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 arccosine.
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 squareds
and we convert them into 1 minus cosine squareds.
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.
Now, 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 sine 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 sine 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-- 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.