Science One Math Jan 28 2019 Last time: tan % sec + For any m - PowerPoint PPT Presentation

Science One Math Jan 28 2019

Last time: ∫ tan % 𝑦 sec + 𝑦 𝑒𝑦 For any m , and n even ∫ tan % 𝑦 sec + 𝑦 𝑒𝑦 = ∫ tan % 𝑦 sec +-. 𝑦 sec . 𝑦 𝑒𝑦 convert powers of secant into powers of tangent using tan . (𝑦) + 1 = sec . (𝑦) then sub 𝑣 = tan (𝑦) For any n , and m odd ∫ tan % 𝑦 sec + 𝑦 𝑒𝑦 = ∫ tan %-3 𝑦 sec +-3 𝑦 sec 𝑦 tan 𝑦 𝑒𝑦 = c onvert powers of tangent into powers of secant by using tan . (𝑦) = sec . (𝑦) − 1 What if n odd and m even? Use integration by parts!

If m is even, n is odd ∫ tan % 𝑦 sec + 𝑦 𝑒𝑦 use IBP and reduction formula Example : ∫ tan . 𝑦 sec @ 𝑦 𝑒𝑦 = ∫(sec . 𝑦 − 1)sec @ 𝑦 𝑒𝑦 = ∫ sec B 𝑦 −sec @ 𝑦 𝑒𝑦 ∫ sec @ 𝑦 𝑒𝑦 = ∫ sec 𝑦 sec . 𝑦 𝑒𝑦 u = sec 𝑦 dv = sec . 𝑦 𝑒𝑦 du = tan 𝑦 sec 𝑦 𝑒𝑦 v = tan 𝑦 = sec 𝑦 tan 𝑦 − ∫ tan 𝑦 tan 𝑦 sec 𝑦 𝑒𝑦 = = sec 𝑦 tan 𝑦 − ∫ tan . 𝑦 sec(𝑦) 𝑒𝑦 [tan . (𝑦) = sec . (𝑦) − 1] = sec 𝑦 tan 𝑦 − ∫ sec @ 𝑦 𝑒𝑦 + ∫ sec(𝑦) 𝑒𝑦 So ∫ sec @ 𝑦 𝑒𝑦 = D E sec 𝑦 tan 𝑦 + D E ∫ sec(𝑦) 𝑒𝑦 it reduces to computing ∫ sec 𝑦 𝑒𝑦 …we need some trickery here… 𝑣 = sec 𝑦 + tan 𝑦 , 𝑒𝑣 = (sec 𝑦 tan 𝑦 + sec . 𝑦)𝑒𝑦 OPQ E JK GHI J LMN J ∫ sec 𝑦 𝑒𝑦 = ∫ sec 𝑦 GHI JKLMNJ RS GHI JKLMNJ 𝑒𝑦 = ∫ 𝑒𝑦 = ∫ S = ln 𝑣 + 𝐷 = ⋯ GHI JKLMNJ

Last time: Trigonometric integrals Integrals of sin + 𝑦 cos % 𝑦 or tan + 𝑦 sec % 𝑦 are computable! Strategies: • 𝑣 -substitution • trig identities • (sometimes) IBP

Last time: Trigonometric integrals Not all integrals with trigonometric functions are computable, e.g. GXN (J) ∫ sin(𝑦 . ) 𝑒𝑦 , ∫ cos(𝑦 . )𝑒𝑦 , ∫ 𝑒𝑦 J We can only approximate these integrals numerically.

When do we see trigonometric integrals? Many computations lead to trigonometric integrals: • Analysis of oscillating systems (waves, alternating current circuits, etc.) • Fourier analysis (widely used technique where a periodic function is decomposed in terms of infinite sums of powers of sines and cosines) • Integrals with roots

Can we compute area of this shape? The shaded area is called a “ lune” The Ducal Palace – Mantua, Italy It reduces to ∫ 1 − 𝑦 . 𝑒𝑦 need a new strategy! Trigonometric substitutions Russ Adams in Pike County, IL

Science One Math Jan 30 2019

[ 3 J D/E KJ Z/E 𝑒𝑦 can be written as ∫ 3 A) ∫ 𝑦 -3/. + 𝑦 -@/. 𝑒𝑦 [ 3 [ 3 3 B) ∫ J D/E + J Z/E 𝑒𝑦 3 C) Either one—they are equivalent 𝟐 D) A and B are equivalent but they are not a correct simplification of 𝒚 𝟐/𝟑 K𝒚 𝟒/𝟑

Today’s goal: Trigonometric substitutions …how to compute integrals with roots.

How to we compute integrals with roots? 1 − 𝑦 . 𝑒𝑦 ∫ ∫ 𝑦 1 − 𝑦 . 𝑒𝑦 start with this one ∫ 𝑦 . 1 − 𝑦 . 𝑒𝑦 ∫ 𝑦 @ 1 − 𝑦 . 𝑒𝑦

@ 𝑣 @/. + C = − @ (1 − 𝑦 . ) @/. + C ∫ 𝑦 1 − 𝑦 . 𝑒𝑦 = ∫ − D 3 3 E 𝑣 𝑒𝑣 = − 𝑣 = 1 − 𝑦 . , 𝑒𝑣 = −2𝑦𝑒𝑦 Does substitution work for the other integrals? 1 − 𝑦 . 𝑒𝑦 ∫ ∫ 𝑦 . 1 − 𝑦 . 𝑒𝑦 ∫ 𝑦 @ 1 − 𝑦 . 𝑒𝑦

@ 𝑣 @/. + C = − @ (1 − 𝑦 . ) @/. + C ∫ 𝑦 1 − 𝑦 . 𝑒𝑦 = ∫ − D 3 3 E 𝑣 𝑒𝑣 = − 𝑣 = 1 − 𝑦 . , 𝑒𝑣 = −2𝑦𝑒𝑦 Does substitution work for the other integrals? 1 − 𝑦 . 𝑒𝑦 no! ∫ ∫ 𝑦 . 1 − 𝑦 . 𝑒𝑦 no! ∫ 𝑦 @ 1 − 𝑦 . 𝑒𝑦 yes! ∫ 𝑦 . 1 − 𝑦 . 𝑦 𝑒𝑦 = ∫ − D E (1 − 𝑣) 𝑣 𝑒𝑣 =…. When u-sub doesn’t work, we need a different type of substitution …

An observation: ∫ 𝑦 @ 1 − 𝑦 . 𝑒𝑦 = ∫ 𝑦 . 1 − 𝑦 . 𝑦 𝑒𝑦 1 − 𝑦 . = 𝑣 . , 𝑦 . = 1 − 𝑣 . − 2 𝑦 𝑒𝑦 = 2 𝑣 𝑒𝑣 . (1 − 𝑣 . ) 𝑣 . 𝑒𝑣 = … 3 3 . (1 − 𝑣 . ) 𝑣 . 𝑣 𝑒𝑣 = ∫ − = ∫ − So maybe to compute the other integrals we could use a substitution of the form 1 − 𝑦 . = (…) 2

An observation: ∫ 𝑦 @ 1 − 𝑦 . 𝑒𝑦 = ∫ 𝑦 . 1 − 𝑦 . 𝑦 𝑒𝑦 1 − 𝑦 . = 𝑣 . , 𝑦 . = 1 − 𝑣 . − 2 𝑦 𝑒𝑦 = 2 𝑣 𝑒𝑣 . (1 − 𝑣 . ) 𝑣 . 𝑒𝑣 = … 3 3 . (1 − 𝑣 . ) 𝑣 . 𝑣 𝑒𝑣 = ∫ − = ∫ − So maybe to compute the other integrals we need to use a substitution of the form 1 − 𝑦 . = (…) 2 1 − sin . 𝜄 = cos . 𝜄

Trigonometric substitutions 𝑦 = sin 𝜄 , d𝑦 = cos 𝜄 𝑒𝜄 then 1 − 𝑦 . = 1 − sin . 𝑦 = cos . 𝑦 1 − 𝑦 . 𝑒𝑦 = ∫ 1 − 𝑡𝑗𝑜 . (𝜄) cos𝜄 𝑒𝜄 = ∫ 𝑑𝑝𝑡 . (𝜄) cos 𝜄 𝑒𝜄 = ∫ 𝑑𝑝𝑡 . 𝜄 𝑒𝜄 = ∫ k k Note 𝑑𝑝𝑡 . (𝜄) = | cos 𝜄 | . We take | cos 𝜄 | = cos 𝜄 , i.e. − . ≤ 𝜄 ≤ . . …and now we compute the trigonometric integral....

“Best” strategy to compute ∫ 𝑑𝑝𝑡 . 𝜄 𝑒𝜄 is… a) By parts b) By using trig identity cos . 𝑦 = 1 − sin . 𝑦 c) By using double-angle identity and then substitution d) By substitution right away e) By parts and then reduction

“Best” strategy to compute ∫ cos . 𝜄 𝑒𝜄 is… a) By parts b) By using trig identity cos . 𝑦 = 1 − sin . 𝑦 c) By using double-angle identify and then substitution d) By substitution right away e) By parts, and then reduction

1 − 𝑦 . 𝑒𝑦 = ∫ cos . 𝜄 𝑒𝜄 = ∫ 3 . [1 + cos 2𝜄 ] 𝑒𝜄 = ∫ 3 3 = . [θ + . sin 2𝜄 ] + C …convert back to the original variable 𝑦 ... Note 𝑦 = sin 𝜄, hence 𝜄 = arcsin (𝑦) 1 − sin . (𝜄) = 1 − 𝑦 . cos 𝜄 = Therefore, recalling sin 2𝜄 = 2sin 𝜄 cos 𝜄, 3 3 3 3 1 − 𝑦 . 𝑒𝑦 = . 𝑦 1 − 𝑦 . + C. . θ + [ 2sin 𝜄 cos 𝜄 + C = . arcsin (𝑦) + ∫

∫ 𝑦 . 1 − 𝑦 . 𝑒𝑦 9 − 𝑦 . 𝑒𝑦 ∫

∫ 𝑦 . 1 − 𝑦 . 𝑒𝑦 𝑦 = sin 𝜄 , d𝑦 = cos 𝜄 𝑒𝜄 = ∫ sin . (𝜄) 1 − sin . (𝜄) cos 𝜄 𝑒𝜄 = ∫ sin . (𝜄) cos . (𝜄) cos𝜄 𝑒𝜄 = = ∫ sin . (𝜄)cos . 𝜄 𝑒𝜄 = ∫(1 − cos . 𝜄 )cos . 𝜄 𝑒𝜄 = = ∫ cos 𝟑 𝜄 𝑒𝜄 − ∫ cos 𝟓 𝜄 𝑒𝜄 = [from last week] 3 3 3 = . θ + [ sin 2𝜄 − @. 12𝜄 + 8 sin 2𝜄 + sin 4𝜄 + 𝐷 = 3 3 = w 𝜄 − @. sin 4𝜄 + 𝐷 = 3 3 = w 𝜄 − @. 2sin 2𝜄 cos(2𝜄) + 𝐷 = 3 3 3x 2sin 𝜄 cos( 𝜄) ( 1 − 2sin . (𝜄)) + 𝐷 = = w 𝜄 − 3 3 1 − 𝑦 . ( 1 − 2𝑦 . ) + 𝐷 = w arcsin (𝑦) − w 𝑦

9 − 𝑦 . 𝑒𝑦 ? What substitution would work for ∫ a) 𝑦 = sin(9𝜄) b) 𝑦 = 9sin𝜄 c) 3𝑦 = sin 𝜄 d) 𝑦 = 3sin𝜄 e) None of the above

9 − 𝑦 . 𝑒𝑦 ? What substitution would work for ∫ a) 𝑦 = sin(9𝜄) b) 𝑦 = 9sin𝜄 c) 3𝑦 = sin 𝜄 d) d) 𝒚 = 𝟒𝒕𝒋𝒐 𝜾 e) None of the above

Using the substitution 𝑦 = 3sin 𝜄 , the integral 9 − 𝑦 . 𝑒𝑦 is equivalent to… ∫ a) ∫ 9 − 9𝑡𝑗𝑜 . (𝑦) 𝑒𝑦 b) ∫ 3 1 − 𝑡𝑗𝑜 . (𝜄) 𝑒𝜄 c) ∫ 3𝑑𝑝𝑡 . (𝜄) 𝑒𝜄 d) ∫ 9𝑑𝑝𝑡 . (𝜄) 𝑒𝜄 e) ∫ 9𝑑𝑝𝑡 @ (𝜄) 𝑒𝜄

Using the substitution 𝑦 = 3sin 𝜄 , the integral 9 − 𝑦 . 𝑒𝑦 is equivalent to… ∫ a) ∫ 9 − 9𝑡𝑗𝑜 . (𝑦) 𝑒𝑦 b) ∫ 3 1 − 𝑡𝑗𝑜 . (𝜄) 𝑒𝜄 c) ∫ 3𝑑𝑝𝑡 . (𝜄) 𝑒𝜄 d) ∫ 𝟘𝒅𝒑𝒕 𝟑 (𝜾) 𝒆𝜾 d) e) ∫ 9𝑑𝑝𝑡 @ (𝜄) 𝑒𝜄

9 + 𝑦 . 𝑒𝑦 ? What substitution would simplify ∫ a) 𝑦 = 3sin𝜄 b) 𝑦 = 3cos𝜄 c) 𝑦 = 3tan 𝜄 d) 𝑦 = 3sec𝜄 e) None of the above

9 + 𝑦 . 𝑒𝑦 ? What substitution would simplify ∫ a) 𝑦 = 3sin𝜄 b) 𝑦 = 3cos𝜄 c) c) 𝒚 = 𝟒𝐮𝐛𝐨 𝜾 d) 𝑦 = 3sec𝜄 e) None of the above

General case: 3 basic substitutions 𝑏 . − 𝑦 . = 𝑏 1 − sin . 𝜄 with 𝑦 = 𝑏 sin 𝜄 , where − k k . ≤ 𝜄 ≤ • . 𝑏 . + 𝑦 . = 𝑏 1 + tan . 𝜄 with 𝑦 = 𝑏 tan 𝜄 , where − k k . < 𝜄 < • . 𝑦 . − 𝑏 . = 𝑏 s𝑓𝑑 . 𝜄 − 1 with 𝑦 = 𝑏 s𝑓𝑑 𝜄 , where k 0 ≤ 𝜄 < • .

RJ (3x-J E ) Z/E ∫ R‰ ∫ 3K(ŠN ‰) E ‰

RJ RJ (3x-J E ) Z/E = ∫ ∫ ] Z/E E 3x Z/E [3- ‹ Œ 𝑦 = 4sin 𝜄 , 𝑒𝑦 = 4 cos 𝜄 𝑒𝜄 3 [ I•G ŽRŽ 3 [ I•G ŽRŽ 3 RŽ 3 = = Q•O Z Ž = Q•O E Ž = 3x tan 𝜄 + C x[ ∫ x[ ∫ 3x ∫ Z [3- GXN Ž E ] E

RJ RJ (3x-J E ) Z/E = ∫ ∫ 3x Z/E (3-(J/[) E ) Z/E 𝑦 = 4sin 𝜄 , 𝑒𝑦 = 4 cos 𝜄 𝑒𝜄 3 [ I•G ŽRŽ 3 [ I•G ŽRŽ 3 RŽ 3 = = Q•O Z Ž = Q•O E Ž = 3x tan 𝜄 + C x[ ∫ x[ ∫ 3x ∫ Z (3- GXN Ž E ) E Trig. Substitutions are a manifestation of Pythagora’s thrm. x 4 𝑦 = 4sin 𝜄 J θ tan 𝜄 = 3x-J E 16 − 𝑦 . Thus RJ 3 J (3x-J E ) Z/E = 3x tan 𝜄 + C = 3x 3x-J E + C ∫