JOEL LEWIS: Hi, welcome back to recitation.
In lecture, Professor Miller did a bunch of examples of
integrals involving trigonometric functions.
So I thought I would give you a couple more, well I guess
three more examples, some of them are a little different
flavor than the ones he did, but some nice examples of some
integrals you can compute now.
One thing you might need for them is you're going to have
to remember some of your trig identities.
Just like you had to remember some of them last time.
So in particular, one identity that you might need today that
you didn't need in lecture, was the angle sum
identity for cosine.
So let me just remind you what that is, it says the cosine of
a plus b is equal to cosine a cosine b minus sine a sine b.
So you're going to need that formula to compute one of
these three integrals.
So why don't you pause the video, take some time
to work these out.
Come back you can check your answers against my work.
Hopefully you had some fun and some luck
working on these integrals.
Let's have a go at them.
The first integral is sine cubed x secant squared x.
So I don't think that Professor Miller did any
integrals that involved the secant function, but one thing
to remember is that the secant function is very closely
related to the cosine function.
It's just cosine to the minus 1.
Sorry, by that I mean 1 over cosine, not the arc cosine.
So in order to, so we can view this first integral here as an
integral of a power of sine times a power of cosine, just
like the ones you had in lecture.
So we can write this as sine cubed of x times--
I'm going to write--
cosine x to the minus 2 dx.
And here we see this is a nice example where the sine occurs
with an odd power.
So when the sine occurs with an odd power, or when one of
them at least occurs with an odd power, life
is relatively simple.
And what we do is just what we did in lecture, which is we
break it up so that we just have one instance of the one
that occurs to an odd power and then a bunch of powers of
the other one.
So in this case, we have sine to an odd power.
So we want to pull out all the even, you know, all the extra
multiples, so in this case we have sine times sine squared.
So we take all those sine squared's and we convert them
into 1 minus cosine squared's.
This is going to be equal to the integral of sine x times,
well, times sine squared x, which is 1 minus cosine
squared x times cosine x to the minus second power dx.
And now you take all these cosines and you multiply them
together and you see what you've got.
So in this case that's equal to the integral--
OK, so 1 times cosine to the minus 2, is just cosine
to the minus 2.
So this first term is cosine x to the minus 2 sine x and then
minus, OK we have cosine squared times cosine to the
minus 2, so that one's even simpler, that's just
minus sine x dx.
So at this point, if you like, you could do a substitution of
a trigonometric function.
You could set u equal to cosine x.
All right?
What's going to happen if you set u equal to cosine x, well
du is going to be minus sine x dx.
Each of these have exactly one appearance of sine x in them,
so this would become--
so if we set u equal cosine x.
So, as I said du equals minus sine x dx.
Because each of these have a sine x dx in them, that's
going to get clumped into the du and the cosines will just
get replaced with u everywhere.
So I'm going to take this over here so I have
a little more space.
So this becomes the integral of u to the minus 2, I guess
it's minus u to the minus 2 plus 1 du.
That minus sine x is just exactly equal to du.
OK.
And now this is easy to finish from here.
This is equal to, well OK, so minus 2, so we have to go up
one to minus 1, so that's just u to the minus 1 plus 1 gives
us a u plus a constant, and then of course we have to
substitute back in.
So this is u is cosine x, so u to the minus 1 is secant x
plus cosine x plus a constant.
So that's the first one.
For the second one, sine x cosine of 2x.
We actually have a slightly more
complicated situation here.
Because the cosine 2x is no longer, it's not just a
trigonometric function of x alone, it's a trigonometric
function of 2x.
So in order to do the sort of the things we did in lecture,
what you're going to have to do is, you're going to have to
expand this out in terms of just sine x and cosine x,
rather than cosine of 2x.
So in order to do that you just need to remember your
double angle formula.
Now if you don't remember your double angle formulas, one way
you can remember them is remember the angle sum formula
and apply it with a and b both equal to x.
If you like.
But, or you could just remember your
double angle formulas.
So let's go over here.
And so in this case, we have the integral of sine x times,
well cosine 2x-- so there are actually, right, there are
several different ways you can write cosine 2x.
So one way you can write it is cosine squared x minus sine
squared x, but we want everything-- again we have an
odd power of sine, so it would be nice if everything else
could be written in terms of just cosine.
So an equivalent form for cosine 2x is to write, is that
it's equal to 2 cosine squared x minus 1 dx.
And here, this is very, very similar to the last question.
Again, you can make the substitution u equals cosine,
if you like.
And so I'm not going to do out the whole substitution for
you, but this is going to work out to 2 cosine squared x sine
x, the integral of that is minus 2/3 cosine cubed x and
then the integral of minus sine x is cosine x plus a
constant of integration.
So that's the second one.
The third one now is of a similar flavor, except you
have here, you have sine 2x and cosine 3x.
So again, when you want to integrate
something where you have--
so here we have both trigonometric functions occur
in multiples.
No if it was the same multiple, if it were 2x and
2x, say, that would be fine.
You know, you could just make a substitution like u equals
2x or something and then proceed as usual.
But because they're different multiples, we need, when we do
these trig integrals, we need to apply the methods that we
used in lecture.
You want the arguments to agree.
So you want it all to be sine of something and cosine of the
same thing.
So here the thing to do, I think, is to try and put
everything in terms of just sine x and cosine x.
So sine 2x is one you should remember,
double angle formula.
Cosine 3x, well in order to figure out what cosine 3x is,
you could just know it, maybe you learned
it once upon a time.
The other thing you can do for cosine 3x, is you can use this
angle sum formula.
So let me just work that out for you quickly.
So cosine of 3x, well how do you use the angle sum formula?
You can write 3x as 2x plus x.
So this is equal to cosine of 2x plus x.
And now this is a sum, so you can use the cosine sum
formula, so you get this is equal to cosine 2x cosine x
minus sine 2x sine x.
OK and now you can use the double angle
formulas here and here.
so in the end, if you do this all out, what you get is
you're going to get 4 cosine cubed x minus 3 cosine x.
You might say, oh what happened to the sine x's?
We're going to get a sign squared x and I want to write
everything in terms of cosine.
So I've replaced the, I've done an extra step here that
I'm not showing, where I replaced the sine squared x
with a 1 minus cosine squared x.
So you make this substitution, and you also make the
substitution for sign x-- sorry-- for sine 2x that you
have. OK, sine 2x is equal to-- you know,
double angle formula--
2 sine x cosine x.
And so when you multiply sine 2x by cosine 3x you again have
an expression that's got powers of cosine with a single
power of sine.
So doing these simplifications, you again get
this nice form.
One of them's odd, the other one, so, one of the trig
functions, sine or cosine appears to just the power of
1, so, OK, so you can do this nice simple substitution.
You don't need to do any of the double angle, or half
angle complication.
I'm not going to finish this one out for you, but I did
write down the answer so you can check your answer.
When you, the third integral that we had, which is integral
sine 2x cosine 3x dx, when you work it all out you should
get, let's see what I got, OK, you should get minus 8/5
cosine to the fifth x plus 2 cosine cubed x plus a constant
of integration.
So if you work out all the details, this is what you
should get.
I'll stop there.