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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Well, because
our subject today is trig integrals
and substitutions, Professor Jerison called in his
substitute teacher for today. That's me. Professor Miller. And I'm going to try to tell
you about trig substitutions and trig integrals. And I'll be here tomorrow to
do more of the same, as well. So, this is about trigonometry,
and maybe first thing I'll do is remind you of some basic
things about trigonometry. So, if I have a
circle, trigonometry is all based on the
circle of radius 1 and centered at the origin. And so if this is an angle
of theta, up from the x-axis, then the coordinates
of this point are cosine theta and sine theta. And so that leads right away
to some trig identities, which you know very well. But I'm going to put them up
here because we'll use them over and over again today. Remember the convention sin^2
theta secretly means (sin theta)^2. It would be more
sensible to write a parenthesis around
the sine of theta and then say you square that. But everybody in the world puts
the 2 up there over the sin, and so I'll do that too. So that follows just because
the circle has radius 1. But then there are some
other identities too, which I think you remember. I'll write them down
here. cos(2theta), there's this double angle
formula that says cos(2theta) = cos^2(theta) - sin^2(theta). And there's also the
double angle formula for the sin(2theta). Remember what that says? 2 sin(theta) cos(theta). I'm going to use
these trig identities and I'm going to use them
in a slightly different way. And so I'd like to pay a little
more attention to this one and get a different way
of writing this one out. So this is actually
the half angle formula. And that says, I'm going to
try to express the cos(theta) in terms of the cos(2theta). So if I know the
cos(2theta), I want to try to express the
cos theta in terms of it. Well, I'll start with a
cos(2theta) and play with that. OK. Well, we know what this is, it's
cos^2(theta) - sin^2(theta). But we also know what the
sin^2(theta) is in terms of the cosine. So I can eliminate the
sin^2 from this picture. So this is equal to cos^2(theta)
minus the quantity 1 - cos^2(theta). I put in what sin^2 is in
terms of cos^2 And so that's 2 cos^2(theta) - 1. There's this cos^2,
which gets a plus sign. Because of these
two minus signs. And there's the one
that was there before, so altogether there
are two of them. I want to isolate
what cosine is. Or rather, what cos^2 is. So let's solve for that. So I'll put the 1
on the other side. And I get 1 + cos(2theta). And then, I want to divide by
this 2, and so that puts a 2 in this denominator here. So some people call that
the half angle formula. What it really is for us is
it's a way of eliminating powers from sines and cosines. I've gotten rid of this
square at the expense of putting in a 2theta here. We'll use that. And, similarly, same calculation
shows that sin^2(theta) = (1 - cos(2theta)) / 2. Same cosine, in that formula
also, but it has a minus sign. For the sin^2. OK. so that's my little
review of trig identities that we'll make use of
as this lecture goes on. I want to talk about trig
identity-- trig integrals, and you know some trig
integrals, I'm sure, already. Like, well, let me write
the differential form first. You know that d sin
theta, or maybe I'll say d sin x, is,
let's see, that's the derivative of sin
x times dx, right. The derivative of
sin x is cos x, dx. And so if I integrate both sides
here, the integral form of this is the integral of cos x dx. Is sin x plus a constant. And in the same way,
d cos x = -sin x dx. Right, the derivative of
the cosine is minus sine. And when I integrate that, I
find the integral of sin x dx is -cos x + c. So that's our starting point. And the game today, for the
first half of the lecture, is to use that basic-- just
those basic integration formulas, together with
clever use of trig identities in order to compute more
complicated formulas involving trig functions. So the first thing,
the first topic, is to think about integrals of
the form sin^n (x) cos^n (x) dx. Where here I have in mind m and
n are non-negative integers. So let's try to integrate these. I'll show you some applications
of these pretty soon. Looking down the
road a little bit, integrals like this show
up in Fourier series and many other subjects
in mathematics. It turns out they're quite
important to be able to do. So that's why we're
doing them now. Well, so there are two
cases to think about here. When you're integrating
things like this. There's the easy case, and
let's do that one first. The easy case is when at
least one exponent is odd. That's the easy case. So, for example, suppose
that I wanted to integrate, well, let's take the case m = 1. So I'm integrating
sin^n (x) cos x dx. I'm taking-- Oh, I
could do that one. Let's see if that's
what I want to take. Yeah. My confusion is that I meant
to have this a different power. You were thinking that. So let's do this
case when m = 1. So the integral I'm
trying to do is any power of the sine times the cosine. Well, here's the trick. Recognize, use this
formula up at the top there to see cos x dx as
something that we already have on the blackboard. So, the way to exploit that
is to make a substitution. And substitution is
going to be u = sin x. And here's why. Then this integral that I'm
trying to do is the integral of u^n, that's already
a simplification. And then there's that cos x dx. When you make a substitution,
you've got to go all the way and replace everything
in the expression by things involving this new
variable that I've introduced. So I'd better get
rid of the cos x dx and rewrite it in terms
of du or in terms of u. And I can do that because du,
according to that formula, is cos x dx. Let me put a box around that. That's our substitution. When you make a
substitution, you also want to compute the
differential of the variable that you substitute in. So the cos x dx that appears
here is just, exactly, du. And I've replaced this trig
integral with something that doesn't involve
trig functions at all. This is a lot easier. We can just plug into
what we know here. This is u^(n+1) /
(n+1) plus a constant, and I've done the integral. But I'm not quite done
with the problem yet. Because to be nice to your
reader and to yourself, you should go back at
this point, probably, go back and get rid of this new
variable that you introduced. You're the one who introduced
this variable, you. Nobody except you,
really, knows what it is. But the rest of the
world knows what they asked for the first
place that involved x. So I have to go back
and get rid of this. And that's not hard to do in
this case, because u = sin x. And so I make this
back substitution. And that's what you get. So there's the answer. OK, so the game was, I use this
odd power of the cosine here, and I could see it appearing as
the differential of the sine. So that's what made
this substitution work. Let's do another example
to see how that works out in a slightly different case. So here's another example. Now I do have an odd power. One of the exponents is odd,
so I'm in the easy case. But it's not 1. The game now is to
use this trig identity to get rid of the largest
even power that you can, from this odd power here. So use sin^2 x = 1 - cos^2 x, to
eliminate a lot of powers from that odd power. Watch what happens. So this is not really a
substitution or anything, this is just a trig identity. This sine cubed is sine
squared times the sine. And the sine squared
is 1 - cos^2 x. And then I have the
remaining sin x. And then I have cos^2 x dx. So let me rewrite that a little
bit to see how this works out. This is the integral
of cos^2 x minus, and then there's the
product of these two. That's cos^4 x times sin x dx. So now I'm really
exactly in the situation that I was in over here. I've got a single power
of a sine or cosine. It happens that
it's a sine here. But that's not going
to cause any trouble, we can go ahead and play the
same game that I did there. So, so far I've just been
using trig identities. But now I'll use a
trig substitution. And I think I want to write
these as powers of a variable. And then this is going to be the
differential of that variable. So I'll take u to be cos x, and
that means that du = -sin x dx. There's the substitution. So when I make that
substitution, what do we get. Cosine squared becomes u^2. Cosine to the 4th becomes u^4,
and sin x dx becomes not quite du, watch for the signum,
watch for this minus sign here. It becomes -du. But that's OK. The minus sign comes outside. And I can integrate
both of these powers, so I get -u^3 / 3. And then this 4th power gives me
a 5th power, when I integrate. And don't forget the constant. Am I done? Not quite done. I have to back
substitute and get rid of my choice of variable, u,
and replace it with yours. Questions? STUDENT: [INAUDIBLE] PROFESSOR: There should indeed. I forgot this minus sign
when I came down here. So these two gang up
to give me a plus. Was that what the other
question was about, too? Thanks. So let's back substitute. And I'm going to
put that over here. And the result is, well, I just
replace the u by cosine of x. So this is - -cos^3(x) / 3 plus,
thank you, cos^5(x) / 5 + c. And there's the answer. By the way, you can remember
one of the nice things about doing an
integral is it's fairly easy to check your answer. You can always differentiate
the thing you get, and see whether you get the
right thing when you go back. It's not too hard
to use the power rules and the
differentiation rule for the cosine to get
back to this if you want to check the work. Let's do one more
example, just to handle an example of this
easy case, which you might have thought of at first. Suppose I just want to
integrate a cube. sin^3 x. No cosine in sight. But I do have an
odd power of a trig function, of a sine or cosine. So I'm in the easy case. And the procedure that I was
suggesting says I want to take out the largest even power
that I can, from the sin^3. So I'll take that out, that's
a sin^2, and write it as 1 - cos^2. Well, now I'm very happy. Because it's just
like the situation we had somewhere
on the board here. It's just like the
situation we had up here. I've got a power of a
cosine times sin x dx. So exactly the same
substitution steps in. You get, and maybe
you can see what happens without doing the work. Shall I do the work here? I make the same substitution. And so this is (1 - u
(1 - u^2) times -du. Which is u - u^3 / 3. But then I want to put
this minus sign in place, and so that gives me -u +
u^3 / 3 plus a constant. And then I back substitute
and get cos x + cos^3 x / 3. So this is the easy case. If you have some odd
power to play with, then you can make use of it and
it's pretty straightforward. OK the harder case is when
you don't have an odd power. So what's the program? I'm going to do the
harder case, and then I'm going to show you an example of
how to integrate square roots. And do an application, using
these ideas from trigonometry. So I want to keep
this blackboard. Maybe I'll come back
and start here again. So the harder case is when
they're only even exponents. I'm still trying to
integrate the same form. But now all the
exponents are even. So we have to do some game. And here the game is use
the half angle formula. Which I just erased, very
sadly, on the board here. Maybe I'll rewrite
them over here so we have them on the board. I think I remember
what they were. So the game is I'm going
to use that half angle formula to start getting
rid of those even powers. Half angle formula written like
this, exactly, talks about-- it rewrites even powers
of sines and cosines. So let's see how that
works out in an example. How about just the cosine
squared for a start. What to do? I can't pull anything out. I could rewrite
this as 1 - sin^2, but then I'd be faced with
integrating the sin^2, which is exactly as hard. So instead, let's use
this formula here. This is really the same
as (1+cos(2theta)) / 2. And now, this is easy. It's got two parts to it. Integrating one half
gives me theta over-- Oh. Miraculously, the x
turned into a theta. Let's put it back as x. I get x/2 by integrating 1/2. So, notice that something
non-trigonometric occurs here when I do these even integrals. x/2 appears. And then the other one, OK, so
this takes a little thought. The integral of the
cosine is the sine, or is it minus the sine. Negative sine. Shall we take a vote? I think it's positive. And so you get sin(2x),
but is that right? Over 2. If I differentiate the
sin(2x), this 2 comes out. And would give me
an extra 2 here. So there's an extra 2
that I have to put in here when I integrate it. And there's the answer. This is not a substitution. I just played with
trig identities here. And then did a
simple trig integral, getting your help to
get the sign right. And thinking about what
this 2 is going to do. It produces a 2 in
the denominator. But it's not applying
any complicated thing. It's just using this identity. Let's do another example
that's a little bit harder. This time, sin^2 times cos^2. Again, no odd powers. I've got to work a
little bit harder. And what I'm going to do
is apply those identities up there. Now, what I recommend
doing in this situation is going over to
the side somewhere. And do some side work. Because it's all just
playing with trig functions. It's not actually doing
any integrals for a while. So, I guess one way to get rid
of the sin^2 and the cos^2 is to use those identities
and so let's do that. So the sine is (1
- cos(2x)) / 2. And the cosine is
(1 + cos(2x)) / 2. So I just substitute them in. And now I can multiply that out. And what I have is a
difference times a sum. So you know a formula for that. Taking the product of these two
things, well there'll be a 4 in the denominator. And then in the numerator,
I get the square of this minus the
square of this. (a-b)(a+b) = a^2 -
b^2. = - So I get that. Well, I'm a little bit
happier, because at least I don't have 4. I don't have 2
different squares. I still have a square, and
want to integrate this. I'm still not in the easy case. I got myself back to
an easier hard case. But we do know what
to do about this. Because I just did it up there. And I could play into
this formula that we got. But I think it's just as easy
to continue to calculate here. Use the half angle
formula again for this, and continue on your way. So I get a 1/4 from this bit. And then minus 1/4 of cos^2(2x). And when I plug in 2x in for
theta, there in the top board, I'm going to get a 4x
on the right-hand side. So it comes out like that. And I guess I could simplify
that a little bit more. This is a 1/4. Oh, but then there's a 2 here. It's half that. So then I can simplify
a little more. It's 1/4 - 1/8, which is 1/8. And then I have 1/8 cos(4x). OK, that's my side work. I just did some trig
identities over here. And rewrote sine
squared times cosine squared as something which
involves just no powers of trig, just cosine by itself. And a constant. So I can take that and
substitute it in here. And now the integration
is pretty easy. 1/8, cos(4x) / 8,
dx, which is, OK the 1/8 is going to give me x/8. The integral or cosine is
plus or minus the sine. The derivative of the
sine is plus the cosine. So it's going to be plus the--
Only there's a minus here. So it's going to be the
sine-- minus sin(4x) / 8, but then I have an additional
factor in the denominator. And what's it going to be? I have to put a 4 there. So we've done that
calculation, too. So any of these-- If you keep
doing this kind of process, these two kinds
of procedures, you can now integrate
any expression that has a power of a sine times
a power of a cosine in it, by using these ideas. Now, let's see. Oh, let me give you an alternate
method for this last one here. I know what I'll do. Let me give an alternate
method for doing, really doing the side work over there. I'm trying to deal
with sin^2 times cos^2. Well that's the
square of sin x cos x. And sin x cos x
shows up right here. In another trig identity. So we can make use of that, too. That reduces the number of
factors of sines and cosines by 1. So it's going in
the right direction. This is equal to 1/2
sin(2x), squared. Sine times cosine is
1/2-- Say this right. It's sin(2x) / 2, and then
I want to square that. So what I get is sin^2(2x) / 4. Which is, well, I'm
not too happy yet, because I still
have an even power. Remember I'm trying to
integrate this thing in the end, even powers are bad. I try to get rid of them. By using that formula,
the half angle formula. So I can apply that
to sin x here again. I get 1/4 of (1 - cos(4x)) / 2. That's what the half angle
formula says for sin^2(2x). And that's exactly the
same as the expression that I got up here, as well. It's the same expression
that I have there. So it's the same
expression as I have here. So this is just an alternate
way to play this game of using the half angle formula. OK, let's do a little
application of these things and change the
topic a little bit. So here's the problem. So this is an
application and example of a real trig substitution. So here's the problem
I want to look at. OK, so I have a circle
whose radius is a. And I cut out from it
a sort of tab, here. This tab here. And the height of
this thing is b. So this length is a number b. And what I want to do is compute
the area of that little tab. That's the problem. So there's an arc over here. And I want to find the
area of this, for a and b, in terms of a and b. So the area, well,
I guess one way to compute the area would be
to take the integral of y dx. You've seen the idea
of splitting this up into vertical strips whose
height is given by a function y(x). And then you integrate that. That's an interpretation
for the integral. The area is given by y dx. But that's a little bit awkward,
because my formula for y is going to be a little strange. It's constant, value of b, along
here, and then at this point it becomes this
arc, of the circle. So working this
out, I could do it but it's a little awkward
because expressing y as a function of x, the
top edge of this shape, it's a little awkward, and
takes two different regions to express. So, a different way to
say it is to say x dy. Maybe that'll work
a little bit better. Or maybe it won't,
but it's worth trying. I could just as well
split this region up into horizontal strips. Whose width is dy,
and whose length is x. Now I'm thinking of
this as a function of y. This is the graph
of a function of y. And that's much better, because
the function of y is, well, it's the square root
of a^2 - y^2, isn't it. That's x x^2 + y^2 = a^2. So that's what x is. And that's what I'm
asked to integrate, then. Square root of (a^2 - y^2), dy. And I can even put in
limits of integration. Maybe I should do that,
because this is supposed to be an actual number. I guess I'm integrating it
from y = 0, that's here. To y = b, dy. So this is what I want to find. This is a integral formula
for the area of that region. And this is a new form. I don't think that
you've thought about integrating expressions
like this in this class before. So, it's a new form and I
want to show you how to do it, how it's related
to trigonometry. It's related to trigonometry
through that exact picture that we have on the blackboard. After all, this a^2 - y^2
is the formula for this arc. And so, what I
propose is that we try to exploit the
connection with the circle and introduce polar coordinates. So, here if I measure
this angle then there are various things you can say. Like the coordinates of this
point here are a cos(theta), a-- Well, I'm sorry, it's not. That's an angle, but I
want to call it theta_0. And, in general you know
that the coordinates of this point are (a cos(theta),
a sin(theta)). If the radius is a, then
the angle here is theta. So x = a cos(theta),
and y = a sin(theta), just from looking at the
geometry of the circle. So let's make that
substitution. y = a sin(theta). I'm using the picture to
suggest that maybe making the substitution is
a good thing to do. Let's follow along
and see what happens. If that's true, what we're
interested in is integrating, a^2 - y^2. Which is a^2-- We're interested
in integrating the square root of a^2 - y^2. Which is the square
root of a^2 minus this. a^2 sin^2(theta). And, well, that's
equal to a cos theta. That's just sin^2 + cos^2
= 1, all over again. It's also x. This is x. And this was x. So there are a lot of different
ways to think of this. But no matter how you say
it, the thing we're trying to integrate, a^2 - y^2 is,
under this substitution it is a cos(theta). So I'm interested in integrating
the square root of (a^2 - y^2), dy. And I'm going to make this
substitution y = a sin(theta). And so under that substitution,
I've decided that the square root of a^2 - y^2
is a cos(theta). That's this. What about the dy? Well, I'd better compute the dy. So dy, just differentiating
this expression, is a cos(theta) d theta. So let's put that in. dy
= a cos(theta) d theta. OK. Making that trig substitution,
y = a sin(theta), has replaced this integral
that has a square root in it. And no trig functions. With an integral that involves
no square roots and only trig functions. In fact, it's not too hard to
integrate this now, because of the work that we've done. The a^2 comes out. This is cos^2(theta) d theta. And maybe we've done that
example already today. I think we have. Maybe we can think it through,
but maybe the easiest thing is to look back at notes
and see what we got before. That was the first example
in the hard case that I did. And what it came out to, I used
x instead of theta at the time. So this is a good step forward. I started with this
integral that I really didn't know how to do by any
means that we've had so far. And I've replaced it
by a trig integral that we do know how to do. And now I've done
that trig integral. But we're still not quite
done, because of the problem of back substituting. I'd like to go back and
rewrite this in terms of the original variable, y. Or, I'd like to go
back and rewrite it in terms of the original
limits of integration that we had in the
original problem. In doing that, it's going
to be useful to rewrite this expression and get
rid of the sin(2theta). After all, the original
y was expressed in terms of sin(theta), not sin(2theta). So let me just do that here,
and say that this, in turn, is equal to a^2
theta / 2 plus, well, sin(2theta) = 2
sin(theta) cos(theta). And so, when there's a 4 in
the denominator, what I'll get is sin(theta) cos(theta) / 2. I did that because I'm getting
closer to the original terms that the problem started with. Which was sin(theta). So let me write down the
integral that we have now. The square root of a^2
- y^2, dy is, so far, what we know is a^2 (theta / 2 +
sin(theta) cos(theta) / 2) + c. But I want to go
back and rewrite this in terms of the original value. The original variable, y. Well, what is theta
in terms of y? Let's see. y in terms of
theta was given like this. So what is theta in terms of y? Ah. So here the fearsome arcsine
rears its head, right? Theta is the angle so
that y = a sin(theta). So that means that theta is
the arcsine, or sine inverse, of y/a. So that's the first
thing that shows up here. arcsin(y/a), all over 2. That's this term. Theta is arcsin(y/a) / 2. What about the other side, here? Well sine and cosine, we knew
what they were in terms of y and in terms of x, if you like. Maybe I'll put the
a^2 inside here. That makes it a
little bit nicer. Plus, and the other term is
a^2 sin(theta) cos(theta). So the a sin(theta) is just y. Maybe I'll write this (a
sin(theta)) (a cos(theta)) / 2 + c. And so I get the same thing. And now here a
sin(theta), that's y. And what's the a cos(theta)? It's x, or, if you like, it's
the square root of a^2 - y^2. And so there I've
rewritten everything, back in terms of the
original variable, y. And there's an answer. So I've done this indefinite
integration of a form-- of this quadratic, this square
root of something which is a constant minus y^2. Whenever you see that, the thing
to think of is trigonometry. That's going to play into
the sin^2 + cos^2 identity. And the way to exploit it
is to make the substitution y = a sin(theta). You could also make a
substitution y = a cos(theta), if you wanted to. And the result would come out
to exactly the same in the end. I'm still not quite done
with the original problem that I had, because
the original problem asked for a definite integral. So let's just go back
and finish that as well. So the area was
the integral from 0 to b of this square root. So I just want to evaluate
the right-hand side here. The answer that we came up
with, this indefinite integral. I want to evaluate
it at 0 and at b. Well, let's see. When I evaluate it at b, I get
a^2 arcsin(b/a) / 2 plus y, which is b, times the
square root of a^2 - b^2, putting y = b, divided by 2. So I've plugged in y
= b into that formula, this is what I get. Then when I plug in y = 0,
well the, sine of 0 is 0, so the arcsine of 0 is 0. So this term goes away. And when y = 0,
this term is 0 also. And so I don't get any
subtracted terms at all. So there's an
expression for this. Notice that this arcsin(b/a),
that's exactly this angle. arcsin(b/a), it's the angle
that you get when y = b. So this theta is
the arcsin(b/a). Put this over here. That is theta_0. That is the angle
that the corner makes. So I could rewrite this as a
a^2 theta_0 / 2 plus b times the square root of
a^2 - b^2, over 2. Let's just think about
this for a minute. I have these two terms in
the sum, is that reasonable? The first term is a^2. That's the radius squared
times this angle, times 1/2. Well, I think that is exactly
the area of this sector. a^2 theta / 2 is the formula
for the area of the sector. And this one, this is
the vertical elevation. This is the horizontal. a^2 - b^2 is this distance. Square root of a^2 - b^2. So the right-hand term is b
times the square root of a^2 - b^2 divided by 2, that's
the area of that triangle. So using a little
bit of geometry gives you the same answer as
all of this elaborate calculus. Maybe that's enough
cause for celebration for us to quit for today.
