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 = convert 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 = ∫ secB 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 ∫ sec 𝑦 𝑒𝑦 = ∫ sec 𝑦 GHI JKLMNJ

GHI JKLMNJ 𝑒𝑦 = ∫ OPQEJK GHI J LMN J GHI JKLMNJ

𝑒𝑦 = ∫

RS S = ln 𝑣 + 𝐷 = ⋯

Last time: Trigonometric integrals

Integrals of sin+ 𝑦 cos% 𝑦

• r 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. ∫ sin(𝑦.) 𝑒𝑦 , ∫ cos(𝑦.)𝑒𝑦 , ∫

GXN (J) J

𝑒𝑦 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 JD/EKJZ/E 𝑒𝑦 [ 3

can be written as A) ∫ 𝑦-3/. + 𝑦-@/.𝑒𝑦

[ 3

B) ∫

3 JD/E + [ 3 3 JZ/E 𝑒𝑦

C) Either one—they are equivalent D) A and B are equivalent but they are not a correct simplification of

𝟐 𝒚𝟐/𝟑K𝒚𝟒/𝟑

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

∫ 𝑦 1 − 𝑦. 𝑒𝑦 = ∫ −D

E 𝑣 𝑒𝑣 = −

3 @ 𝑣@/. + C = − 3 @ (1 − 𝑦.)@/. + C

𝑣 = 1 − 𝑦., 𝑒𝑣 = −2𝑦𝑒𝑦 Does substitution work for the other integrals? ∫ 1 − 𝑦. 𝑒𝑦 ∫ 𝑦. 1 − 𝑦. 𝑒𝑦 ∫ 𝑦@ 1 − 𝑦. 𝑒𝑦

∫ 𝑦 1 − 𝑦. 𝑒𝑦 = ∫ −D

E 𝑣 𝑒𝑣 = −

3 @ 𝑣@/. + C = − 3 @ (1 − 𝑦.)@/. + C

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 =∫ −

3 . (1 − 𝑣.) 𝑣. 𝑣 𝑒𝑣 = ∫ − 3 . (1 − 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 =∫ −

3 . (1 − 𝑣.) 𝑣. 𝑣 𝑒𝑣 = ∫ − 3 . (1 − 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 Note 𝑑𝑝𝑡.(𝜄) = |cos 𝜄 |. We take |cos 𝜄 | = cos 𝜄, i.e. −

k . ≤ 𝜄 ≤ k . .

…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 . [θ + 3 . sin 2𝜄 ] + C …convert back to the original variable 𝑦...

Note 𝑦 = sin 𝜄, hence 𝜄 = arcsin (𝑦) cos 𝜄 = 1 − sin.(𝜄) = 1 − 𝑦. Therefore, recalling sin 2𝜄 = 2sin 𝜄 cos 𝜄, ∫ 1 − 𝑦. 𝑒𝑦 =

3 . θ + 3 [ 2sin 𝜄 cos 𝜄 + C = 3 . arcsin

(𝑦) +

3 . 𝑦 1 − 𝑦.+ C.

∫ 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 = = ∫ cos𝟑 𝜄 𝑒𝜄 − ∫ cos𝟓 𝜄 𝑒𝜄 = [from last week] =

3 . θ + 3 [ sin 2𝜄 − 3 @. 12𝜄 + 8 sin 2𝜄 + sin 4𝜄

+ 𝐷 = =

3 w 𝜄 − 3 @. sin 4𝜄 + 𝐷 =

=

3 w 𝜄 − 3 @. 2sin 2𝜄 cos(2𝜄) + 𝐷 =

=

3 w 𝜄 − 3 3x 2sin 𝜄 cos( 𝜄) ( 1 − 2sin.(𝜄)) + 𝐷 =

=

3 w arcsin

(𝑦) −

3 w 𝑦

1 − 𝑦. ( 1 − 2𝑦.) + 𝐷

What substitution would work for ∫ 9 − 𝑦. 𝑒𝑦 ?

a) 𝑦 = sin(9𝜄) b) 𝑦 = 9sin𝜄 c) 3𝑦 = sin 𝜄 d) 𝑦 = 3sin𝜄 e) None of the above

What substitution would work for ∫ 9 − 𝑦. 𝑒𝑦 ?

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) d) ∫ 𝟘𝒅𝒑𝒕𝟑(𝜾) 𝒆𝜾 e) ∫ 9𝑑𝑝𝑡@(𝜄) 𝑒𝜄

What substitution would simplify ∫ 9 + 𝑦. 𝑒𝑦 ?

a) 𝑦 = 3sin𝜄 b) 𝑦 = 3cos𝜄 c) 𝑦 = 3tan 𝜄 d) 𝑦 = 3sec𝜄 e) None of the above

What substitution would simplify ∫ 9 + 𝑦. 𝑒𝑦 ?

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 . ≤ 𝜄 ≤ k .

• 𝑏. + 𝑦. = 𝑏 1 + tan.𝜄 with 𝑦 = 𝑏 tan 𝜄, where −

k . < 𝜄 < k .

• 𝑦. − 𝑏. = 𝑏 s𝑓𝑑.𝜄 − 1 with 𝑦 = 𝑏 s𝑓𝑑 𝜄, where

0 ≤ 𝜄 <

k .

∫

RJ (3x-JE)Z/E

∫

R‰ ‰ 3K(ŠN ‰)E

∫

RJ (3x-JE)Z/E = ∫

RJ 3xZ/E[3- ‹

ΠE

]Z/E

𝑦 = 4sin 𝜄, 𝑒𝑦 = 4 cos 𝜄 𝑒𝜄 =

3 x[ ∫ [ I•G ŽRŽ [3- GXN Ž E]

Z E

=

3 x[ ∫ [ I•G ŽRŽ Q•OZŽ = 3 3x ∫ RŽ Q•OEŽ = 3 3x tan 𝜄 + C

∫

RJ (3x-JE)Z/E = ∫

RJ 3xZ/E(3-(J/[)E)Z/E

𝑦 = 4sin 𝜄, 𝑒𝑦 = 4 cos 𝜄 𝑒𝜄 =

3 x[ ∫ [ I•G ŽRŽ (3- GXN Ž E)

Z E

=

3 x[ ∫ [ I•G ŽRŽ Q•OZŽ = 3 3x ∫ RŽ Q•OEŽ = 3 3x tan 𝜄 + C

• Trig. Substitutions are a manifestation of Pythagora’s thrm.

x 4 𝑦 = 4sin 𝜄 θ tan 𝜄 =

J 3x-JE

16 − 𝑦. Thus ∫

RJ (3x-JE)Z/E = 3 3x tan 𝜄 + C = J 3x 3x-JE + C

∫

R‰ ‰ 3K(ŠN ‰)E

𝑣 = ln 𝑧, 𝑒𝑣 =

3 ‰ 𝑒𝑧

∫

R‰ ‰ 3K(ŠN ‰)E = ∫ RS 3KSE

𝑣 = tan 𝜄, 𝑒𝑣 = sec. 𝜄 dθ, 1 + 𝑣. = 1 + tan. 𝜄 =

• sec. 𝜄

∫

RS 3KSE = ∫ GHIE Ž ’“ GHI Ž

= ∫ sec 𝜄 𝑒𝜄 = ln sec 𝜄 + tan 𝜄 + 𝐷 = =ln 1 + 𝑣. + 𝑣 + 𝐷 = ln 1 + (ln 𝑧). + ln 𝑧 + 𝐷

Compute the area of a circular segment (shaded region) A = 2 ∫ 𝑏. − 𝑦.

”

• 𝑒𝑦 = 2a ∫

1 −

J ” . ”

• 𝑒𝑦

b a 𝑦 = 𝑏 sin 𝜄 2𝑏. ∫ 𝑑𝑝𝑡.𝜄 𝑒𝜄

J–” J–•

= 𝑏.(𝜄 + D

Esin(2𝜄)) J–”

J–•

= 𝑏. arcsin (𝑦/𝑏) +

J ”

1 − (𝑦/𝑏).

J–” J–•

= 𝑏.

k . + 0 − arcsin

• ” −
• ”

1 −

• ”

.

=

k . 𝑏. − 𝑏. arcsin

• ” − 𝑐 𝑏. − 𝑐.