Science One Math Feb 4, 2019 Today Some more practice with - PowerPoint PPT Presentation

Science One Math Feb 4, 2019

Today • Some more practice with trigonometric substitutions • Partial fractions and other strategies to integrate rational functions

Last time: Trigonometric substitutions When to use them? • when the integrand has a square root • when the argument of the radical can be reduced to either one of % − 1 % , % , # # # 1 − 1 + $ $ $ Let’s do some more practice… 𝑓 ) 1 − 𝑓 %) * 𝑒𝑦 ∫ + ,- %

𝑓 ) 1 − 𝑓 %) * 𝑒𝑦 is equivalent to ∫ + ,- % * A. ∫ 𝑣 1 − 𝑣 % 𝑒𝑦 + ,- % * B. ∫ 1 − 𝑣 % 𝑒𝑣 + ,- % 5 C. ∫ 𝑣 1 − 𝑣 % 𝑒𝑣 5/% 5 D. ∫ 1 − 𝑣 % 𝑒𝑣 5/% E. Either B or D

𝑓 ) 1 − 𝑓 %) * 𝑒𝑦 is equivalent to ∫ + ,- % * A. ∫ 𝑣 1 − 𝑣 % 𝑒𝑦 + ,- % * B. ∫ 1 − 𝑣 % 𝑒𝑣 + ,- % 5 C. ∫ 𝑣 1 − 𝑣 % 𝑒𝑣 5/% 𝟐 D. D. ∫ 𝟐 − 𝒗 𝟑 𝒆𝒗 𝟐/𝟑 E. Either B or D When you make a substitution, remember to change the limits of integration!

𝑓 ) 1 − 𝑓 %) * 𝑣 = 𝑓 ) , 𝑒𝑣 = 𝑓 ) 𝑒𝑦 𝑒𝑦 u-substitution ∫ + ,- % 5 1 − 𝑣 % 𝑒𝑣 𝑣 = sin 𝜄 , 𝑒𝑣 = cos 𝜄𝑒𝜄 trig substitution (inverse sub.) ∫ 5/% F/% 5 5 5 F F 5 O 5 F O % sin 2𝜄 ] MNF/% 𝑑𝑝𝑡 % 𝜄𝑒𝜄 = % [θ + MNF/G = % − G − = O − ∫ F/G % % % % P Note: Trig. substitutions are useful even if there is no root in the integrand. Q) ∫ 5R) S S

Trig. substitutions are useful even if there is no root in the integrand TUV S MQM TUV S MQM Q) TUV Z M = ∫ 𝑑𝑝𝑡 % 𝜄 𝑒𝜄 = ⋯ 5R) S S = ∫ 5RWXY S M S = ∫ ∫ 𝑦 = tan 𝜄

Some practice (homework) ) S ) S 1) Find the area of the region enclosed by the ellipse X S + ^ S = 1 2) Find the volume of the solid generated by revolving about the 𝑦 - axis the region bounded by the curve 𝑧 = 4/(𝑦 2 + 4), the 𝑦 - axis, the lines 𝑦 = 0 and 𝑦 = 2 . ) f 3) ∫ ) S R% 𝑒𝑦 5 4) ∫ ) g+) S 𝑒𝑦

Which of the following is the most effici cient Q) strategy for computing ∫ 5RP) S ? 5 a) substitute 𝑦 = % tan𝜄 b) substitute 𝑦 = 2 tan𝜄 c) substitute 𝑦 = 2 sec𝜄 5 d) substitute 𝑦 = % cos𝜄 e) None of the above

Which of the following is the most effici cient Q) strategy for computing ∫ 5RP) S ? 5 a) 𝑦 = % 𝑢𝑏𝑜𝜄 ☚ also correct but a little longer to compute b) 𝑦 = 2 𝑢𝑏𝑜𝜄 c) 𝑦 = 2 𝑡𝑓𝑑𝜄 5 d) 𝑦 = % 𝑑𝑝𝑡𝜄 e) None of the above 𝑣 = 2𝑦, 𝑒𝑣 = 2𝑒𝑦 Q) 5 Qn 5 5RP) S = 5Rn S = % arctan(2x) + C ….don’t forget the basics! ∫ % ∫

Another useful trick: Completing the square …useful when we don’t have a perfect square. 𝑦 % − 2𝑦 + 2 𝑒𝑦 ∫ 𝑦 % − 2𝑦 + 2 = (𝑦 % −2𝑦 + 1) + 1 = 𝑦 − 1 % + 1 , now trig. sub. 𝑦 − 1 = tan 𝜄 2𝑦 − 𝑦 % 𝑒𝑦 ∫ −𝑦 % + 2𝑦 + 0 = (−𝑦 % +2𝑦 − 1) + 1 = − 𝑦 − 1 % + 1 , now trig. sub. 𝑦 − 1 = sin 𝜄 we’ll use these tricks again later today…

OK, enough with trig substitutions…. ...back to trigonometric integrals. Last week we mentioned ∫sec 𝑦 𝑒𝑦 = ln | sec 𝑦 + tan𝑦 | + C …famous problem in cartography... Mercator’s map (1569) “…In making this representation the world, we had to spread on a plane the surface of the sphere in such a way that the positions of places shall correspond on all sides with each other both in true direction and in distance. With this intention we had to employ a new proportion and a new arrangement of the meridians with reference to the parallels. For these reasons we have progressively increased the length of latitude towards each pole in proportion to the lengthening of parallels with reference to the equator.”

λ = π y( φ ) On the sphere, the equation has axis y 180W 180E length 2𝜌𝑆 and the parallel at parallel through P N P' P(x,y) P latitude 𝜚 has length 2𝜌𝑆 cos 𝜚 . x true scale on equator φ λ 0 x λ =0 On the map, the equator and S the parallel have same length. x = − x = π π central meridian meridian through P central Thus the parallel has been stretched meridian 5 by a factor tuv w = sec𝜚 . We want proportions and directions to be conserved ( spherical square ⟺ square on map). Problem :Let 𝐺(𝜚) be distance of point P from equation on the map. So 𝐺 0 = 0. Find a formula for 𝐺 𝜚 . %Fy tuv w = {| y∆w {| %Fy Proportions are conserved ⟹ %F ⟹ ∆w = %Fy tuv w = sec𝜚 w Hence 𝐺 𝜚 = ∫ sec𝜎 𝑒𝜎 . *

Some history of ∫ sec 𝑦 𝑒𝑦 no, you don’t need to remember M M F (1645, first conjecture) ∫ sec 𝑢𝑒𝑢 = ln tan % + these for the exam! * P (logarithms were invented in 1614 by Napier) • (modern proof) 𝑣 = sec 𝑦 + tan 𝑦 , 𝑒𝑣 = (sec 𝑦 tan 𝑦 + sec % 𝑦)𝑒𝑦 sec 𝑦 + tan𝑦 𝑒𝑦 = ~sec % 𝑦 + sec𝑦 tan𝑦 ~sec 𝑦 𝑒𝑦 = ~ sec𝑦 sec 𝑦 + tan𝑦 𝑒𝑦 = ~𝑒𝑣 𝑣 = ln sec 𝑦 + tan𝑦 + 𝐷 sec 𝑦 + tan𝑦 • (one of the earliest proofs, 1670) 𝑣 = sin 𝑦 , 𝑒𝑣 = cos 𝑦 𝑒𝑦 need strategies tuv ) Qn ∫ sec 𝑦 𝑒𝑦 = ∫ V€T S ) 𝑒𝑦 = ∫ 5+n S …now what? to integrate rational functions 5+n S ) %Qn • (a classic, sneaky substitution, mid 1700) 𝑣 = tan % , cos 𝑦 = 5Rn S , d𝑦 = 5Rn S 5Rn S 5 5+n S %Qn 5 ∫sec 𝑦 𝑒𝑦 = ∫ tuv ) 𝑒𝑦 = ∫ 5Rn S = 2∫ 5+n S 𝑒𝑣 …now what?

FYI: Tangent half-angle substitution ) 𝑣 = tan (no, these won’t be on the exam—no need to memorize them ) % 5+n S cos 𝑦 = 5Rn S %n sin 𝑦 = 5Rn S %Qn d𝑦 = 5Rn S

Q) How to integrate ∫ 5+) S ? We use a new technique based on the fact that we know how to integrate these Q) Qn Q) Qn )+5 = ∫ n =ln 𝑦 − 1 + 𝐷 )R5 = ∫ n =ln 𝑦 + 1 + 𝐷 ∫ ∫ 𝑣 = 𝑦 − 1 𝑣 = 𝑦 + 1 5 In general, we can integrate any rational functions of the form X)R^ .

Ne New technique: Partial Fractions Start with an observation: 5/% 5/% *.g )R5 +*.g )+5 5 )+5 − )R5 = = )+5 )R5 ()+5)()R5) This means we can break the integrand up into “pieces” that are easy to integrate 5 5 5/% 5/% 5 5 ∫ ) S +5 𝑒𝑦 = ∫ ()+5)()R5) 𝑒𝑦 = ∫ )+5 − )R5 𝑒𝑦 = % ln 𝑦 − 1 − % ln 𝑦 + 1 + 𝐷 partial fractions ☛ easy to integrate

Other examples 5 5 ) • +P) 𝑒𝑦 = ∫ )()+%)()R%) 𝑒𝑦 ∫ Q) 5 What does the integral above have in common with ∫ 5+) S = ∫ (5+))(5R)) 𝑒𝑦 ? What form is its partial fraction decomposition?

Other examples 5 5 ) • +P) 𝑒𝑦 = ∫ )()+%)()R%) 𝑒𝑦 ∫ Q) 5 What does the integral above have in common with ∫ 5+) S = ∫ (5+))(5R)) 𝑒𝑦 ? What form is its partial fraction decomposition? Common features : the numerator is a degree-zero poly, the denominator has only linear factors Strategy : decompose the integrand into partial fractions , one fraction for each factor.

5 5 seek decomposition of the form ‚ ƒ „ ) • +P) 𝑒𝑦 = ∫ )()+%)()R%) 𝑒𝑦 ) + )+% + ∫ )R% ( one fraction for each factor) 3 factors We want ‚ ƒ „ 5 ) + )+% + )R% = )()+%)()R%) Strategies for finding A, B, C: Method 1 : add partial fractions, choose coefficients so that numerators match (recall two polynomials are equal for all 𝑦 if the like terms are equal) Method 2: “cover-up” the factor (𝑦 − 𝑏) and then substitute 𝑦 = 𝑏 . Repeat for each factor. ⇒ 𝐵 = −1/4 , 𝐶 = 1/8 , 𝐷 = 1/8 ˆ ln 𝑦 % − 4 + 𝐷 )()+%)()R%) 𝑒𝑦 = − 5 5 Q) ) + 5 )+% 𝑒𝑦 + 5 5 )R% 𝑒𝑦 = − 5 5 P ln 𝑦 + 5 ∫ P ∫ ˆ ∫ ˆ ∫

Partial Fractions: when to use The degree of numerator < degree of denominator and • the numerator is a degree-zero poly, the denominator has only linear factors Q) 5 g 5 5+) S , ∫ ) • +P) 𝑒𝑦 , ∫ %) S +g)+O 𝑒𝑦 , ∫ X S )R^ S … X ‹ )R^ ‹ 𝑒𝑦 ∫ X ‰ )R^ ‰ • the numerator is a linear poly, the denominator has only linear factors )R% g)R5 ) • +) 𝑒𝑦 %)R5 ()+5) 𝑒𝑦 ∫ ∫ • the numerator is either a degree-zero or linear poly, the denominator has at least one (irreducible) quadratic factor

Partial Fractions: when to use The degree of numerator < degree of denominator and • the numerator is a degree-zero poly, the denominator has only linear factors Q) 5 g 5 5+) S , ∫ ) • +P) 𝑒𝑦 , ∫ %) S +g)+O 𝑒𝑦 , ∫ X S )R^ S … X ‹ )R^ ‹ 𝑒𝑦 ∫ X ‰ )R^ ‰ • the numerator is a linear poly, the denominator has only linear factors )R% g)R5 V)RQ ) • +) 𝑒𝑦 , ∫ %)R5 ()+5) 𝑒𝑦 , ∫ X S )R^ S … X ‹ )R^ ‹ 𝑒𝑦 ∫ X ‰ )R^ ‰ • the numerator is either a degree-zero or linear poly, the denominator has at least one (irreducible) quadratic factor

𝑦 O − 𝑦 = 𝑦 𝑦 + 1 )R% ) • +) 𝑒𝑦 𝑦 − 1 ∫ ‚ ƒ „ )R% ) + )R5 + )+5 = ) • +) 𝐵 = −2, 𝐶 = 1/2, 𝐷 = 3/2 )R% +% 5/% O/% 5 O ) • +) 𝑒𝑦 = ∫ ) + )R5 + )+5 𝑒𝑦 = −2 ln 𝑦 + % ln 𝑦 + 1 + % ln 𝑦 − 1 + 𝐷 . ∫

g)R5 %)R5 ()+5) 𝑒𝑦 ∫ ‚ ƒ g)R5 (%)R5)()+5) ⇒ 𝐵 = 1 , 𝐶 = 2 (%)R5) + )+5 = g)R5 Q) % 5 %)R5 ()+5) 𝑒𝑦 = ∫ %)R5 + ∫ )+5 𝑒𝑦 = % ln 2𝑦 + 1 + 2 ln 𝑦 − 1 + 𝐷 ∫