Science One Math. Jan

Transcription

1 Science One Math Jan

2 Announcements Midterm : February 13 th, 2018 Details on topics and list of practice problems to appear on course webpage soon.

3 What integration techniques have we learned so far?

4 What integration techniques have we learned so far? By inspection By substitution By parts

5 Can we compute the following integrals? sec % x dx cos x sin x dx x cos x dx sin % x cos, sin, x dx tan, x sec / x dx x dx

6 Can we compute the following integrals? sec % x dx = [by inspection] = tan(x) + C

7 Can we compute the following integrals? sec % x dx = [by inspection] = tan(x) + C cos x sin x dx =

8 Can we compute the following integrals? sec % x dx = [by inspection] = tan(x) + C cos x sin x dx = [sub: u=cos(x)] = u du = 5678 (9) + C %

9 Can we compute the following integrals? sec % x dx= [by inspection] = tan(x) + C cos x sin x dx = [sub: u=cos(x)] = u du = 5678 (9) x cos x dx = % + C

10 Can we compute the following integrals? sec % x dx = [by inspection] = tan(x) + C cos x sin x dx = [sub: u=cos(x)] = u du = 5678 (9) + C x cos x dx = [IBP] = x sin(x) sin (x)dx = x sin(x) + cos(x) + C %

11 Can we compute the following integrals? sec % x dx = [by inspection] = tan(x) + C cos x sin x dx = [sub: u=cos(x)] = u du = 5678 (9) + C x cos x dx = [IBP] = x sin(x) sin (x)dx = x sin(x) + cos(x) + C % sin % x cos, sin, x dx tan, x sec / x dx x dx

12 Can we compute the following integrals? sec % x dx [by inspection] = tan(x) + C cos x sin x dx [sub: u=cos(x)] = u du = 5678 (9) + C x cos x dx [IBP] = x sin(x) sin (x)dx = x sin(x) + cos(x) + C % sin % x cos, x dx =??? sin, x dx =??? tan, x sec / x dx =???

13 Today: Trigonometric Integrals Goal: develop strategies to integrate any product of sines and cosines, any product of tangents and secants.

14 Useful stuff. Basic antiderivatives: sin (x)dx = cos(x) + C cos (x)dx = sin(x) + C sec % x dx = tan(x) + C tan x sec x dx = sec(x) + C Basic trig identity sin % (x) + cos % (x) = 1 sin % (x) = : % Double-angle formulae tan % (x) + 1 = sec % (x) cos % (x) = : % [1 cos 2x ] [1 + cos 2x ]

15 Simple cases 1) sin x cos % x dx 2) sin, (x) cos(x)dx 3) sin % x dx 4) sin, x dx

16 Simple cases 1) sin x cos % x dx = [sub: u=cos(x)] = u % du = BCDE 9, + C

17 Simple cases 1) sin x cos % x dx [sub: u=cos(x)] = u % du = BCDE 9 2) sin, x cos x dx, + C A. u=cos(x) B. u=sin(x) C. neither

18 Simple cases 1) sin x cos % x dx [sub: u=cos(x)] = u % du = BCDE 9 2) sin, (x) cos(x)dx [sub: u=sin(x)] = u, du = DFGH (9) I, + C + C

19 Simple cases 1) sin x cos 2 x dx [sub: u=cos(x)] Product of sine (or cosine) and 2) sin 3 (x) cos(x)dx [sub: u=sin(x)] some power of cosine (or sine)

20 Simple cases 1) sin x cos % x dx [sub: u=cos(x)] Product of sine (or cosine) and 2) sin, (x) cos(x)dx [sub: u=sin(x)] some power of cosine (or sine) General case: For any n For any m sin R (x) cos(x)dx cos S (x) sin(x)dx u= sin(x) u=cos(x) du = cos(x)dx du = -sin(x)dx

21 Simple cases 3) sin % x dx 4) sin, x dx

22 Simple cases 3) sin % x dx = [use sin % (x) = : % [1 cos 2x ] ] = : [1 cos 2x ] dx = : [x : sin 2x ] + C % % %

23 Simple cases 3) sin % x dx [use sin % (x) = : % [1 cos 2x ] ] = : % [1 cos 2x ] dx = : % [x : % sin 2x ] + C 4) sin, x dx = sin % x sin x dx = [use sin % (x) = 1 cos % (x) ] = (1 cos % x )sin x dx = sin (x)dx cos % (x) sin(x)dx = = cos(x) u % du = cos(x) + BCDE 9, + C

24 Simple cases 3) sin 2 x dx powers of sine 4) sin 3 x dx no cosine term

25 Simple cases 3) sin 2 x dx [double-angle formula] powers of sine 4) sin 3 x dx [trig. identity + u-sub] no cosine term

26 Simple cases 3) sin 2 x dx [double-angle formula] powers of sine even power 4) sin 3 x dx [trig. Identity + u-sub] no cosine term odd power

27 Simple cases 3) sin 2 x dx [double-angle formula] powers of sine even power 4) sin 3 x dx [trig. identity + u-sub] no cosine term odd power (same strategy works for powers of cosine, no sine term)

28 Simple cases 3) sin 2 x dx [double-angle formula] powers of sine even power 4) sin 3 x dx [trig. identity + u-sub] no cosine term odd power General case For even n > 1, sin R x dx sin % (x) = U [1 cos 2x ] 8 cos R x dx cos % (x) = U [1 + cos 2x ] 8 For odd n >1 sin R x dx isolate sin(x), convert the rest to powers of cosine, u=cos(x) sin % (x) = 1 cos % (x) cos R x dx isolate cos(x), convert the rest to powers of sine, u=sin(x) cos % (x) = 1 sin % (x)

29 Simple cases 3) sin 2 x dx [double-angle formula] powers of sine even power 4) sin 3 x dx [trig. Identity + u-sub] no cosine term odd power General case For even n > 1, alternatively: sin R x dx sin % (x) = U [1 cos 2x ] Integration by parts 8 cos R x dx cos % (x) = U [1 + cos 2x ] 8 For odd n >1 sin R x dx isolate sin(x), convert the rest to powers of cosine, u=cos(x) sin % (x) = 1 cos % (x) cos R x dx isolate cos(x), convert the rest to powers of sine, u=sin(x) cos % (x) = 1 sin % (x)

30 sin, x dx let s use integration by parts. A. u = sin x, dv = sin % x dx B. u = sin % (x), dv = sin x dx C. u = sin, (x), dv = dx

31 Example: cos I x dx

32 Example: cos I x dx Method 1: cos I x dx = [ U 8 [:\BCD %9 ]% dx = U [1 + 2 cos 2x + H cos% (2x)]dx Reduce cos % (2x) to cos (4x) by using double-angle formula..tedious if high powers

33 Example: cos I x dx Method 1: cos I x dx = [ U 8 (:\BCD %9 ]% dx = = U [1 + 2 cos 2x + H cos% (2x)]dx = Reduce cos % (2x) to cos (4x) by using double-angle formula again. Method 2: By IBP, cos I x dx =

34 Example: cos I x dx Method 1: cos I x dx = [ U 8 (:\BCD %9 ]% dx = = U [1 + 2 cos 2x + H cos% (2x)]dx = Reduce cos % (2x) to cos (4x) by using double-angle formula again. Method 2: By IBP, cos I x dx = cos, x cos (x)dx u = cos, (x) dv = cos x dx du = 3cos % x sin x dx v = sin x = cos, x sin x sin x 3cos % x sin (x )dx = = cos, x sin x + 3 cos % x sin % x dx How do we do the last integral?

35 sin % x cos % x dx [this is neither sin R (x) cos(x)dx nor cos R (x) sin(x)dx ]

36 sin % x cos % x dx Strategy: use trig identity, convert integrand to powers of sine or cosine only (1 cos % x )cos % x dx

37 sin % x cos % x dx Strategy: use trig identity, convert integrand to powers of sine or cosine only (1 cos % x )cos % x dx = [cos % x cos I x ]dx = cos 2 x dx cos 4 x dx now use double-angle formulae for cos % x

38 .back to cos I x dx Method 2: By IBP u = cos, (x) dv = cos x dx du = 3cos % x sin x dx v = sin x cos 4 x dx = cos, x cos (x)dx = cos, x sin x sin x 3cos % x sin (x )dx = cos, x sin x + 3 cos % x sin % x dx = = cos, x sin x + 3 cos 2 x dx 3 cos 4 x dx cos I x dx = U H cos, x sin x + E H cos 2 x dx we reduced the power of cosine. There are handy reduction formulas for integrating powers of sine and cosine

39 (homework) 1) Prove the reduction formula a cos S x dx = 1 m cossc: x sin x + m 1 m a cossc% x dx 2) Find a similar formula for sin R x dx.

40 Other cases 1) We just learned how to compute sin % x cos % x dx. 2) Can we compute sin, x cos % x dx?

41 sin, x cos % x dx = = sin % x cos % x sin x dx [Strategy: reduce the odd power]

42 sin, x cos % x dx = [Strategy: reduce the odd power] = sin % x cos % x sin x dx [Convert to powers of cosines, then substitute u = cos (x)] = (1 cos % x )cos % x sin x dx = = cos % x sin x dx cos I x sin x dx = u % du + u I du =

43 sin, x cos % x dx = [Strategy: reduce the odd power] = sin % x cos % x sin x dx [Convert to powers of cosines, then substitute u = cos (x)] = (1 cos % x )cos % x sin x dx = = cos % x sin x dx cos I x sin x dx = u % du + u I du = Alternatively, sin, x cos % x dx = sin % x cos % x sin x dx [convert to powers of sine, compute integrals of powers of sine] = sin % x (1 sin % x )sin x dx = = sin, x dx sin / x dx odd powers of sine reduce power, doable, longer! convert to cosine, u-substitution...

44 Practice: cos, x sin % x dx The most efficient strategy to compute the above integral above is to convert it to which of the following integrals? A. cos % x sin % x cos(x) dx B. cos, x sin x sin(x) dx C. cos, x (1 cos % x )dx

45 Practice cos, x sin % x dx = cos % x sin % x cos x dx = = (1 sin % x )sin % x cos x dx = = sin % x cos x dx sin I x cos x dx = = u % du u I du =

46 Recap A) sin R (x) cos(x)dx or cos R (x) sin(x)dx use u-substitution

47 Recap A) sin R (x) cos(x)dx or cos R (x) sin(x)dx use u-substitution B) sin R x dx or cos R x dx if n is even use double-angle formulae if n is odd reduce the power, use trig identify

48 Recap: A) sin R (x) cos(x)dx or cos R (x) sin(x)dx use u-substitution B) sin R x dx or cos R x dx if n is even use double-angle formulae if n is odd reduce the power, use trig identify C) sin R x cos S x dx for both m > 1 and n > 1 if one power is odd, reduce the odd power, use trig identity, reduce to A) if both powers are even, use trig identify, reduce to B)

49 Recap: Every sin n x cos m x can be integrated! A) sin R (x) cos(x)dx or cos R (x) sin(x)dx use u-substitution B) sin R x dx or cos R x dx if n is even use double-angle formulae if n is odd split the power, use trig identify C) sin R x cos S x dx for both m > 1 and n > 1 if one power is odd, split the odd power, use trig identity, reduce to A) if both powers are even, use trig identify, reduce to B)

50 sin n x cos m x : what if powers are negative? When we allow negative powers n and m, logarithms may appear. tan x dx

51 sin n x cos m x : what if powers are negative? When we allow negative powers n and m, logarithms may appear. tan x dx = DFG 9 BCD 9 dx

52 sin n x cos m x : what if powers are negative? When we allow negative powers n and m, logarithms may appear. tan x dx = DFG 9 dx [sub u = cos x] BCD 9 = cgh = ln cos x = ln sec x +C h

53 sin n x cos m x : what if powers are negative? When we allow negative powers n and m, logarithms may appear. tan x dx = DFG 9 dx [sub u = cos x] BCD 9 = cgh = ln cos x = ln sec x +C h sec(x) dx

54 sin n x cos m x : what if powers are negative? When we allow negative powers n and m, logarithms may appear. tan x dx = DFG 9 dx [sub u = cos x] BCD 9 = cgh = ln cos x + C = ln sec x +C h sec(x) dx= [sub u = sec x + tan x ] = ln sec x + tan x + C

55 What about sec n x tan m x? Basic examples sec 2 x dx = tan x + C For any m, tan S x sec % x dx = [sub u=tan(x)] U rsu tans\: x + C

56 What about sec n x tan m x? Basic examples sec % x dx = tan x + C For any m, tan S x sec % x dx [sub u=tan(x)] = U rsu tans\: x + C For any m, and n even tan S x sec R x dx = tan S x sec Rc% x sec % x dx convert powers of secant into powers of tangent using tan % (x) + 1 = sec % (x) then sub u=tan(x)

57 Example: tan / x sec I x dx

58 Example: tan / x sec I x dx Keep sec % x, convert the other secant factor into powers of tangent = tan / x sec % x sec % x dx = = tan / x (tan % x +1)sec % x dx= = tan z x sec % x dx + tan / x sec % x dx [sub u=tan(x)] = u z du + u / du =

59 What about sec n x tan m x? Basic examples tan x sec x dx = sec(x) + C For any n, tan (x) sec R x dx = sec Rc: x sec x tan x dx = U { secr x + C [sub u =sec(x)]

60 What about sec n x tan m x? Basic examples tan x sec x dx = sec(x) + C For any n, tan (x) sec R x dx = sec Rc: x sec x tan x dx [u =sec(x)] = U { secr x + C For any n, and m odd tan S x sec R x dx = tan Sc: x sec Rc: x sec x tan x dx = convert powers of tangent into powers of secant by using tan % (x) = sec % (x) 1

61 If m is even, n is odd tan S x sec R x dx use IBP and reduction formula Example: sec, x dx = sec x sec % x dx u = sec x dv = sec % x dx du = tan x sec x dx v = tan x = sec x tan x tan x tan x sec x dx = = sec x tan x tan % x sec(x) dx [tan % (x) = sec % (x) 1] = sec x tan x sec, x dx + sec(x) dx So sec, x dx = U sec x tan x + U sec(x) dx 8 8