Like Nancy Kerrigan sez: Why me? Why now? Why inverse … · Why? Like Nancy Kerrigan sez: "Why me?...

21
Why? Like Nancy Kerrigan sez: ”Why me? Why now? Why inverse trig?” It’s all motivated by trig substitution: like Wednesday. For an in- tegral like R dx 1+x 2 , you set x = tan θ. That isn’t exactly something that leaps to mind, but check out what happens: with x like that, 1+ x 2 = 1 + (tan θ) 2 = sec 2 θ. Then dx = sec 2 θ dθ also, so Z dx 1+ x 2 = Z sec 2 θ dθ sec 2 θ = Z = θ + C Now the problem is, getting back the variables that started the problem: getting back to x. I have to take the equation x = tan θ and solve it for θ = something(x). that’s what the inverse trig functions do: if you take x = tan θ and solve for θ, you get θ = tan -1 x. This lecture has four parts: a) Talking about what it means to write θ = tan -1 x. Of course, this is just one of the inverse trig functions: since you also got sin θ, cos θ, sec θ, csc θ, cot θ, there’s gonna be sin -1 x, cos -1 x, sec -1 x, csc -1 x, cot -1 x. b) Finding out the usual things about the inverse trig functions: domain range derivative increasing decreasing concavity graph – oh my. c) Every new derivative is a new friend – a new integral. So, more integrals to remember, put on cheat sheet. d) Integrals to die for: a whole new collection.

Transcript of Like Nancy Kerrigan sez: Why me? Why now? Why inverse … · Why? Like Nancy Kerrigan sez: "Why me?...

Why?

Like Nancy Kerrigan sez: ”Why me? Why now? Why inverse trig?”

It’s all motivated by trig substitution: like Wednesday. For an in-tegral like

∫dx

1+x2 , you set x = tan θ. That isn’t exactly somethingthat leaps to mind, but check out what happens: with x like that,1 + x2 = 1 + (tan θ)2 = sec2 θ. Then dx = sec2 θ dθ also, so∫

dx

1 + x2=

∫sec2 θ dθ

sec2 θ=

∫dθ = θ + C

Now the problem is, getting back the variables that started theproblem: getting back to x. I have to take the equation x = tan θand solve it for θ = something(x). that’s what the inverse trigfunctions do: if you take x = tan θ and solve for θ, you get θ =tan−1 x.

This lecture has four parts:a) Talking about what it means to write θ = tan−1 x. Of course,this is just one of the inverse trig functions: since you also got

sin θ, cos θ, sec θ, csc θ, cot θ, there’s gonna be sin−1 x, cos−1 x,

sec−1 x, csc−1 x, cot−1 x.

b) Finding out the usual things about the inverse trig functions:domain range derivative increasing decreasing concavity graph – ohmy.c) Every new derivative is a new friend – a new integral. So, moreintegrals to remember, put on cheat sheet.d) Integrals to die for: a whole new collection.

Graphical View: Idea of An Inverse Function

I want to look at inverse trig with three different views:a) Graphicalb) Solving an equationc) The secret language of functions.

I’m gonna work through this first with x = sin θ. Graphical: in anormal function, you pick a θ, on the horizontal axis (θ-axis: bluedot in pic, below left). Then you go up to the graph of x = sin θ,and that gives you a height (blue dot, middle pic). Finally, youcarry that over to the vertical axis (blue dot x-axis: below right) toget the x value.

Every gives you an x. Now in the inverse-function thing, everyx gives me a θ: I start with a red dot on the x axis, go to the graph,get a dot on the graph, then lead that back down to a dot on the θaxis.

θ

EquationsView: Definition of An Inverse Function

The second view is the ’solving equations’ view: you have an equa-tion like sin θ = .5 and you want to solve for θ. On a calculatorlike the TI-85, you’d press the buttons ”2nd” (yellow button) then”SIN” then ”.5” This gives you sin−1 like on the calculator. Youget back 30o.

The third view is the ’functions’ view: this is a philosophy thatstarted in the fifteenth to seventeenth centuries*, and received astrong push in the nineteen twenties. In this view, you don’t thinkin pictures; you don’t think of equations: all there is, is functions.

The function view asks, what does it mean to say that θ = 30o is thesolution to the equation sin θ = .5? Well, I’d say it means somethinglike, when you plug θ = 30o back into the equation, you get .5 backagain: sin(30o) = .5. In function language, 30o = sin−1(.5), andplugging in gives sin(30o) = sin

(sin−1(.5)

). And getting back .5

means sin(sin−1(.5)

)= .5

Functions View: Definition of An Inverse Function

* The philosophy that started at this time was that of of ”compu-tation”. Though humans had been computing for millenia, goingback in the archaelogical record to Sumeria, the formal concept of’computation’ as an entity in itself needed a language to talk aboutit. The language that Westerners used was the language of variablesand algebra. This was brought to Western mathematicians throughArabic thinkers, particularly the Muslim mathematician Muham-mad Ibn Musa al-Khwarizmi, who wrote the book Kitab al-jabr waal-muqabalah, shortened to al-jabr or algebra. The late developmentof algebra in the west was a reflection of the division of the medievalMediterranean into Christian and Muslim spheres of influence.

Inverse Function: Domain

In equation terms, you’d say ”the equation sin θ = 1.3 has no so-lution if x > 1”. That’s not way surprising; take an equationlike x2 = −5; that doesn’t have solutions, neither does ex = −1.Same way, x = sin θ has a solution only for −1 ≤ x ≤ 1. Asusual, you can try it on your generic calculator; mine gives mesin−1(1.3) = not a number.

The function view looks at this differently: you’re supposed to beable to plug x into sin−1(x) and get something out. It isn’t working.Well, hey: that happens with

√−5 too, but for

√x you just say, -5

isn’t in the domain of the square root function. Ditto here: x = 1.3isn’t in the domain of the sin−1(x) function. If you want to be morehelpful, say what is in the domain: the domain of sin−1(x) is [−1, 1].

One of the first things you see, if you’re trying to do the inversefunction by pictures, is that you can’t go too high: in the picturebelow, I tried to start x = sin θ with x = 1.3 Even though I tookthe orange arrows left and right, they never hit the graph. D’uh!The graph of x = sin θ never gets higher than x = 1 or lower thanx = −1, so the orange arrows aren’t gonna intersect it. In graphterms: You’ll only hit the graph if you start with −1 ≤ x ≤ 1.

Inverse Function: Range, I

The second thing you notice, if you’re doing the graph thing, is thatyer typical orange line hits the graph of x = sin θ not just at oneplace, but at an infinite number of places: try it for x = -.5, in thepic below:

The second thing you notice, if you’re doing the graph thing, is thatyer typical orange line hits the graph of x = sin θ not just at oneplace, but at an infinite number of places: try it for x = -.5, in thepic below: the line hits at −150o, −30o, 210o, 330o. Ah . . whichsolution am I supposed to choose?

If you try this with the calculator, you see something interesting: itgives sin−1(−.5) = −30o. Out of all the choices.

Again, squares and square roots make a good comparison. You tryand solve the equation x2 = 3; it has two solutions, x = +

√3 and

x = −√

3. Which to choose? Again, if you try it with a calculator,press the SQRT button and then 3, it gives you something like√

3 = 1.78... It makes a choice.

That’s the thing with the function concept: it’s modelled after theidea of a computation, of pushing buttons on a calculator: you putin ”3” you’re supposed to get just one answer out. The square rootfunction makes a choice: always take the positive square root.

Ditto with the sin−1(x) function: well, oops it takes the −30o so-lution, which isn’t positive. What it does instead is a little morecomplicated: it takes the solution closest to the origin. In the pic-ture . . . check it out.

Inverse Function: Range, II

What you see, is for a generic x, there are two theta’s, one on eitherside of the bump. The one closer to the origin is the one sin−1(x)likes to take. Now the ’bump’ is the local maximum of sin θ; ithappens at θ = 90o, or θ = π

2 . This means you can always find asolution to sin θ = x with θ ≤ π

2 .

Same deal on the left; there’s a local minimum for sin θ at θ = −π2

and you can always find a solution to sin θ = x with −π2 ≤ θ.

In equation terms, the equation sin θ = x always has a solution with−π

2 ≤ θ ≤ π2 .

In function terms, when you say where θ has to be, θ = sin−1(x), soyou’re saying −π

2 ≤ sin−1(x) ≤ π2 . That’s the range of the inverse

sin function: you’re saying that sin−1(x) has range [−π2 ,

π2 ].

Summary: Inverse Sine

Executive summary:

In equation terms:

Theequation sin θ = x has a solution only for −1 ≤ x ≤ 1.

If there is a solution θ, you can always find it so that −π2 ≤ θ ≤ π

2 .

In function terms: the sin−1(x) function is defined

by three requirements:

sin(sin−1(x)

)= x

the domain of sin−1(x) is [−1, 1]the range of sin−1(x) is [−π

2 ,π2 ].

One of the things this doesn’t say is that sin−1 (sin(θ)) = θ. And,if you try it, it doesn’t work. sin(π) = 0 and sin(0) = 0 so givena choice, sin−1(0) is gonna pick 0 as the solution; it’s closer to theorigin that π. Then:

sin−1 (sin(π)) = sin−1 (0) = 0 6= π

The only way you get sin−1 (sin(θ)) = θ is if starts in [−π2 ,

π2 ].

Inverse Cosine, I

There’s a lesson to be learned from sin−1(x) that helps to understandthe other inverse trig functions. First, the domain of inverse sine isthe same as the range of sine. Second, the range of inverse sine is apiece of the domain of sine: it has to be big enough so that sine cango through all its possible values, but small enough so that it takeson those values only once.

I’ll start with cos−1(x) first; here’s the graph of x = cos(θ).

Inverse Cosine, II

For domain of cos−1(x), you’ve got the range of cos(θ), which is[−1, 1] just like with sin. But for the range, . . . right away yousee that cos has its bump at the origin; for any x, there are θ’s oneither side of the origin, that satisfy x = cos(θ) and are each just asclose to the origin.

What worked for sin was to take the θ closest to the origin; here,you gotta do something different. The easy decision is to take thepositive θ′s; to cover everything, you need 0 ≤ θ ≤ 2π. On thegraph above, I did that part of the graph in red.

In function terms: the cos−1(x) function is defined bythree requirements:cos

(cos−1(x)

)= x

the domain of cos−1(x) is [−1, 1]the range of cos−1(x) is [0, 2π].

As an ad for the function method, I want to note how painless thiswas; I just modelled everything after sin and it went really fast.

Inverse Tangent, I

Now fer tan−1(x). Again, start with the graph of x = tan(θ). Sincetan has asymptotes when cos(θ) = 0, that is, at ±π

2 , ± 3π2 , . . . , I

put the asymptotes in with nice green lines.

Inverse Tangent, II

The domain of tan−1(x) is the range of tan(θ), which is a littledifferent from sin, cos: tan has asymptotes, and like the graph shows,the function takes on all possible values. This makes the domainof tan−1(x) into (−∞, ∞). Occassionally you’ll see people writetan−1(∞) = π

2 , and of course this really means that as θ → π2

+,then tan(θ) → +∞.

The range . . . tan doesn’t have a bump; you can just take θ valuesclose to the origin, as long as you don’t mess with the asymptotes.Since there are asymptotes at ±π

2 , you can stop there: the range oftan−1(x) is (−π

2 , +π2 ). Again I showed this part of the tan graph

in red, above.

This is a little different from what happened in the cases of sinand cos, and again it happens because of asymptotes. The domainof sin−1(x) is [−1, 1], the range [−π

2 ,π2 ]. The closed intervals here

reflect the fact that sin(π2 ) = 1 and that x = 1 is a value that sinactually takes on.

It’s different for tan−1(x), with a domain of (−∞, ∞) and rangeof (−π

2 , +π2 ). The open interval reflects the fact that ∞ is not

a specific value that tan takes on; you can’t say tan(π2 ) = +∞;instead, tan(π2 ) does not exist.

In function terms: the tan−1(x) function is defined bythree requirements:tan

(tan−1(x)

)= x

the domain of tan−1(x) is (−∞, ∞)the range of tan−1(x) is (−π

2 , +π2 ). 2

Inverse Secant, I

Finally, sec−1(x) (I say finally because I am not gonna go throughinverse csc and inverse cot; it’s Sunday and I been working on thislecture fer three days and I ain’t even half done and. . .)

As I wuz saying, sec−1(x); to understand it, we have to look at thegraph of x = sec(θ), and of course the graph comes fromsec(θ) = 1

cos(θ) , so that sec(θ) has an asymptote, again, at

θ = ±π2 , ± 3π

2 , . . . Again those are in lime-green, reflecting the factthat it’d be nice if this were summer. But there’s something newhere: since −1 ≤ cos(θ) ≤ 1, which, when you take 1

over , you getsec(θ) ≤ −1; sec(θ) ≥ 1. I put the x = ±1 bars in the graph too.Just in case.

Inverse Secant, II

The domain of sec−1(x) is the range of sec(θ), which is a little dif-ferent; like I said, sec(θ) ≤ −1; sec(θ) ≥ 1, so sec−1(x) has domain(−∞, −1], [1, ∞). And you can try this yerself; try punching in ona calculator sec−1(.5), and see what happens. You should get backsome error message, saying that sec−1(x) really doesn’t appreciatebeing given numbers like .5 to work with.The range . . . it’s always the range. Like cos, from which it comes,sec has a bump at θ = 0, and you have to make the positive-negativechoice. Like with cos, I choose positive, and then to go through allthe values of sec, you find yourself in the unusual position of havingto cross an asymptote. I show it in red, in the graph above, and whatit means is, like the domain of sec−1(x) is broken into two pieces,so the range is broken into two pieces: [0, π

2 ), (π2 , π]. Just like withtan−1(x), the open versus the closed intervals represent what valuesyou can and can’t plug into sec(θ). The domain has [1, ∞) and therange has (π2 , π]. The closed parts reflect the fact that sec(π) = 1so you see the invervals closed off at x = 1 for domain and θ = πfor range. On the other hand, sec(π2 ) does not exist, so x = ∞ inthe domain and θ = π

2 in the range are both represented by openinterval signs, ”)”.

In function terms: the sec−1(x) function is defined bythree requirements:sec

(sec−1(x)

)= x

the domain of sec−1(x) is (−∞, −1], [1, ∞)the range of sec−1(x) is [0, π

2 ), (π2 , π]. 2

Some Inverse Function Values

I though it’d be nice to give some values, for sin−1(x) and the gang.

sin−1(0) = 0

sin−1(.5) =π

6

sin−1(.707) = sin−1(1√2) =

π

4

sin−1(.86) = sin−1(

√3

2) =

π

3

sin−1(1) =π

2

cos−1(1) = 0

cos−1(.86) = cos−1(

√3

2) =

π

6

cos−1(.707) = cos−1(1√2) =

π

4

cos−1(.5) =π

3

cos−1(0) =π

2

tan−1(0) = 0

tan−1(.57) = tan−1(1√3) = tan−1(

12√3

2

) =π

6

tan−1(.707) = tan−1(1√2) =

π

4

tan−1(1.73) = tan−1(√

3) = tan−1(

√3

212

) =π

3

tan−1(∞) =π

2

One of the things that these tables do is make it really clear that

tan−1(x) 6= 1

tanx

For example, tan−1(0) = 0, but 1tan 0 = 1

0 .

And that’s nasty, because tan2(x) really is equal to (tan(x))2. But

once again, tan−1(x) does not mean (tan(x))−1

.

’Cuz of this, some calculus books don’t even use the tan−1(x) sym-bol; they write arctan(x) instead.

Inverse Cosine and Inverse Sine: Separated at Birth?

I did some trig substitiutions in class, and they gave me:∫1√

1 − x2dx = sin−1(x) + C∫

1

1 + x2dx = tan−1(x) + C∫

1

|x|√x2 − 1

dx = sec−1(x) + C

If you try the same thing with the co-functions, you get an inter-esting fact: Say try

∫1√

1−x2dx where you let x = cos(θ) instead

of the usual x = sin(θ). Then dx = − sin(θ) dθ and√

1 − x2 =√1 − (cos(θ))2 =

√(sin(θ))2 = sin(θ). Put it all together, and∫1√

1 − x2dx =

∫1

sin(θ)[− sin(θ)] dθ

= −∫

dθ = −θ + C = − cos−1(x) + C

What is weird about this is that you got∫

1√1−x2

dx = sin−1(x) +

C = − cos−1(x) + C so that sin−1(x) = − cos−1(x) + C. But thenthat is actually true: in fact, sin−1(x) = π

2 − cos−1(x), which isanother way of saying something I know already: cos(θ) = sin(π2−θ).

It is really tempting to say, ”well goody for sine and cosine”, butthere’s something here: I don’t need inverse cosine! It is just a minusand a shift by π

2 of inverse sine. So all the integration formulae you

see me do – they’ll all be sin−1(x) and tan−1(x) and sec−1(x) – theco-functions won’t even show.And – by the way: remember the range of sin−1(x) was [−π

2 ,π2 ]?

And the range of cos−1(x) was [0, π]? Now it makes sense: youget the sine values from cosine by flipping (gives ya [−π, 0]) thenadding π

2 (gives ya [−π + π2 , 0 + π

2 ] = [−π2 ,

π2 ], check).

Derivatives of Inverse Triggies

Another way to think of these is as derivatives;[sin−1(x)

]′=

1√1 − x2

[tan−1(x)

]′=

1

1 + x2[sec−1(x)

]′=

1

|x|√x2 − 1

Except this isn’t the way people usually do these derivatives. Usu-ally, you differentiate like-a-this:

sin[sin−1(x)

]= x definition of inverse sine(

sin[sin−1(x)

])′= (x)

′differentiate both sides(

cos[sin−1(x)

]) (sin−1(x)

)′= 1 chain rule(

sin−1(x))′

=1(

cos[sin−1(x)

]) divide

=1√

1 −(sin

[sin−1(x)

])2=

1√1 − x2

Here we used the idea that cos(θ) =√

1 − sin2(θ), but of course

that isn’t really right: it’s ±√

1 − sin2(θ). What lets me get rid of

the minus in ± is the fact that θ = sin−1(x), and that has values in[−π

2 ,π2 ]. So cos(θ) = cos([−π

2 ,π2 ]). But if you look at the values

of cosine from −π2 to π

2 , cosine is always positive or maybe onceand awhile it’s zero, but it isn’t negative for those values of θ. So,

while it’s totally true that cos(θ) = ±√

1 − sin2(θ), it’s also true

that cos[sin−1(x)

]= +

√1 −

(sin

[sin−1(x)

])2= +

√1 − x2.

Now are you sorry you didn’t believe me? This is here cuz I couldn’ta done it with out knowing the range of sin−1(x), by the way. Whichwas why I had to do the whole first part of this lecture.

Basic Inverse Tangent Integration Formula

OK, now: I got some new integrals:∫1√

1 − x2dx = sin−1(x) + C

∫1

1 + x2dx = tan−1(x) + C

but what I really want is these:∫1√

a2 − x2dx = sin−1(

x

a) + C

∫1

a2 + x2dx =

1

atan−1(

x

a) + C

where a > 0 is just a constant. Like, say you have∫

12+x2 dx and you

compare with∫

1a2+x2 dx; you’d have to say, a2 = 2 and so a = ±

√2

except I’m supposed to have a > 0 so a =√

2 and∫

12+x2 dx =

1√2

tan−1( x√2) + C, and you just get the answer, like that.

Here’s how you get it:∫1

a2 + x2dx =

∫1

a2(1 + x2

a2 )dx =

1

a2

∫1

1 + (xa )2dx

Now let u = xa so that du = 1

adx and adu = dx. Then:

1

a2

∫1

1 + (xa )2dx =

1

a2

∫1

1 + u2a du =

1

a

∫1

1 + u2du

Which gives me 1a tan−1(u) + C = 1

a tan−1(xa ) + C like the for-mula says. Since the integral for inverse sine has a square rootin the denominator, when you pull the same trick, you get just1a

∫1√

1−( xa )2

dx so that when you do the u-substitution, the a′s

cancel and so there’s no outside factor of 1a in the inverse sine equa-

tion.

Stupid Inverse Tan Tricks: Completing the Square

More integrals. A whole week of more integrals. Can life get better?New trix: ∫

1

1 + 3x2dx =

∫1

1 + (√

3 x)2dx

So u =√

3 x an’ du =√

3 dx and 1√3du = dx. Then:∫

1

1 + (√

3 x)2dx =

∫1

1 + u2

1√3du =

1√3

∫1

1 + u2du

=1√3

tan−1(u) + C =1√3

tan−1(√

3 x) + C

More.∫1

x2 + 2x + 2dx =

∫1

[x2 + 2x + 1] + 1dx =

∫1

[x + 1]2 + 1dx

Then if u = [x + 1] you got du = dx and∫1

[x + 1]2 + 1dx =

∫1

u2 + 1du = tan−1(u)+C = tan−1(x+1)+C

That’s called completing the square. You can do it with otherquadratics, too: say you got x2 +x+1; rewrite it as [x2 +x+ 1

4 ]+ 34

then the first piece is [x+ 12 ]2 + 3

4 . If it was in an integral, you could

let a2 = 34 so a =

√3

2 and then∫1

x2 + x + 1dx =

∫1

[x + 12 ]2 + 3

4

dx =

∫1

[u]2 + 34

du

=

∫1

[u]2 + (√

32 )2

du =1√

32

tan−1(u√

32

) + C

=2√3

tan−1(2(x + 1

2 )√3

) + C =2√3

tan−1(2x + 1√

3) + C

Inverse Tan and Quadratics That Factor

That trick works with any quadratic htat doesn’t factor. If it factors– say x2 − 1, they you won’t be able to write it as u2 + a2 because –well, x2−1 can be zero, but u2 +a2 is always positive, so they can’tbe equal. This isn’t just a trick failing, it means that integrals like∫

1x2+1 dx come out very different from integrals like

∫1

x2−1 dx. Infact,∫

1

x2 + 1dx = tan−1(x) + C

∫1

x2 − 1dx =

1

2ln |x− 1

x + 1| + C

So it makes a big difference, whether the answer comes out an inversetangent, or a natural log. Here’s the rule: if you have a quadraticin the denominator: ∫

anything

ax2 + bx + cdx

check to see if the ax2 + bx + c factors.a) If it doesn’t factor, you can complete the square and get an in-verse tangent in the answer.b) If it does factor,you gotta go and do the method of partial frac-tions, which I’ll talk about on Friday. You’ll get logs in the answer.Oh, right: how to tell if ax2 + bx + c factors? If b2 − 4ac < 0, itdoesn’t factor; if b2 − 4ac ≥ 0, it does factor: that’s the quadraticformula in action, before yer very eyes.

Inverse Tan and Sums in the Numerator

Man it is 8pm on Sunday night. I am never gonna cancel class onWednesday again. Ever.

Last trick is an old trick: sums in numerators split:∫x + 1

x2 + 1dx =

∫x

x2 + 1dx +

∫1

x2 + 1dx

The second integral is just tan−1(x) + C, but the first integral isripe fer a u-ey: let u = x2 + 1; du = 2x dx or 1

2du = dx so you got:∫x

x2 + 1dx +

∫1

x2 + 1dx =

∫1

u

1

2du + tan−1(x) + C

=1

2ln |u| + tan−1(x) + C =

1

2ln |1 + x2| + tan−1(x) + C

and this is an important trick to remember(oh, kathy, don’t be coy,say it: you’ll need to remember this for the final). Try it out yourselfon ∫

3x− 1

x2 + 1dx and

∫x− 1

(x + 2)2 + 1dx

In the second integral, do the u-subbie first, then split the numeratorinto two integrals.