¼ cos 2x + c 3 k dx = kx + c u n du = u n+1 + c n+1 e u du = e u + c Rules of Integration du = ln u + c u sin u du = cos u + c cos u du = sin u + c sec u tan u du = sec u + c sec 2 u du = tan u + c

13 (x+1) 2 x dx u = du = (x+1) 2 x dx (x+1)(2 x) ½ ( ) dx u = 2 x du = dx Note: x+1 is neither part of u nor du, but we need to convert all x to u before continuing. The conversion of x+1 to a function of u must be consistent with u=2 x: u= 2 x; so x= 2 u, and x+1= 2 u+1 = 3 u (3 u)(u) ½ du = (3u ½ u 3/2 ) du = 3u 3/2 + u 5/2 + c 3/2 5/2 = 2(2 x) 3/2 + 2(2 x) 5/2 + c

14 # 8 x 2 e x 3 dx u = du = x 2 e x 3 dx u = x 3 1/3 e x 3 3x 2 dx du = 3x 2 dx 1/3 e u du e u + c = e 3 3 x 3 + c

15 # 32 x x dx u = du = # 32 ½ x dx u = x x x x ½ du u dx du = 2x dx ½ ln u + c ½ ln x c = ½ ln (x 2 + 4) + c

16 # 6 1 x e x 2 dx u = du = # 6 1 x e x 2 dx x ½ e 2 ( 2x) dx u = x 2 du = 2x dx 1 Limit x u= x 2 LL 1 ½ e u du UL 1 1 ½ e u du = ½ e u = ½ [e e 1 ] 1 1 = ½ [1 1/e] = e 1 2e

17 # 82 G: A bacteria population starts with 4 bacteria and grows at the rate in bacteria/hr as noted below. r(t) = (45.268)e t F: How many bacteria will there be after 3 hours? Amt after 3 hours = original amt ± net change # 82 G: A bacteria population starts with 4 bacteria and grows at the rate in bacteria/hr as noted below. r(t) = (45.268)e t F: How many bacteria will there be after 3 hours? Amt after 3 hours = original amt ± net change A in 3hr = 1 + net change 3 u = t (45.268)e t dt du = dt e t ( )dt e u du = e u = A in 3hr = = bacteria after 3 hours

18 # 6 1 x e x 2 dx x ½ e 2 ( 2x) dx u = x 2 du = 2x dx 1 Limit x u= x 2 LL 1 ½ e u du UL 1 1 ½ e u du = ½ e u = ½ [e e 1 ] 1 1 = ½ [1 1/e] = e 1 2e

19 217 19

20 Solve the diff eq: 217 2

21 Solve the diff eq: Solve the diff eq, and find the equations thru (2,7):